Proof of functional equation of $\psi(x)=\sum_{n=1}^{\infty}e^{-n^2\pi x}$ without using Poisson Summation Formula

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The function $$\psi(x)=\sum_{n=1}^{\infty}e^{-n^2\pi x}$$ satisfies the functional equation $$\dfrac{1+2\psi(x)}{1+2\psi(x^{-1})}=\dfrac{1}{\sqrt{x}}$$


I was trying to find a simple proof of this. I found one using the Poisson Summation Formula if $w(x)=\sum \limits_{n=1}^\infty e^{-n^2\pi x}$ then how to show that $\frac {1+2w(x)} {1+2w(\frac{1}{x})}=\frac{1}{\sqrt{x}}$ . Is there some other proof of this?

Best Answer

I provide here a brief outline of how Ramanujan proved the functional equation in question.

Let us then start with a real variable $x\in(0,1)$ and define variables $q, y, z$ dependent on $x$ as follows \begin{align} z&={}_2F_1\left(\frac{1}{2},\frac{1}{2};1;x\right)\tag{1}\\ y&=\pi\cdot\dfrac{{}_2F_1\left(\dfrac{1}{2},\dfrac{1}{2};1;1-x\right)} {{}_2F_1\left(\dfrac{1}{2},\dfrac{1}{2};1;x\right)}\tag{2}\\ q&=e^{-y} \tag{3} \end{align} where $${}_2F_1(a,b;c;x)=1+\frac{a\cdot b} {c} \cdot\frac{x} {1!}+\frac{a(a+1)\cdot b(b+1)}{c(c+1)}\cdot\frac{x^2}{2!}+\dots$$ is Gauss hypergeometric function. You can get a more detailed explanation of the notation for hypergeometric functions in this answer of mine.

The relation between $x$ and $z$ can also be written in terms of elliptic integrals as $$z=\frac{2}{\pi}\int_0^{\pi/2}\frac{dt}{\sqrt{1-x\sin^2t}}\tag{4}$$ Using these definitions Ramanujan proved that $$\lim_{x\to 0^+}\frac{q}{x}=\frac{1}{16}\tag{5}$$ Using this result and some properties of hypergeometric functions he inverted the relation between $q$ and $x$ to obtain $$x=1-\frac{\phi^4(-q)}{\phi^4(q)}, z=\phi^2(q)\tag{6}$$ where $$\phi(q) =\sum_{n\in\mathbb{Z}} q^{n^2}\tag{7}$$ is a theta function (in Ramanujan's notation, Jacobi used $\vartheta_3(q)$ to denote the same function). This theta function is related to the function $\psi$ mentioned in question as $$\phi(q) =1+2\psi(y/\pi)\tag{8}$$ Let the functional dependence of $q$ on $x$ be denoted as $q=F(x) $ and then one can note that $\log F(x) =-y$ and $$\log F(x) \cdot \log F(1-x)=\pi^2\tag{9}$$ Using this equation we get $\log F(1-x)=-\pi^2/y$ or $$F(1-x)=e^{-\pi^2/y}\tag{10}$$ Let us also observe that the second equation in $(6)$ can be written as $$\phi^2(F(x)) ={}_2F_1\left(\frac{1}{2},\frac{1}{2};1;x\right)\tag{11}$$ so that we also have $$\phi^2(F(1-x))={}_2F_1\left(\frac{1}{2},\frac{1}{2};1;1-x\right)\tag{12}$$ Let $\alpha, \beta$ be positive real numbers with $\alpha\beta=\pi$ and let $y=\alpha^2$ so that $\pi^2/y=\beta^2$. Dividing $(12)$ by $(11)$ we get $$\frac{\phi^2(F(1-x))}{\phi^2(F(x))}=\frac{y}{\pi}=\frac{\alpha}{\beta}$$ or $$\alpha \phi^2(e^{-\alpha^2})=\beta\phi^2(e^{-\beta^2})$$ Taking square roots we get the desired identity in question.


The crux of Ramanujan's approach lies in the proof of identities $(5),(6)$ using the definitions $(1),(2),(3),(7)$ and identity $(4)$. The proofs have been provided with all relevant details in my blog posts: post 1, post 2.

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