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$$f(z) = \int\limits_{-\infty}^{\infty} \frac{x}{\sin(x z)} dx$$

What is the function $f(z)$? This doesn't converge for real $z$ but when $z$ is purely imaginary then $\sin$ becomes $\sinh$. So for example $f(i)=-\frac{i}{2}\pi^2$. But that's as far as I got.

zooby
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1 Answers1

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For $z \in \mathbb{C}$, let $f(z)$ be defined as follows:

\begin{equation} f(z)=2\int\limits_{0}^{+\infty} \frac{x}{\sin(zx)}\,\mathrm{d}x \end{equation}

Using the complex exponential definition of $\sin(zx)$:

\begin{equation} f(z)=4i\int\limits_{0}^{+\infty} \frac{x}{e^{izx}-e^{-izx}}\,\mathrm{d}x \end{equation}

\begin{equation} f(z)=4i\int\limits_{0}^{+\infty} \frac{xe^{-izx}}{1-e^{-2izx}}\,\mathrm{d}x \end{equation}

Using the geometric series for $e^{-2izx}$, you get the following:

\begin{equation} f(z)=4i\int\limits_{0}^{+\infty} xe^{-izx}\sum_{k=0}^{+\infty}e^{-2izxk}\,\mathrm{d}x \end{equation}

\begin{equation} f(z)=4i\sum_{k=0}^{+\infty}\int\limits_{0}^{+\infty} xe^{-izx(1+2k)}\,\mathrm{d}x \end{equation}

With the substitution $s=izx(1+2k)$, you get that:

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

The integral is just $\Gamma(2)$, thus:

\begin{equation} f(z)=\frac{4i}{z^{2}}\sum_{k=0}^{+\infty}\frac{1}{(1+2k)^{2}} \end{equation}

The last sum is known to converge to $\pi^{2}/8$, so you can conclude that:

\begin{equation} \int\limits_{-\infty}^{+\infty} \frac{x}{\sin(zx)}\,\mathrm{d}x= \frac{i\pi^{2}}{2z^{2}} \end{equation}

From the result, one can indeed see that $f(i)=-i\pi^{2}/2$.

Luis Sierra
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    the answer is not quite correct - note that $f(-z)=-f(z)$, but it can be easily corrected; expanding the series for $\Im z>0$ requires the series in $e^{ixz(1+2k)}$ for $x>0$ to have convergence at $\infty$; also the change of variable is generally incorrect since $s$ is not real unless $z$ is purely imaginary but you can argue as follows - $f$ is analytic on both $\Im z>0$ and $\Im z<0$ and for $z=iy, y >0$ the computation is almost correct (the series has to change), $s=yx(1+2k)$ is a valid change of variable, so one gets $f(iy)=-i\pi^2/(2y^2)$ so $f(z)=\frac{i\pi^{2}}{2z^{2}}, \Im z >0$ – Conrad Jul 29 '20 at 02:12