Introduction
This post is a follow-up to my post on infinite series wherein I discussed the Riemann zeta function as the analytic continuation of the Dirichlet series $D(s)=\sum^\infty_{n=1}a_n/n^s$ in the $a_n=1$ case. Since then I've been asked to write a little bit more about the zeta function and analytic continuation, so in this post I'm going to go through Riemann's second proof of the analytic continuation of the zeta function (based primarily on this presentation, though there are many others on the Internet) and hopefully make it a little bit more accessible. There are a number of other proofs, but I chose this one because it's quite straightforward and doesn't introduce too many techniques foreign to a physics-minded person like myself.
I have a soft spot for the gamma function, so you'll have to indulge me going into a little extra detail here. We'll start by considering an operator we've seen many times before, the factorial operator, which is given by $n!=n\times (n-1)\times (n-2)\times \ldots \times 3\times 2\times 1$ for all non-negative integers $n$ (with $0!=1$ by definition).$^1$ Now, from this definition it is clear that it makes no sense to ask what, say, $1.5!$ or $\pi !$ are, but if we plot $n!$ we notice something interesting -- they seem to follow a sensible 'path' (fig. 1) and as such it may be possible that $n!$ may be sensibly interpolated (simply put, we may be able to join the dots in a consistent and useful way).The gamma function
 |
| Figure 1: $n!$ for $n=0,1,\ldots,6$. Note the logarithmic scale on the vertical axis. The plot strongly suggests that $n!$ is interpolable. |
 |
| Figure 2: a) The Gamma function $\Gamma(s)$ along the real axis (shown in blue) and the factorial operation $(n-1)!$ for integers $n$ (shown in red) overlaid. The Gamma function as the interpolation of the factorial operation is clear from this plot. b) The absolute value of the Gamma function $\Gamma(s)$ over $\sigma\in [-4,4]$ and $t\in [-1,1]$ where $s=\sigma+it$. The locations of the poles at the negative integers are clear to see here. |
In fact, the gamma function is not only defined for the real numbers but can itself be analytically continued over the whole complex plane. It is meromorphic, which means it is everywhere analytic or holomorphic (everywhere differentiable) apart from at one or more singularities (poles)$^2$ (see fig. 2b). The poles of the gamma function are simple poles located at the non-positive integers, which means that at $s=0, -1, -2, \ldots$ it behaves like $1/s$ behaves as $s\rightarrow 0$. This fact will become important shortly.
The gamma function is given by the (improper) integral
\begin{equation}
\Gamma(s):=\int^\infty_0 t^{s-1}e^{-t}\mathrm{d}t,\quad\mathrm{Re}(s)>0.
\end{equation}
It is easy to show using integration by parts that the gamma function can be analytically continued using the functional (implicit) equation $\Gamma(s+1)=s\Gamma(s)$. The formula for integration by parts is
\begin{equation}
\int_a^b u(t)v'(t)\mathrm{d}t=\left[u(t)v(t)\right]_a^b-\int_a^b u'(t)v(t)\mathrm{d}t,
\end{equation}
where a prime denotes differentiation with respect to $t$. For this case, $\Gamma(s+1)=\int^\infty_0 t^s e^{-t}\mathrm{d}t$ so we can choose $u(t)=t^s$ and $v'(t)=e^{-t}$ to give
\begin{align}
\Gamma(s+1)&=\int^\infty_0 t^s e^{-t}\mathrm{d}t\nonumber \\
&=\left[-t^s e^{-t}\right]_0^\infty-\int_0^\infty -st^{s-1}e^{-t}\mathrm{d}t\nonumber \\
&=0+s\int_0^\infty t^{s-1}e^{-t}\mathrm{d}t\nonumber \\
&=s\Gamma(s).
\end{align}
Now we will consider a different but equivalent definition of the gamma function:$^3$
\begin{equation*}
\Gamma(s):=\frac{1}{s}\prod_{n=1}^\infty\frac{\left(1+\frac{1}{n}\right)^s}{1+\frac{s}{n}}.
\end{equation*}
Using the functional equation (eqn. 3) for $-s$ we find
\begin{align}
\Gamma(s)\big(\Gamma(1-s)\big)&=\Gamma(s)\big(-s\Gamma(-s)\big)\nonumber\\
&=-s\frac{1}{s}\prod_{n=1}^\infty\frac{\left(1+\frac{1}{n}\right)^s}{1+\frac{s}{n}}\frac{1}{-s}\prod_{k=1}^\infty\frac{\left(1+\frac{1}{k}\right)^{-s}}{1-\frac{s}{k}}\nonumber\\
&=\frac{1}{s}\prod_{n=1}^\infty \frac{1}{1-\frac{s^2}{n^2}}\nonumber\\
&=\frac{1}{s}\frac{\pi s}{\sin(\pi s)}\nonumber\\
&=\frac{\pi}{\sin(\pi s)}\nonumber\\
\Rightarrow \frac{1}{\Gamma(s)}&=\frac{1}{\pi}\sin(\pi s)\Gamma(1-s)
\end{align}
where in the third line we have combined the two infinite products under a single infinite product sign and simplified (this is acceptable because we have a product of products and the ordering of terms is not important; this is analogous to combining two infinite sums added together under a single infinite summation) and in the fourth line we have used the identity$^4$ $\sin(\pi x)/\pi x=\prod_{n=1}^\infty (1-x^2/n^2)$.
The final line is the important result here; the gamma function is meromorphic, that is it is analytic except for simple poles at $s=0, -1, -2, \ldots$. which means that $\Gamma(1-s)$ will also be meromorphic but with simple poles at $s=1,2,3,\ldots$. However, for those values of $s$, $\sin(\pi s)=0$, which removes the poles, so in the end $1/\Gamma(s)$ is entire, that is, analytic over the whole complex plane.
The Fourier transform
The Fourier transform is an integral transform we have seen before in the context of quantum mechanics, but is closely related to Fourier series, which we have also seen in some depth. To give a heuristic understanding, in the discrete case the Fourier transform allows us to take a function $f(x)$ which is a sum of waves of certain frequencies with heights given by the corresponding Fourier coefficients, and form a new function $\mathcal{F}\{ f(x)\} =\tilde{f}(\xi)$ made up of a sum of "peaks" with heights given by the Fourier coefficients at locations given by the corresponding wave frequencies$^5$.
The transform itself is given by$^6$
\begin{equation*}
\tilde{f}(\xi):=\int^\infty_\infty f(x)e^{-2\pi i\xi x}\mathrm{d}x.
\end{equation*}
\tilde{f}(\xi):=\int^\infty_\infty f(x)e^{-2\pi i\xi x}\mathrm{d}x.
\end{equation*}
It is important to note that there is nothing about this definition which restricts its use to 'discrete' Fourier sum cases only. In fact, though the above motivating example uses the case of the Fourier series, any function (even non-periodic functions) can be Fourier transformed, provided of course that they are integrable.
This is exploited by the Poisson summation formula, which states that the infinite sum of values of any continuous function at integer points is equal to the infinite sum of values of the function's Fourier transform at integer points, or, mathematically,
\begin{equation}
\sum^\infty_{n=-\infty}f(n)=\sum^\infty_{k=-\infty}\tilde{f}(k).
\end{equation}
\sum^\infty_{n=-\infty}f(n)=\sum^\infty_{k=-\infty}\tilde{f}(k).
\end{equation}
This is easy to prove, but as the proof is not essential it is included in the Notes.$^7$
Let us also consider a couple of examples which will prove useful. First, the Fourier transform of some function $g(x)=f(ax)$ for $a>0$. In this case,
\begin{align*}
\tilde{g}(\xi)=\int^\infty_{-\infty}g(x)e^{-2\pi i\xi x}\mathrm{d}x&=\int^\infty_{-\infty}f(ax)e^{-2\pi i\xi x}\mathrm{d}x\\
&=\int^\infty_{-\infty}f(x)e^{-2\pi i\xi \frac{x}{a}}\frac{\mathrm{d}x}{a}\\
&=\frac{1}{a}\int^\infty_{-\infty}f(x)e^{-2\pi i\frac{\xi}{a}x}\mathrm{d}x
\end{align*}
where in the second line we have made the change of variables $x\mapsto x'=x/a$. However we can simplify further:
\begin{align}
\tilde{f}\left(\frac{\xi}{a}\right)&=\int^\infty_{-\infty}f(x)e^{2\pi i\frac{\xi}{a}x}\mathrm{d}x\nonumber\\
\therefore \tilde{g}(\xi)&=\frac{1}{a}\tilde{f}\left(\frac{\xi}{a}\right).
\end{align}
The second example is for the case of $f(x)=e^{-\pi x^2}$. We start by differentiating
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}\xi}\tilde{f}(\xi)&=\frac{\mathrm{d}}{\mathrm{d}\xi}\int^\infty_{-\infty}f(x)e^{2\pi i\xi x}\mathrm{d}x\\
&=\int^\infty_{-\infty}e^{-\pi x^2}\frac{\mathrm{d}}{\mathrm{d}\xi}\left(e^{2\pi i\xi x}\right)\mathrm{d}x\\
&=\int^\infty_{-\infty}2\pi ixe^{-\pi x^2}e^{2\pi i\xi x}\mathrm{d}x
\end{align*}
where in the second line we have substituted in the definition of $f(x)$ and used the fact that it does not include a $\xi$ term to take it outside of the derivative.$^8$ The resulting integral is certainly daunting, so we will use integration by parts (eqn. 2) to tackle it. If we choose to take $u(x)=e^{2\pi i\xi x}$ and $v'(x)=xe^{-\pi x^2}$ (where the prime denotes differentiation with respect to $x$) it follows that $u'(x)=2\pi i\xi e^{2\pi i\xi x}$ and $v(x)=-e^{-\pi x^2}/2\pi+c$ where $c$ is a constant of integration which we will set to $0$. Substituting these in and evaluating gives$^9$
\begin{align*}
\frac{\mathrm{d}\tilde{f}}{\mathrm{d}\xi}&=2\pi i\Bigg(\left[e^{2\pi i\xi x}\frac{e^{-\pi x^2}}{-2\pi}\right]_{-\infty}^\infty-\int_{-\infty}^\infty 2\pi i\xi e^{2\pi i\xi x}\frac{e^{-\pi x^2}}{-2\pi}\mathrm{d}x\Bigg) \\
&=-2\pi i\int_{-\infty}^\infty 2\pi i\xi e^{2\pi i\xi x}\frac{e^{-\pi x^2}}{-2\pi}\mathrm{d}x \\
&=-2\pi\xi\int_{-\infty}^\infty e^{2\pi i\xi x}e^{-\pi x^2}\mathrm{d}x \\
&=-2\pi\xi\int_{-\infty}^\infty e^{2\pi i\xi x}f(x)\mathrm{d}x=-2\pi\xi\tilde{f}(\xi) \\
\end{align*}
where we have taken out the common constant factor of $2\pi i$ at the beginning. The final result is especially pleasing because it allows us to set up and solve a standard differential equation:$^{10}$
\begin{align*}
\frac{\mathrm{d}\tilde{f}(\xi)}{\mathrm{d}\xi}&=-2\pi\xi\tilde{f}(\xi) \\
\frac{\mathrm{d}\tilde{f}(\xi)}{\tilde{f}(\xi)}\frac{1}{\tilde{f}(\xi)}&=-2\pi\xi \\
\Rightarrow\tilde{f}(\xi)&=Ce^{-\pi\xi^2}
\end{align*}
for some constant $C$. It is plain to see that $\tilde{f}(\xi=0)=Ce^0=C$ so from the definition of the Fourier transform we then have $C=\tilde{f}(0)=\int_{-\infty}^{\infty}e^{0}f(x)\mathrm{d}x=\int_{-\infty}^{\infty}e^{-\pi x^2}\mathrm{d}x=1$. Thus we finally and rather anticlimactically find that
\begin{equation*}
\tilde{f}(\xi)=e^{-\pi\xi^2}=f(\xi).
\end{equation*}
The integration by parts and differential equation are treated in much more detail in the Notes.
Let us also consider a couple of examples which will prove useful. First, the Fourier transform of some function $g(x)=f(ax)$ for $a>0$. In this case,
\begin{align*}
\tilde{g}(\xi)=\int^\infty_{-\infty}g(x)e^{-2\pi i\xi x}\mathrm{d}x&=\int^\infty_{-\infty}f(ax)e^{-2\pi i\xi x}\mathrm{d}x\\
&=\int^\infty_{-\infty}f(x)e^{-2\pi i\xi \frac{x}{a}}\frac{\mathrm{d}x}{a}\\
&=\frac{1}{a}\int^\infty_{-\infty}f(x)e^{-2\pi i\frac{\xi}{a}x}\mathrm{d}x
\end{align*}
where in the second line we have made the change of variables $x\mapsto x'=x/a$. However we can simplify further:
\begin{align}
\tilde{f}\left(\frac{\xi}{a}\right)&=\int^\infty_{-\infty}f(x)e^{2\pi i\frac{\xi}{a}x}\mathrm{d}x\nonumber\\
\therefore \tilde{g}(\xi)&=\frac{1}{a}\tilde{f}\left(\frac{\xi}{a}\right).
\end{align}
The second example is for the case of $f(x)=e^{-\pi x^2}$. We start by differentiating
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}\xi}\tilde{f}(\xi)&=\frac{\mathrm{d}}{\mathrm{d}\xi}\int^\infty_{-\infty}f(x)e^{2\pi i\xi x}\mathrm{d}x\\
&=\int^\infty_{-\infty}e^{-\pi x^2}\frac{\mathrm{d}}{\mathrm{d}\xi}\left(e^{2\pi i\xi x}\right)\mathrm{d}x\\
&=\int^\infty_{-\infty}2\pi ixe^{-\pi x^2}e^{2\pi i\xi x}\mathrm{d}x
\end{align*}
where in the second line we have substituted in the definition of $f(x)$ and used the fact that it does not include a $\xi$ term to take it outside of the derivative.$^8$ The resulting integral is certainly daunting, so we will use integration by parts (eqn. 2) to tackle it. If we choose to take $u(x)=e^{2\pi i\xi x}$ and $v'(x)=xe^{-\pi x^2}$ (where the prime denotes differentiation with respect to $x$) it follows that $u'(x)=2\pi i\xi e^{2\pi i\xi x}$ and $v(x)=-e^{-\pi x^2}/2\pi+c$ where $c$ is a constant of integration which we will set to $0$. Substituting these in and evaluating gives$^9$
\begin{align*}
\frac{\mathrm{d}\tilde{f}}{\mathrm{d}\xi}&=2\pi i\Bigg(\left[e^{2\pi i\xi x}\frac{e^{-\pi x^2}}{-2\pi}\right]_{-\infty}^\infty-\int_{-\infty}^\infty 2\pi i\xi e^{2\pi i\xi x}\frac{e^{-\pi x^2}}{-2\pi}\mathrm{d}x\Bigg) \\
&=-2\pi i\int_{-\infty}^\infty 2\pi i\xi e^{2\pi i\xi x}\frac{e^{-\pi x^2}}{-2\pi}\mathrm{d}x \\
&=-2\pi\xi\int_{-\infty}^\infty e^{2\pi i\xi x}e^{-\pi x^2}\mathrm{d}x \\
&=-2\pi\xi\int_{-\infty}^\infty e^{2\pi i\xi x}f(x)\mathrm{d}x=-2\pi\xi\tilde{f}(\xi) \\
\end{align*}
where we have taken out the common constant factor of $2\pi i$ at the beginning. The final result is especially pleasing because it allows us to set up and solve a standard differential equation:$^{10}$
\begin{align*}
\frac{\mathrm{d}\tilde{f}(\xi)}{\mathrm{d}\xi}&=-2\pi\xi\tilde{f}(\xi) \\
\frac{\mathrm{d}\tilde{f}(\xi)}{\tilde{f}(\xi)}\frac{1}{\tilde{f}(\xi)}&=-2\pi\xi \\
\Rightarrow\tilde{f}(\xi)&=Ce^{-\pi\xi^2}
\end{align*}
for some constant $C$. It is plain to see that $\tilde{f}(\xi=0)=Ce^0=C$ so from the definition of the Fourier transform we then have $C=\tilde{f}(0)=\int_{-\infty}^{\infty}e^{0}f(x)\mathrm{d}x=\int_{-\infty}^{\infty}e^{-\pi x^2}\mathrm{d}x=1$. Thus we finally and rather anticlimactically find that
\begin{equation*}
\tilde{f}(\xi)=e^{-\pi\xi^2}=f(\xi).
\end{equation*}
The integration by parts and differential equation are treated in much more detail in the Notes.
The theta function
The Jacobi theta function is a function of two complex variables, $z$ across the whole complex plane and $\tau$ across the upper-half of the complex plane (the complex numbers with positive imaginary part), which is frequently used in the theory of (amongst others) elliptic functions. As most of its uses are well beyond the scope of this proof, I will gloss over them here. The theta function is (in generality)$^{11}$ given by
\begin{equation*}
\vartheta(z;\tau):=\sum^\infty_{n=-\infty}e^{i\pi n^2 \tau+2\pi inz}.
\end{equation*}
However, we will only consider the simplified function given by restricting the domain to $z=0$, $\tau=it$
\begin{equation}
\vartheta(0;it)\equiv\theta(t)=\sum^\infty_{n=-\infty}e^{-\pi n^2 t}
\end{equation}
where we have introduced the notation $\theta(t)$ for the sake of readability for this proof.
Now let's define functions $f(n)=e^{-\pi n^2}$ and $g(n)=e^{-\pi n^2 t}$ for some fixed $t>0$. We can see that $g(n)=f(n\sqrt{t})$. This allows us to use the first example from the above section on Fourier transforms (eqn. 6) to give the functional equation
\begin{equation*}
\tilde{g}(k)=\frac{1}{\sqrt{t}}\tilde{f}\left(\frac{k}{\sqrt{t}}\right)
\end{equation*}
We now use the Poisson summation formula (eqn. 5) to find
\begin{equation}
\sum_{k=-\infty}^\infty\tilde{g}(k)=\sum_{n=-\infty}^\infty g(n)=\theta(t).
\end{equation}
From equation 8 it immediately follows that
\begin{align}
\theta(t)&=\sum_{k=-\infty}^\infty\tilde{g}(k)=\sum_{k=-\infty}^\infty\frac{1}{\sqrt{t}}\tilde{f}\left(\frac{k}{\sqrt{t}}\right)\nonumber\\
&=\frac{1}{\sqrt{t}}\sum_{n=-\infty}^\infty\tilde{f}\left(\frac{n}{\sqrt{t}}\right)\nonumber\\
&=\frac{1}{\sqrt{t}}\theta\left(\frac{1}{t}\right)
\end{align}
where in the second line we have made a trivial index relabelling of $k\rightarrow n$ to keep our notation consistent.
We will now consider the behaviour of $\theta(t)$ for $t$ approaching $0$ from above (from the positive-$t$ direction, also denoted $t\rightarrow 0^+$). While it is not immediately obvious why, we will need to exploit it for the proof. We start with the expression
\begin{equation*}
\left|\theta(t)-\frac{1}{\sqrt{t}}\right|=\left|\frac{1}{\sqrt{t}}\left(\theta\left(\frac{1}{t}\right)-1\right)\right|
\end{equation*}
where we have made use of equation 9. Now let's consider the summand in the definition of the theta function, $e^{-\pi n^2 t}$. Due to the squaring of $n$, negative values of $n$ give the same result as positive values, while for $n=0$ we simply have $e^0=1$. This means we can rewrite the definition of the theta function given in equation 7 as the equivalent form
\begin{equation}
\theta(t)=1+2\sum_{n=1}^{\infty}e^{-\pi n^2t}.
\end{equation}
Using this form, we see that
\begin{equation*}
\left|\theta(t)-\frac{1}{\sqrt{t}}\right|=\left|\frac{1}{\sqrt{t}}\left(1+2\sum_{n=1}^{\infty}e^{-\pi n^2/t}-1\right)\right|=\frac{2}{\sqrt{t}}\sum_{n=1}^{\infty}e^{-\pi n^2/t}.
\end{equation*}
As we want to know the behaviour for very small $t$, we can assume that $1/\sqrt{t}<e^{1/t}/4$ (see fig. 3a). This allows us to state that
\begin{equation*}
\left|\theta(t)-\frac{1}{\sqrt{t}}\right|=\frac{2}{\sqrt{t}}\sum_{n=1}^{\infty}e^{-\pi n^2/t}<\frac{1}{2}e^{1/t}\sum_{n=1}^{\infty}e^{-\pi n^2/t}.
\end{equation*}
In turn we can likewise state that $e^{-3\pi/t}<1/2$ (see fig. 3b). Following on from this we find
\begin{align*}
\left|\theta(t)-\frac{1}{\sqrt{t}}\right|&<\frac{1}{2}e^{1/t}\sum_{n=1}^{\infty}e^{-\pi n^2/t}\\
&<\frac{1}{2}e^{1/t}\left(e^{-\pi/t}+e^{-4\pi/t}+e^{-9\pi/t}+\ldots\right)\\
&<\frac{1}{2}e^{-(\pi-1)/t}\left(1+e^{-3\pi/t}+e^{-8\pi/t}+\ldots\right)\\
&<\frac{1}{2}e^{-(\pi-1)/t}\left(1+\frac{1}{2}+\frac{1}{4}+\ldots\right)
\end{align*}
Of course, while we could have chosen a great number of values for these inequalities, as we can see, these choices have allowed us to simplify our calculations significantly. Note that the final two lines are not equal, but rather we have made use of the fact that $e^{-\pi n^2/t}<e^{-3(n-1)\pi/t}$ for $n\geq 2$ and therefore $e^{-\pi n^2/t}<(1/2)^{n-1}$ from $e^{-3\pi/t}<1/2$, thus preserving the overall inequality.
At this point, we note that, famously,
\begin{equation*}
\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots=\sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^n=1.
\end{equation*}
Thus our inequality becomes
\begin{align}
\left|\theta(t)-\frac{1}{\sqrt{t}}\right|&<\frac{1}{2}e^{-(\pi-1)/t}\left(1+\frac{1}{2}+\frac{1}{4}+\ldots\right)\nonumber\\
&<\frac{1}{2}e^{-(\pi-1)/t}(1+1)\nonumber\\
&<e^{-(\pi-1)/t}
\end{align}
This is a very useful result because we know that $e^{-(\pi-1)/t}$ is positive and finite for $t\rightarrow 0^+$, and so it provides a useful upper bound. Furthermore, it is easy to show that in this limit $e^{-(\pi-1)/t}\rightarrow 0$. Do note, however, that at $t=0$, $e^{-(\pi-1)/t}$ is not defined!
\begin{equation*}
\vartheta(z;\tau):=\sum^\infty_{n=-\infty}e^{i\pi n^2 \tau+2\pi inz}.
\end{equation*}
However, we will only consider the simplified function given by restricting the domain to $z=0$, $\tau=it$
\begin{equation}
\vartheta(0;it)\equiv\theta(t)=\sum^\infty_{n=-\infty}e^{-\pi n^2 t}
\end{equation}
where we have introduced the notation $\theta(t)$ for the sake of readability for this proof.
Now let's define functions $f(n)=e^{-\pi n^2}$ and $g(n)=e^{-\pi n^2 t}$ for some fixed $t>0$. We can see that $g(n)=f(n\sqrt{t})$. This allows us to use the first example from the above section on Fourier transforms (eqn. 6) to give the functional equation
\begin{equation*}
\tilde{g}(k)=\frac{1}{\sqrt{t}}\tilde{f}\left(\frac{k}{\sqrt{t}}\right)
\end{equation*}
We now use the Poisson summation formula (eqn. 5) to find
\begin{equation}
\sum_{k=-\infty}^\infty\tilde{g}(k)=\sum_{n=-\infty}^\infty g(n)=\theta(t).
\end{equation}
From equation 8 it immediately follows that
\begin{align}
\theta(t)&=\sum_{k=-\infty}^\infty\tilde{g}(k)=\sum_{k=-\infty}^\infty\frac{1}{\sqrt{t}}\tilde{f}\left(\frac{k}{\sqrt{t}}\right)\nonumber\\
&=\frac{1}{\sqrt{t}}\sum_{n=-\infty}^\infty\tilde{f}\left(\frac{n}{\sqrt{t}}\right)\nonumber\\
&=\frac{1}{\sqrt{t}}\theta\left(\frac{1}{t}\right)
\end{align}
where in the second line we have made a trivial index relabelling of $k\rightarrow n$ to keep our notation consistent.
We will now consider the behaviour of $\theta(t)$ for $t$ approaching $0$ from above (from the positive-$t$ direction, also denoted $t\rightarrow 0^+$). While it is not immediately obvious why, we will need to exploit it for the proof. We start with the expression
\begin{equation*}
\left|\theta(t)-\frac{1}{\sqrt{t}}\right|=\left|\frac{1}{\sqrt{t}}\left(\theta\left(\frac{1}{t}\right)-1\right)\right|
\end{equation*}
where we have made use of equation 9. Now let's consider the summand in the definition of the theta function, $e^{-\pi n^2 t}$. Due to the squaring of $n$, negative values of $n$ give the same result as positive values, while for $n=0$ we simply have $e^0=1$. This means we can rewrite the definition of the theta function given in equation 7 as the equivalent form
\begin{equation}
\theta(t)=1+2\sum_{n=1}^{\infty}e^{-\pi n^2t}.
\end{equation}
Using this form, we see that
\begin{equation*}
\left|\theta(t)-\frac{1}{\sqrt{t}}\right|=\left|\frac{1}{\sqrt{t}}\left(1+2\sum_{n=1}^{\infty}e^{-\pi n^2/t}-1\right)\right|=\frac{2}{\sqrt{t}}\sum_{n=1}^{\infty}e^{-\pi n^2/t}.
\end{equation*}
As we want to know the behaviour for very small $t$, we can assume that $1/\sqrt{t}<e^{1/t}/4$ (see fig. 3a). This allows us to state that
\begin{equation*}
\left|\theta(t)-\frac{1}{\sqrt{t}}\right|=\frac{2}{\sqrt{t}}\sum_{n=1}^{\infty}e^{-\pi n^2/t}<\frac{1}{2}e^{1/t}\sum_{n=1}^{\infty}e^{-\pi n^2/t}.
\end{equation*}
In turn we can likewise state that $e^{-3\pi/t}<1/2$ (see fig. 3b). Following on from this we find
\begin{align*}
\left|\theta(t)-\frac{1}{\sqrt{t}}\right|&<\frac{1}{2}e^{1/t}\sum_{n=1}^{\infty}e^{-\pi n^2/t}\\
&<\frac{1}{2}e^{1/t}\left(e^{-\pi/t}+e^{-4\pi/t}+e^{-9\pi/t}+\ldots\right)\\
&<\frac{1}{2}e^{-(\pi-1)/t}\left(1+e^{-3\pi/t}+e^{-8\pi/t}+\ldots\right)\\
&<\frac{1}{2}e^{-(\pi-1)/t}\left(1+\frac{1}{2}+\frac{1}{4}+\ldots\right)
\end{align*}
 |
| Figure 3: a) Plot of $1/\sqrt{t}$ (blue) and $e^{1/t}$ (yellow). Clearly for small $t$, $1/\sqrt{t}<e^{1/t}$. b) Plot of $e^{-3\pi/t}$. Note the values on the vertical axis -- for small $t$, the function is much smaller than $1/2$. |
At this point, we note that, famously,
\begin{equation*}
\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots=\sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^n=1.
\end{equation*}
Thus our inequality becomes
\begin{align}
\left|\theta(t)-\frac{1}{\sqrt{t}}\right|&<\frac{1}{2}e^{-(\pi-1)/t}\left(1+\frac{1}{2}+\frac{1}{4}+\ldots\right)\nonumber\\
&<\frac{1}{2}e^{-(\pi-1)/t}(1+1)\nonumber\\
&<e^{-(\pi-1)/t}
\end{align}
This is a very useful result because we know that $e^{-(\pi-1)/t}$ is positive and finite for $t\rightarrow 0^+$, and so it provides a useful upper bound. Furthermore, it is easy to show that in this limit $e^{-(\pi-1)/t}\rightarrow 0$. Do note, however, that at $t=0$, $e^{-(\pi-1)/t}$ is not defined!
The Mellin transform
The final tool we need to add to our toolbox is the Mellin transform $\mathcal{M}\{f(t)\}=\hat{f}(s)$. The Mellin transform is an integral transform closely related to the Fourier transform (amongst others) and finds uses all over mathematics, from number theory to statistics to, naturally, complex analysis. It is given by
\begin{equation*}
\hat{f}(s):=\int^\infty_0 f(t)t^{s-1}\mathrm{d}t.
\end{equation*}
\begin{equation*}
\hat{f}(s):=\int^\infty_0 f(t)t^{s-1}\mathrm{d}t.
\end{equation*}
Let us consider the Mellin transform of $f(t)=e^{-ct}$:
\begin{equation*}
\hat{f}(s)=\int^\infty_0 e^{-ct}t^{s-1}\mathrm{d}t.
\end{equation*}
We can make the substitution of variables $ct=u$ and $c\mathrm{d}t=\mathrm{d}u$ to give
\begin{align}
\hat{f}(s)&=\int^\infty_0 e^{-u}\left(\frac{u}{c}\right)^{s-1}\frac{\mathrm{d}u}{c}\nonumber\\
&=c^{-s}\int^\infty_0 e^{-u}u^{s-1}\mathrm{d}u\nonumber\\
&=c^{-s}\Gamma(s)
\end{align}
where in the final line we have used the integral definition of the gamma function (eqn. 1), noting that the variable of integration is a dummy variable (it doesn't matter if it is $u$ or $t$ as $\Gamma(s)$ does not depend on it).$^{12}$
\begin{equation*}
\hat{f}(s)=\int^\infty_0 e^{-ct}t^{s-1}\mathrm{d}t.
\end{equation*}
We can make the substitution of variables $ct=u$ and $c\mathrm{d}t=\mathrm{d}u$ to give
\begin{align}
\hat{f}(s)&=\int^\infty_0 e^{-u}\left(\frac{u}{c}\right)^{s-1}\frac{\mathrm{d}u}{c}\nonumber\\
&=c^{-s}\int^\infty_0 e^{-u}u^{s-1}\mathrm{d}u\nonumber\\
&=c^{-s}\Gamma(s)
\end{align}
where in the final line we have used the integral definition of the gamma function (eqn. 1), noting that the variable of integration is a dummy variable (it doesn't matter if it is $u$ or $t$ as $\Gamma(s)$ does not depend on it).$^{12}$
The proof itself
We begin by considering the Mellin transform of the theta function:
\begin{equation*}
\int_0^\infty \theta(t)t^{s-1}\mathrm{d}t=\int_0^\infty \sum_{n=-\infty}^\infty e^{-\pi n^2 t}t^{s-1}\mathrm{d}t.
\end{equation*}
As $t\rightarrow\infty$, $\theta(t)\rightarrow 1$ since all terms in the sum approach $0$ except for $n=0$. However, as $t\rightarrow 0^+$, we can see from equation 11 that $\theta(t)-1/\sqrt{t}\rightarrow 0$ and therefore $\theta(t)$ behaves like $1/\sqrt{t}$, which approaches infinity as $t\rightarrow 0^+$. In order to get convergence at both ends of the integral, we want the integrand to go to zero at the limits, and therefore we want to modify our integral to include correction terms. To do this we break the integral into two parts, the upper integral with the integrand $\theta(t)-1$ and the lower integral with the integrand $\theta(t)-1/\sqrt{t}$. To ensure the integrands are continuous over the full domain of integration, we make the break at $t=1$, where both integrands are equal to $\theta(t=1)-1$. So, mathematically, our integral becomes
\begin{equation*}
\int_0^1 \left(\theta(t)-\frac{1}{\sqrt{t}}\right)t^{s-1}\mathrm{d}t+\int_1^\infty (\theta(t)-1)t^{s-1}\mathrm{d}t.
\end{equation*}
Before we move on, we will make the substitution $s\mapsto s/2$. This will ensure the zeta function is in terms of $s$ rather than $2s$, which we choose simply for aesthetic appeal. Doing so gives us
\begin{equation*}
\phi(s)=\int_0^1 \left(\theta(t)-\frac{1}{\sqrt{t}}\right)t^{\frac{s}{2}-1}\mathrm{d}t+\int_1^\infty (\theta(t)-1)t^{\frac{s}{2}-1}\mathrm{d}t.
\end{equation*}
Now we calculate the first integral, using the definition of the theta function given by equation 10 and assuming $\mathrm{Re}(s)>1$ to avoid the possibility of taking the divergent integral of $1/t$.
\begin{align*}
\int_0^1 \left(\theta(t)-\frac{1}{\sqrt{t}}\right)t^{\frac{s}{2}-1}\mathrm{d}t&=\int_0^1 \left(1+2\sum_{n=1}^\infty e^{-\pi n^2 t}-t^{-\frac{1}{2}}\right)t^{\frac{s}{2}-1}\mathrm{d}t\\
&=\int_0^1 t^{\frac{s}{2}-1}\mathrm{d}t+2\int_0^1\sum_{n=1}^\infty e^{-\pi n^2 t}t^{\frac{s}{2}-1}\mathrm{d}t-\int_0^1 t^{\frac{s-3}{2}}\mathrm{d}t\\
&=\left[\frac{2}{s}t^{\frac{s}{2}}\right]^1_0+2\int_0^1\sum_{n=1}^\infty e^{-\pi n^2 t}t^{\frac{s}{2}-1}\mathrm{d}t-\left[\frac{2}{s-1}t^{\frac{s-1}{2}}\right]^1_0\\
&=\frac{2}{s}+2\int_0^1\sum_{n=1}^\infty e^{-\pi n^2 t}t^{\frac{s}{2}-1}\mathrm{d}t+\frac{2}{1-s}
\end{align*}
Using the same definition of the theta function, the second integral simplifies to
\begin{align*}
\int_1^\infty (\theta(t)-1)t^{\frac{s}{2}-1}\mathrm{d}t&=\int_1^\infty \left(1+2\sum_{n=1}^\infty e^{-\pi n^2 t}-1\right)t^{\frac{s}{2}-1}\mathrm{d}t\\
&=2\int_1^\infty \sum_{n=1}^\infty e^{-\pi n^2 t}t^{\frac{s}{2}-1}\mathrm{d}t.
\end{align*}
Therefore the overall integral becomes
\begin{align*}
\phi(s)&=\frac{2}{s}+2\int_0^1\sum_{n=1}^\infty e^{-\pi n^2 t}t^{\frac{s}{2}-1}\mathrm{d}t+\frac{2}{1-s}+2\int_1^\infty \sum_{n=1}^\infty e^{-\pi n^2 t}t^{\frac{s}{2}-1}\mathrm{d}t\\
&=2\sum_{n=1}^\infty \int_0^\infty e^{-\pi n^2 t}t^{\frac{s}{2}-1}\mathrm{d}t+\frac{2}{s}+\frac{2}{1-s}
\end{align*}
where we still have $\mathrm{Re}(s)>1$. Now we can employ equation 12 to find
\begin{align}
\frac{1}{2}\phi(s)&=\sum_{n=1}^\infty \left(\pi n^2\right)^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)+\frac{1}{s}+\frac{1}{1-s}\nonumber\\
&=\pi^{-\frac{s}{2}}\zeta(s)\Gamma\left(\frac{s}{2}\right)+\frac{1}{s}+\frac{1}{1-s}\nonumber\\
\Rightarrow\zeta(s)&=\frac{\pi^{\frac{s}{2}}}{\Gamma\left(\frac{s}{2}\right)}\left(\frac{1}{2}\phi(s)-\frac{1}{s}-\frac{1}{1-s}\right).
\end{align}
Note that in the second line we have identified the Dirichlet series $D(s)=\sum^\infty_{n=1}1/n^s$ with the zeta function $\zeta(s)$. We can do this because we are still working under the $\mathrm{Re}(s)>1$ condition, and so this is true by definition.
Now we are on the home stretch. We know from equation 4 that $1/\Gamma(s/2)$ is entire, and we ensured that $\phi(s)$ was everywhere convergent by hand. This means that the only possible poles in equation 13 can come from the $1/s$ and $1/(1-s)$ terms. For the $1/s$ term, we expect a possible pole at $s=0$, but in fact when we consider the whole term, $\pi^{s/2}/\big(s\Gamma(s/2)\big)$, we find
\begin{align*}
\frac{\pi^{\frac{s}{2}}}{s\Gamma\left(\frac{s}{2}\right)}&=\frac{\pi^{\frac{s}{2}}}{2\frac{s}{2}\Gamma\left(\frac{s}{2}\right)}\\
&=\frac{\pi^{\frac{s}{2}}}{2\Gamma\left(\frac{s}{2}+1\right)}\\
&\overset{s\mapsto 0}{=}\frac{1}{2\Gamma(1)}=\frac{1}{2}
\end{align*}
where we have again made use of the gamma function's functional equation (eqn. 3). It is easy to show that this trick does not work for the $1/(1-s)$ term at $s=1$, which is genuinely singular and produces a simple pole.
Thus equation 13 gives us an expression for $\zeta(s)$ which we know is equal to the Dirichlet series for $\mathrm{Re}(s)>1$ but is also meromorphic over the complex plane with a simple pole at $s=1$. Therefore, at long last, we have completed the analytic continuation of the Riemann zeta function.
Before we finish, one question remains unanswered. If $\zeta(s)$ can be analytically continued, how can we find its values for $\mathrm{Re}(s)<1$, where the Dirichlet series is no longer convergent? The simplest answer to this question is that the zeta function satisfies a functional equation (in fact it satisfies a number of them) which can be used to find the values in the $\mathrm{Re}(s)<1$ region. The seminar that this post is based on gives the proof for the functional equation $\Lambda(s)=\Lambda(1-s)$ where $\Lambda(s):=\pi^{\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\zeta(s)$. I strongly encourage the interested reader to attempt to prove this functional equation themselves as an exercise -- all of the required techniques have already been presented in this post, so it's certainly achievable! I hope this post was interesting and useful, thank you for reading along and best of luck with the exercise!
\begin{equation*}
\int_0^\infty \theta(t)t^{s-1}\mathrm{d}t=\int_0^\infty \sum_{n=-\infty}^\infty e^{-\pi n^2 t}t^{s-1}\mathrm{d}t.
\end{equation*}
As $t\rightarrow\infty$, $\theta(t)\rightarrow 1$ since all terms in the sum approach $0$ except for $n=0$. However, as $t\rightarrow 0^+$, we can see from equation 11 that $\theta(t)-1/\sqrt{t}\rightarrow 0$ and therefore $\theta(t)$ behaves like $1/\sqrt{t}$, which approaches infinity as $t\rightarrow 0^+$. In order to get convergence at both ends of the integral, we want the integrand to go to zero at the limits, and therefore we want to modify our integral to include correction terms. To do this we break the integral into two parts, the upper integral with the integrand $\theta(t)-1$ and the lower integral with the integrand $\theta(t)-1/\sqrt{t}$. To ensure the integrands are continuous over the full domain of integration, we make the break at $t=1$, where both integrands are equal to $\theta(t=1)-1$. So, mathematically, our integral becomes
\begin{equation*}
\int_0^1 \left(\theta(t)-\frac{1}{\sqrt{t}}\right)t^{s-1}\mathrm{d}t+\int_1^\infty (\theta(t)-1)t^{s-1}\mathrm{d}t.
\end{equation*}
Before we move on, we will make the substitution $s\mapsto s/2$. This will ensure the zeta function is in terms of $s$ rather than $2s$, which we choose simply for aesthetic appeal. Doing so gives us
\begin{equation*}
\phi(s)=\int_0^1 \left(\theta(t)-\frac{1}{\sqrt{t}}\right)t^{\frac{s}{2}-1}\mathrm{d}t+\int_1^\infty (\theta(t)-1)t^{\frac{s}{2}-1}\mathrm{d}t.
\end{equation*}
Now we calculate the first integral, using the definition of the theta function given by equation 10 and assuming $\mathrm{Re}(s)>1$ to avoid the possibility of taking the divergent integral of $1/t$.
\begin{align*}
\int_0^1 \left(\theta(t)-\frac{1}{\sqrt{t}}\right)t^{\frac{s}{2}-1}\mathrm{d}t&=\int_0^1 \left(1+2\sum_{n=1}^\infty e^{-\pi n^2 t}-t^{-\frac{1}{2}}\right)t^{\frac{s}{2}-1}\mathrm{d}t\\
&=\int_0^1 t^{\frac{s}{2}-1}\mathrm{d}t+2\int_0^1\sum_{n=1}^\infty e^{-\pi n^2 t}t^{\frac{s}{2}-1}\mathrm{d}t-\int_0^1 t^{\frac{s-3}{2}}\mathrm{d}t\\
&=\left[\frac{2}{s}t^{\frac{s}{2}}\right]^1_0+2\int_0^1\sum_{n=1}^\infty e^{-\pi n^2 t}t^{\frac{s}{2}-1}\mathrm{d}t-\left[\frac{2}{s-1}t^{\frac{s-1}{2}}\right]^1_0\\
&=\frac{2}{s}+2\int_0^1\sum_{n=1}^\infty e^{-\pi n^2 t}t^{\frac{s}{2}-1}\mathrm{d}t+\frac{2}{1-s}
\end{align*}
Using the same definition of the theta function, the second integral simplifies to
\begin{align*}
\int_1^\infty (\theta(t)-1)t^{\frac{s}{2}-1}\mathrm{d}t&=\int_1^\infty \left(1+2\sum_{n=1}^\infty e^{-\pi n^2 t}-1\right)t^{\frac{s}{2}-1}\mathrm{d}t\\
&=2\int_1^\infty \sum_{n=1}^\infty e^{-\pi n^2 t}t^{\frac{s}{2}-1}\mathrm{d}t.
\end{align*}
Therefore the overall integral becomes
\begin{align*}
\phi(s)&=\frac{2}{s}+2\int_0^1\sum_{n=1}^\infty e^{-\pi n^2 t}t^{\frac{s}{2}-1}\mathrm{d}t+\frac{2}{1-s}+2\int_1^\infty \sum_{n=1}^\infty e^{-\pi n^2 t}t^{\frac{s}{2}-1}\mathrm{d}t\\
&=2\sum_{n=1}^\infty \int_0^\infty e^{-\pi n^2 t}t^{\frac{s}{2}-1}\mathrm{d}t+\frac{2}{s}+\frac{2}{1-s}
\end{align*}
where we still have $\mathrm{Re}(s)>1$. Now we can employ equation 12 to find
\begin{align}
\frac{1}{2}\phi(s)&=\sum_{n=1}^\infty \left(\pi n^2\right)^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)+\frac{1}{s}+\frac{1}{1-s}\nonumber\\
&=\pi^{-\frac{s}{2}}\zeta(s)\Gamma\left(\frac{s}{2}\right)+\frac{1}{s}+\frac{1}{1-s}\nonumber\\
\Rightarrow\zeta(s)&=\frac{\pi^{\frac{s}{2}}}{\Gamma\left(\frac{s}{2}\right)}\left(\frac{1}{2}\phi(s)-\frac{1}{s}-\frac{1}{1-s}\right).
\end{align}
Note that in the second line we have identified the Dirichlet series $D(s)=\sum^\infty_{n=1}1/n^s$ with the zeta function $\zeta(s)$. We can do this because we are still working under the $\mathrm{Re}(s)>1$ condition, and so this is true by definition.
Now we are on the home stretch. We know from equation 4 that $1/\Gamma(s/2)$ is entire, and we ensured that $\phi(s)$ was everywhere convergent by hand. This means that the only possible poles in equation 13 can come from the $1/s$ and $1/(1-s)$ terms. For the $1/s$ term, we expect a possible pole at $s=0$, but in fact when we consider the whole term, $\pi^{s/2}/\big(s\Gamma(s/2)\big)$, we find
\begin{align*}
\frac{\pi^{\frac{s}{2}}}{s\Gamma\left(\frac{s}{2}\right)}&=\frac{\pi^{\frac{s}{2}}}{2\frac{s}{2}\Gamma\left(\frac{s}{2}\right)}\\
&=\frac{\pi^{\frac{s}{2}}}{2\Gamma\left(\frac{s}{2}+1\right)}\\
&\overset{s\mapsto 0}{=}\frac{1}{2\Gamma(1)}=\frac{1}{2}
\end{align*}
where we have again made use of the gamma function's functional equation (eqn. 3). It is easy to show that this trick does not work for the $1/(1-s)$ term at $s=1$, which is genuinely singular and produces a simple pole.
Thus equation 13 gives us an expression for $\zeta(s)$ which we know is equal to the Dirichlet series for $\mathrm{Re}(s)>1$ but is also meromorphic over the complex plane with a simple pole at $s=1$. Therefore, at long last, we have completed the analytic continuation of the Riemann zeta function.
Before we finish, one question remains unanswered. If $\zeta(s)$ can be analytically continued, how can we find its values for $\mathrm{Re}(s)<1$, where the Dirichlet series is no longer convergent? The simplest answer to this question is that the zeta function satisfies a functional equation (in fact it satisfies a number of them) which can be used to find the values in the $\mathrm{Re}(s)<1$ region. The seminar that this post is based on gives the proof for the functional equation $\Lambda(s)=\Lambda(1-s)$ where $\Lambda(s):=\pi^{\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\zeta(s)$. I strongly encourage the interested reader to attempt to prove this functional equation themselves as an exercise -- all of the required techniques have already been presented in this post, so it's certainly achievable! I hope this post was interesting and useful, thank you for reading along and best of luck with the exercise!
Notes
$1$. The reasoning behind this may not be obvious, but the convention is chosen as such for a number of good reasons that I won't go into here. Just to give one motivating example though, consider the power series expansion of the exponential function
\begin{equation}
e^x=\sum^\infty_{n=0}\frac{x^n}{n!}.\nonumber
\end{equation}
e^x=\sum^\infty_{n=0}\frac{x^n}{n!}.\nonumber
\end{equation}
If $0!$ were not defined, or defined differently, this elegant expression would not be possible, and the case exists likewise for a large number of other unrelated examples.
$2$. First, let's consider a meromorphic function $m(s)$ with a single pole at $s=s_0$. We can express this function as $m(s)=h(s)/(s-s_0)^n$ for some entire (everywhere holomorphic) function $h(s)$ -- effectively the holomorphic function is 'punctured' by a singularity acting like $1/s^n$ located at $s_0$. The behaviour of the pole is determined by its order $n$, with "simple" poles corresponding to poles of order $n=1$. This negative-degree polynomial behaviour is important to ensure differentiability in the neighbourhood of the pole. Note the similarity of this definition to that of the Laurent series (mentioned in Note 2 of my Fourier Fun 2 post), which is defined on an annulus, rather than by a Taylor series as we might naively expect from real analysis. In fact the Laurent series $m(s)=\sum_{k=-\infty}^\infty a_k (s-s_0)^k$ about a pole at $s_0$ can be broken into two parts: the "regular part" for $k\geq 0$ and the "principal part" for $k<0$. The principal part will always (necessary and sufficient) have $a_k=0$ for values of $k<-n$ and $a_n\neq 0$, where $n$ is still the order of the pole. For meromorphic functions with multiple poles, we simply make the replacement $m(s)=h(s)/z(s)$ for appropriately chosen $z(s)$ (appropriately chosen in the sense that the behaviour of the poles is the same as in the single pole case and the poles are located in the correct positions).
$3$. I have omitted the proof that this definition is equivalent to the integral representation in equation 1 because it is somewhat non-trivial and is not particularly relevant to the overall proof. If the reader is unsatisfied with taking the definition for granted and would like to research it themselves, this formula is known as the Euler (infinite product) representation of the gamma function. If they are up for a challenge, they may wish to prove it themselves!
$4$. As in the case of Note 3, proving this identity is challenging and not particularly important for the overall proof, so I am omitting it. Coincidentally, as in the case of Note 3, this infinite product definition of the sine function is also due to Euler. Again, I encourage the interested reader to read more about it or attempt a proof of their own.
$5$. This will be covered in more depth in my upcoming post to round out the Fourier Fun series, Part 3.
$6$. This definition, as well as the definition for the inverse Fourier transform which results from it, is a matter of convention. For example, this definition is unitary and uses spatial frequency $\xi$ rather than angular spatial frequency (wavenumber) $k$ which is arguably more common in a physics context. Neither of these properties are necessary, although the corresponding definitions will be slightly different (e.g., factors of $1/\sqrt{2}$).
$7$. Define a function $f(x)=\sum^\infty_{n=-\infty}g(x+n)$. As the summation runs over the integers $n$, the function $f(x)$ must have period $1$ (if this is not clear, consider as an example what $f(0)$ and $f(1)$ would look like if the summation were expanded out term by term -- as the summation is infinite, they must be identical) and therefore can be expressed as a Fourier series: $f(x)=\sum^\infty_{n=-\infty}c_n e^{2\pi inx}$ where we have used the polar/exponential form rather than sines and cosines. Interestingly, the Fourier coefficients are given by
\begin{align*}
c_n&=\int^1_0 f(x)e^{2\pi inx}\mathrm{d}x\\
&=\int^1_0 \sum^\infty_{n=-\infty}g(x+n) e^{2\pi inx}\mathrm{d}x\\
&=\int^\infty_{-\infty} g(x) e^{2\pi inx}\mathrm{d}x\\
&=\tilde{g}(n)
\end{align*}
where in going from the second to the third line we have exploited the periodicity of $g$ and taken the integral-of-a-sum as a sum-of-integrals (note the change in the limits of integration). Following on from this, we have
\begin{align*}
\sum^\infty_{n=-\infty}g(n)=f(0)&=\sum^\infty_{n=-\infty}c_n e^{2\pi in0}\\
&=\sum^\infty_{n=-\infty}c_n=\sum^\infty_{n=-\infty}\tilde{g}(n) \nonumber
\end{align*}
thus proving the formula.
$8$. Furthermore, in the first line we have interchanged the order of integration and differentiation. The more keenly rigour-minded readers may be somewhat concerned about this. While it is certainly true that there are functions for which the order of integration and differentiation cannot be interchanged (that is, for some $g(x)=\int f(x,y) \mathrm{d}y$, $\mathrm{d}g/\mathrm{d}x\neq\int\mathrm{d}y\partial f/\partial x$) these functions are often "pathological", or at the very least discontinuous in $f$ or $\partial f/\partial x$ over the region defined by the limits of integration. The case is similar for interchange of the order of integration (for multiple integrals) or interchange of the order of differentiation (for multiple partial derivatives). We will not have to worry about such concerns in this, or indeed for virtually any of the posts on this blog except, of course, where otherwise noted.
$9$. This part of the derivation is especially tricky so I want to give it a more solid treatment in the notes. First, we start with the function $v'(x)=xe^{-\pi x^2}$. In order to find $v(x)$ we integrate both sides with respect to $x$ to give
\begin{equation*}
\int\frac{\mathrm{d}v(x)}{\mathrm{d}x}\mathrm{d}x=\int xe^{-\pi x^2}\mathrm{d}x
\end{equation*}
Now we simplify the left-hand side and make the substitution of variables $u=-\pi x^2$ and $\mathrm{d}u=-2\pi x\mathrm{d}x$:
\begin{align*}
\int\mathrm{d}v(u)&=\int\frac{e^u}{-2\pi}\mathrm{d}u\\
v(u)&=\frac{e^u}{-2\pi}+c\\
v(x)&=\frac{e^{-\pi x^2}}{-2\pi}+c
\end{align*}
where in the second line we have made use of the fact that the integral of $e^u$ is itself and in the third line substituted $-\pi x^2$ back in for $u$. In this final line, $c$ is a constant of integration.
As the computation of $u(x)$ is straightforward, we are now in a position to substitute in $u(x)$, $v(x)$, $u'(x)$ and $v'(x)$ into the integration by parts formula to yield
\begin{equation*}
\frac{\mathrm{d}\tilde{f}}{\mathrm{d}\xi}=2\pi i\Bigg(\left[e^{2\pi i\xi x}\left(\frac{e^{-\pi x^2}}{-2\pi}+c\right)\right]_{-\infty}^\infty-\int_{-\infty}^\infty 2\pi i\xi e^{2\pi i\xi x}\left(\frac{e^{-\pi x^2}}{-2\pi}+c\right)\mathrm{d}x\Bigg)
\end{equation*}
Momentary consideration of the first term inside the brackets shows why we must set the constant of integration to zero -- we have the expression $ce^{2\pi i\xi x}|^\infty_{-\infty}$ appearing. The evaluations of $x$ at $\pm\infty$ cannot be dealt with sensibly and so must be removed by setting $c=0$. (As a side-note, in general this constant of integration does not need to be considered -- for all definite integrations by parts it will not play a relevant part. I have included it here for completeness but will not do so for other cases.) The other part of the first term, $e^{2\pi i\xi x}e^{-\pi x^2}|^\infty_{-\infty}$, is automatically zero because $e^{-\pi x^2}\rightarrow 0$ as $x\rightarrow +\infty$ or $-\infty$.
This only leaves what was the second (integral) term inside the brackets, simplified by taking $c=0$. Since the first term was zero, we can remove the brackets to give
\begin{equation*}
\frac{\mathrm{d}\tilde{f}}{\mathrm{d}\xi}=-2\pi i\int_{-\infty}^\infty 2\pi i\xi e^{2\pi i\xi x}\frac{e^{-\pi x^2}}{-2\pi}\mathrm{d}x
\end{equation*}
After cancelling off factors of $2\pi$ and being very careful with signs, substituting $e^{-\pi x^2}=f(x)$ causes the main result to neatly fall out:
\begin{equation*}
\frac{\mathrm{d}\tilde{f}}{\mathrm{d}\xi}=-2\pi\xi\int_{-\infty}^\infty e^{2\pi i\xi x}f(x)=-2\pi\xi\tilde{f}(\xi)\mathrm{d}x
\end{equation*}
$10$. This differential equation is very easily to solve manually in case you can't be bothered memorising the standard solutions to first order ODEs. I presume that if the reader has made it this far they are reasonably comfortable with them but I have provided the solution below regardless, perhaps in case of a memory lapse.
\begin{align*}
\frac{\mathrm{d}\tilde{f}(\xi)}{\mathrm{d}\xi}&=-2\pi\xi\tilde{f}(\xi) \\
\frac{\mathrm{d}\tilde{f}(\xi)}{\mathrm{d}\xi}\frac{1}{\tilde{f}(\xi)}&=-2\pi\xi \\
\int\frac{\mathrm{d}\tilde{f}(\xi)}{\mathrm{d}\xi}\frac{1}{\tilde{f}(\xi)}\mathrm{d}\xi&=-2\pi\int\xi\mathrm{d}\xi \\
\int\frac{\mathrm{d}\tilde{f}(\xi)}{\tilde{f}(\xi)}&=-2\pi\int\xi\mathrm{d}\xi \\
\log_e(\tilde{f}(\xi))+c_1&=-\pi\xi^2+c_2 \\
\log_e(\tilde{f}(\xi))&=-\pi\xi^2+c_3 \\
\tilde{f}(\xi)&=Ce^{-\pi\xi^2}
\end{align*}
where $c_1$ and $c_2$ are constants of integration, $c_3=c_2-c_1$ and $C=e^{c_3}$.
$11$. Similar to the case of the Bessel functions, there are also three modified versions of the theta function which are known as auxiliary theta functions. When considered in this context, the theta function as described here is referred to as $\vartheta_{00}(z;\tau)$ to distinguish it from the first auxiliary theta function. Additionally, when the theta function and the three auxiliary theta functions are taken together as a function of "nomes" $q=e^{i\pi\tau}$ instead of $\tau$, they are labelled $\theta_1\ldots\theta_4$ in which case the theta function in this proof is given by $\theta_3$. These are only notational subtleties, however, which is why I do not raise them in the main body. I only mention them here to avoid confusion for any interested readers who may wish to find out more about the topic elsewhere.
$12$. Note also that this implies that $\mathcal{M}\{e^{-t}\}=\Gamma(s)$. Furthermore, this method can be used to prove one of the elementary properties of the Mellin transform, that $\mathcal{M}\{f(ct)\}=c^{-s}\hat{f}(s)$. As this is the generalised form of what we showed, we could also have assumed it from the start to gain the result immediately. Interestingly, this generalised form can also be proven using another related integral transform, the two-sided Laplace transform.
$3$. I have omitted the proof that this definition is equivalent to the integral representation in equation 1 because it is somewhat non-trivial and is not particularly relevant to the overall proof. If the reader is unsatisfied with taking the definition for granted and would like to research it themselves, this formula is known as the Euler (infinite product) representation of the gamma function. If they are up for a challenge, they may wish to prove it themselves!
$4$. As in the case of Note 3, proving this identity is challenging and not particularly important for the overall proof, so I am omitting it. Coincidentally, as in the case of Note 3, this infinite product definition of the sine function is also due to Euler. Again, I encourage the interested reader to read more about it or attempt a proof of their own.
$5$. This will be covered in more depth in my upcoming post to round out the Fourier Fun series, Part 3.
$6$. This definition, as well as the definition for the inverse Fourier transform which results from it, is a matter of convention. For example, this definition is unitary and uses spatial frequency $\xi$ rather than angular spatial frequency (wavenumber) $k$ which is arguably more common in a physics context. Neither of these properties are necessary, although the corresponding definitions will be slightly different (e.g., factors of $1/\sqrt{2}$).
$7$. Define a function $f(x)=\sum^\infty_{n=-\infty}g(x+n)$. As the summation runs over the integers $n$, the function $f(x)$ must have period $1$ (if this is not clear, consider as an example what $f(0)$ and $f(1)$ would look like if the summation were expanded out term by term -- as the summation is infinite, they must be identical) and therefore can be expressed as a Fourier series: $f(x)=\sum^\infty_{n=-\infty}c_n e^{2\pi inx}$ where we have used the polar/exponential form rather than sines and cosines. Interestingly, the Fourier coefficients are given by
\begin{align*}
c_n&=\int^1_0 f(x)e^{2\pi inx}\mathrm{d}x\\
&=\int^1_0 \sum^\infty_{n=-\infty}g(x+n) e^{2\pi inx}\mathrm{d}x\\
&=\int^\infty_{-\infty} g(x) e^{2\pi inx}\mathrm{d}x\\
&=\tilde{g}(n)
\end{align*}
where in going from the second to the third line we have exploited the periodicity of $g$ and taken the integral-of-a-sum as a sum-of-integrals (note the change in the limits of integration). Following on from this, we have
\begin{align*}
\sum^\infty_{n=-\infty}g(n)=f(0)&=\sum^\infty_{n=-\infty}c_n e^{2\pi in0}\\
&=\sum^\infty_{n=-\infty}c_n=\sum^\infty_{n=-\infty}\tilde{g}(n) \nonumber
\end{align*}
thus proving the formula.
$8$. Furthermore, in the first line we have interchanged the order of integration and differentiation. The more keenly rigour-minded readers may be somewhat concerned about this. While it is certainly true that there are functions for which the order of integration and differentiation cannot be interchanged (that is, for some $g(x)=\int f(x,y) \mathrm{d}y$, $\mathrm{d}g/\mathrm{d}x\neq\int\mathrm{d}y\partial f/\partial x$) these functions are often "pathological", or at the very least discontinuous in $f$ or $\partial f/\partial x$ over the region defined by the limits of integration. The case is similar for interchange of the order of integration (for multiple integrals) or interchange of the order of differentiation (for multiple partial derivatives). We will not have to worry about such concerns in this, or indeed for virtually any of the posts on this blog except, of course, where otherwise noted.
$9$. This part of the derivation is especially tricky so I want to give it a more solid treatment in the notes. First, we start with the function $v'(x)=xe^{-\pi x^2}$. In order to find $v(x)$ we integrate both sides with respect to $x$ to give
\begin{equation*}
\int\frac{\mathrm{d}v(x)}{\mathrm{d}x}\mathrm{d}x=\int xe^{-\pi x^2}\mathrm{d}x
\end{equation*}
Now we simplify the left-hand side and make the substitution of variables $u=-\pi x^2$ and $\mathrm{d}u=-2\pi x\mathrm{d}x$:
\begin{align*}
\int\mathrm{d}v(u)&=\int\frac{e^u}{-2\pi}\mathrm{d}u\\
v(u)&=\frac{e^u}{-2\pi}+c\\
v(x)&=\frac{e^{-\pi x^2}}{-2\pi}+c
\end{align*}
where in the second line we have made use of the fact that the integral of $e^u$ is itself and in the third line substituted $-\pi x^2$ back in for $u$. In this final line, $c$ is a constant of integration.
As the computation of $u(x)$ is straightforward, we are now in a position to substitute in $u(x)$, $v(x)$, $u'(x)$ and $v'(x)$ into the integration by parts formula to yield
\begin{equation*}
\frac{\mathrm{d}\tilde{f}}{\mathrm{d}\xi}=2\pi i\Bigg(\left[e^{2\pi i\xi x}\left(\frac{e^{-\pi x^2}}{-2\pi}+c\right)\right]_{-\infty}^\infty-\int_{-\infty}^\infty 2\pi i\xi e^{2\pi i\xi x}\left(\frac{e^{-\pi x^2}}{-2\pi}+c\right)\mathrm{d}x\Bigg)
\end{equation*}
Momentary consideration of the first term inside the brackets shows why we must set the constant of integration to zero -- we have the expression $ce^{2\pi i\xi x}|^\infty_{-\infty}$ appearing. The evaluations of $x$ at $\pm\infty$ cannot be dealt with sensibly and so must be removed by setting $c=0$. (As a side-note, in general this constant of integration does not need to be considered -- for all definite integrations by parts it will not play a relevant part. I have included it here for completeness but will not do so for other cases.) The other part of the first term, $e^{2\pi i\xi x}e^{-\pi x^2}|^\infty_{-\infty}$, is automatically zero because $e^{-\pi x^2}\rightarrow 0$ as $x\rightarrow +\infty$ or $-\infty$.
This only leaves what was the second (integral) term inside the brackets, simplified by taking $c=0$. Since the first term was zero, we can remove the brackets to give
\begin{equation*}
\frac{\mathrm{d}\tilde{f}}{\mathrm{d}\xi}=-2\pi i\int_{-\infty}^\infty 2\pi i\xi e^{2\pi i\xi x}\frac{e^{-\pi x^2}}{-2\pi}\mathrm{d}x
\end{equation*}
After cancelling off factors of $2\pi$ and being very careful with signs, substituting $e^{-\pi x^2}=f(x)$ causes the main result to neatly fall out:
\begin{equation*}
\frac{\mathrm{d}\tilde{f}}{\mathrm{d}\xi}=-2\pi\xi\int_{-\infty}^\infty e^{2\pi i\xi x}f(x)=-2\pi\xi\tilde{f}(\xi)\mathrm{d}x
\end{equation*}
$10$. This differential equation is very easily to solve manually in case you can't be bothered memorising the standard solutions to first order ODEs. I presume that if the reader has made it this far they are reasonably comfortable with them but I have provided the solution below regardless, perhaps in case of a memory lapse.
\begin{align*}
\frac{\mathrm{d}\tilde{f}(\xi)}{\mathrm{d}\xi}&=-2\pi\xi\tilde{f}(\xi) \\
\frac{\mathrm{d}\tilde{f}(\xi)}{\mathrm{d}\xi}\frac{1}{\tilde{f}(\xi)}&=-2\pi\xi \\
\int\frac{\mathrm{d}\tilde{f}(\xi)}{\mathrm{d}\xi}\frac{1}{\tilde{f}(\xi)}\mathrm{d}\xi&=-2\pi\int\xi\mathrm{d}\xi \\
\int\frac{\mathrm{d}\tilde{f}(\xi)}{\tilde{f}(\xi)}&=-2\pi\int\xi\mathrm{d}\xi \\
\log_e(\tilde{f}(\xi))+c_1&=-\pi\xi^2+c_2 \\
\log_e(\tilde{f}(\xi))&=-\pi\xi^2+c_3 \\
\tilde{f}(\xi)&=Ce^{-\pi\xi^2}
\end{align*}
where $c_1$ and $c_2$ are constants of integration, $c_3=c_2-c_1$ and $C=e^{c_3}$.
$11$. Similar to the case of the Bessel functions, there are also three modified versions of the theta function which are known as auxiliary theta functions. When considered in this context, the theta function as described here is referred to as $\vartheta_{00}(z;\tau)$ to distinguish it from the first auxiliary theta function. Additionally, when the theta function and the three auxiliary theta functions are taken together as a function of "nomes" $q=e^{i\pi\tau}$ instead of $\tau$, they are labelled $\theta_1\ldots\theta_4$ in which case the theta function in this proof is given by $\theta_3$. These are only notational subtleties, however, which is why I do not raise them in the main body. I only mention them here to avoid confusion for any interested readers who may wish to find out more about the topic elsewhere.
$12$. Note also that this implies that $\mathcal{M}\{e^{-t}\}=\Gamma(s)$. Furthermore, this method can be used to prove one of the elementary properties of the Mellin transform, that $\mathcal{M}\{f(ct)\}=c^{-s}\hat{f}(s)$. As this is the generalised form of what we showed, we could also have assumed it from the start to gain the result immediately. Interestingly, this generalised form can also be proven using another related integral transform, the two-sided Laplace transform.
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