Calculating the integral $\int_{0}^\infty x^{s-1} \frac{\cos(x^2/ \pi)}{\sinh x}dx$ where s is a complex variable and $1<Re(s)$.

Question

I want to evaluate the above integral, I know this is related to Mellin transform. But I am unable to calculate the integral using the definitions of integral or any other method. Any help is highly appreciated.

Answer 1

Here is my attempt, however, it does not lead to a nice closed form solution. Under the conditions defined above, let $I_{s}$ be defined as:

\begin{equation} I_{s}=\int\limits_{0}^{+\infty} \frac{x^{s-1}\cos\left(x^{2}/\pi\right)}{\sinh(x)}\,\mathrm{d}x \end{equation}

Note that for all $x$:

\begin{equation} \cos\left(\frac{x^{2}}{\pi}\right)=\sum_{k=0}^{+\infty}\frac{(-1)^{k}\pi^{-2k}x^{4k}}{(2k)!} \end{equation}

Plugging this last expression in $I_{s}$:

\begin{equation} I_{s}=\int\limits_{0}^{+\infty} \frac{x^{s-1}}{\sinh(x)}\sum_{k=0}^{+\infty}\frac{(-1)^{k}\pi^{-2k}x^{4k}}{(2k)!}\,\mathrm{d}x \end{equation}

\begin{equation} I_{s}=\sum_{k=0}^{+\infty}\frac{(-1)^{k}\pi^{-2k}}{(2k)!}\int\limits_{0}^{+\infty} \frac{x^{s-1+4k}}{\sinh(x)}\,\mathrm{d}x \end{equation}

Introducing the exponential definition of $\sinh(x)$:

\begin{equation} I_{s}=2\sum_{k=0}^{+\infty}\frac{(-1)^{k}\pi^{-2k}}{(2k)!}\int\limits_{0}^{+\infty} \frac{x^{s-1+4k}}{e^{x}-e^{-x}}\,\mathrm{d}x \end{equation}

\begin{equation} I_{s}=2\sum_{k=0}^{+\infty}\frac{(-1)^{k}\pi^{-2k}}{(2k)!}\int\limits_{0}^{+\infty} \frac{x^{s-1+4k}e^{-x}}{1-e^{-2x}}\,\mathrm{d}x \end{equation}

Let's deal with the integral first and set it equal to $I$:

\begin{equation} I=\int\limits_{0}^{+\infty} \frac{x^{s-1+4k}e^{-x}}{1-e^{-2x}}\,\mathrm{d}x \end{equation}

In the interval $[0,\infty)$, it holds that: $0\leq e^{-2x}\leq1$, so we can use the geometric series for $e^{-2x}$:

\begin{equation} I=\int\limits_{0}^{+\infty} x^{s-1+4k}e^{-x}\sum_{n=0}^{+\infty}e^{-2nx}\,\mathrm{d}x \end{equation}

\begin{equation} I=\sum_{n=0}^{+\infty}\int\limits_{0}^{+\infty} x^{s-1+4k}e^{-x(1+2n)}\,\mathrm{d}x \end{equation}

With the substitution $z=x(1+2n)$, you get that:

\begin{equation} I=\sum_{n=0}^{+\infty}\frac{1}{(1+2n)^{s+4k}}\int\limits_{0}^{+\infty} z^{(s+4k)-1}e^{-z}\,\mathrm{d}z \end{equation}

The last integral evaluates to $\Gamma(s+4k)$, thus:

\begin{equation} I=\Gamma(s+4k)\sum_{n=0}^{+\infty}\frac{1}{(1+2n)^{s+4k}} \end{equation}

It is known that for $\mathrm{Re}(z)>1$, the following holds:

\begin{equation} \sum_{n=0}^{+\infty}\frac{1}{(1+2n)^{z}}=(1-2^{-z})\zeta(z) \end{equation}

Thus:

\begin{equation} I=\int\limits_{0}^{+\infty} \frac{x^{s-1+4k}e^{-x}}{1-e^{-2x}}\,\mathrm{d}x=\Gamma(s+4k)(1-2^{-(s+4k)})\zeta(s+4k) \end{equation}

Getting back to our $I_{s}$ and plugging this in, you can conclude that:

\begin{equation} \boxed{\int\limits_{0}^{+\infty} \frac{x^{s-1}\cos\left(x^{2}/\pi\right)}{\sinh(x)}\,\mathrm{d}x=2\sum_{k=0}^{+\infty}\frac{(-1)^{k}\pi^{-2k}\Gamma(s+4k)(1-2^{-(s+4k)})\zeta(s+4k)}{(2k)!}} \end{equation}