4
$\begingroup$

Due to this question, I found myself trying to take the following integral:

$$\int_0^x\vartheta_3(0,t)\ dt=\ ?$$

However, I know not of how to do this. As per this post, I find the evaluation at $x=1$ to be $\frac\pi{\tanh\pi}$. It is equivalent to trying to evaluate the following series:

$$\sum_{n=0}^\infty\frac{x^{n^2}}{n^2+1}$$

My end problem is that I want to evaluate the following integral:

$$\int_0^1\frac1x\int_0^x\vartheta_3(0,t)\ dt\ dx=\sum_{n=0}^\infty\frac1{(n^2+1)^2}$$

  • 0
    by partial fraction decomposition of $\frac{1}{((x+i)(x-i))^2}$ it becomes the Hurwitz zeta at $(s,a) = (1,\pm i)$ and $ (2,\pm i)$ – 2017-01-01
  • 3
    If you know the formula $$\sum_{n=0}^{\infty} \frac{1}{n^2 + a} = \frac{1}{2a} + \frac{\pi}{2\sqrt{a}} \coth(\pi\sqrt{a}), $$ then you can differentiate both sides to evaluate the series. But for the integral of the theta function, I have no idea. – 2017-01-01
  • 0
    @SangchulLee I see, with the [usual partial fraction decomposition](http://math.stackexchange.com/questions/1959285/sum-k-infty-infty-frac1k-alpha2-frac-pi2-sin2-pi-alpha/1960134#1960134) of trigonometric functions $\sum_{n=-\infty}^\infty \frac{1}{n^2+a} = \frac{\pi}{\sqrt{a}} \coth(\pi \sqrt{a})$ – 2017-01-01
  • 3
    @user1952009, Yes, that is one way of proving this. There are other ways, such as log-differentiating the infinite product formula for $\sinh$, computing the Fourier series of $\cosh ax$, applying Poisson summation formula to a certain integral representation, etc. – 2017-01-01
  • 0
    @SangchulLee You are right my friend, why did I not think of this? If you could post as an answer, I'd be happy. – 2017-01-01
  • 0
    No worries, that happens to everyone! Also, I think the question you raised may also be an interesting problem independent of your end goal. As I have no idea on it, maybe we can be patient and wait to see if someone else is willing to try this. – 2017-01-01

1 Answers 1

3

About the last point, we have: $$ \mathcal{L}^{-1}\left(\frac{1}{(x^2+1)^2}\right)=\frac{\sin s-s\cos s}{2} $$ hence: $$\sum_{n\geq 0}\frac{1}{(n^2+1)^2} = 1+\int_{0}^{+\infty}\frac{\sin s-s\cos s}{2(e^s-1)}\,ds$$ and the last integral can be computed through the residue theorem. As an alternative, starting from $$ \frac{\sinh(\pi x)}{\pi x}=\prod_{n\geq 1}\left(1+\frac{x^2}{n^2}\right)$$ and considering $\frac{d}{dx}\log(\cdot)$ of both sides, we get: $$ -\frac{1}{x}+\pi\coth(\pi x) = \sum_{n\geq 1}\frac{2x}{n^2+x^2} $$ and by differentiating again: $$ \frac{1}{x^2}-\frac{\pi^2}{\sinh^2(\pi x)}=2\sum_{n\geq 1}\frac{n^2-x^2}{(n^2+x^2)^2}$$ so that: $$\boxed{ \sum_{n\geq 1}\frac{1}{(n^2+x^2)^2} = \color{red}{\frac{1}{4x^4}\left(-2+\pi x\coth(\pi x)+\left(\frac{\pi x}{\sinh(\pi x)}\right)^2\right)}}$$ We may also exploit the Poisson summation formula to get the same, since the Fourier transform of $\frac{1}{(1+x^2)^2}$ is a multiple of $(1+|x|)\,e^{-|x|}$.