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Calculate the integral: $$\int_{0}^{\infty}\left(\frac{x}{e^{x}-e^{-x}}-\frac{1}{2}\right)\frac{\,dx}{x^2}$$

My solution is not very glamorous.

Write the integrand as:

$$1/2\int_{0}^{\infty}\left(\frac{1}{\sinh(x)}-\frac{1}{x}\right)\cdot \frac{1}{x}dx$$

Consider the Dirichlet Lambda Function.

For $s>1$, the DLF is defined as :

$$\lambda(s) = \sum_{k=0}^{\infty}\frac{1}{(2k+1)^{s}}=\left(1-\frac{1}{2^{s}}\right)\zeta(s)=\frac{1}{2\Gamma(s)}\int_{0}^{\infty}\frac{x^{s-1}}{\sinh(x)}dx$$.

But, in the domain $0<s<1$, it can also be written as:

$$\lambda(s)=\frac{1}{2\Gamma(s)}\int_{0}^{\infty}\left(\frac{1}{\sinh(x)}-\frac{1}{x}\right)^{s-1}dx................[1]$$

Now, we can write:

$$2\Gamma(s)\lambda(s)=2\Gamma(s)\left(1-\frac{1}{2^{s}}\right)\zeta(s)$$

As can be seen from [1] as well, if we take the limit of this as $s\to 0$, then we find it converges to $$-\ln(2)$$.

Don't forget the 1/2 from in front of the integral sign in the original integral, then we finally have our result: $$\boxed{-1/2\ln(2)}$$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I found a rather obscure identity in my notes. I have not derived it. It works in a jiffy here:

$$\int_{0}^{\infty}\left(\frac{1}{\sinh(x)}-\frac{1}{x}\right)\frac{x}{x^{2}+4\pi^{2}s^{2}}=1/2\left[\psi\left(s+1/2\right)-\psi(s+1)\right]$$

Now, let $s=0$ and we have.......................$-\ln(2)$

Cool C! I didn't notice the connection between the integral and Dirichlet's Lambda Function.

\begin{align*}\int_0^{\infty} \left(\frac{x}{e^{x}-e^{-x}} - \frac{1}{2}\right)\frac{\,dx}{x^2}&= \int_0^{\infty}\int_0^{\infty} \left(\frac{x}{e^{x}-e^{-x}} - \frac{1}{2}\right)se^{-sx}\,ds\,dx\\&= \frac{1}{2}\int_0^{\infty} \int_0^{\infty} \frac{2xe^{-x} + e^{-2x} - 1}{1-e^{-2x}}se^{-sx}\,dx\,ds\\&= \frac{1}{2}\int_0^{\infty} \int_0^{\infty} s(2xe^{-(1+s)x} + e^{-(2+s)x} - e^{-sx})\sum\limits_{n=0}^{\infty} e^{-2nx}\,dx\,ds\\&= \frac{1}{2}\int_0^{\infty} s\sum\limits_{n=0}^{\infty}\left(\frac{2}{(1+s+2n)^2} + \frac{1}{2+s+2n} - \frac{1}{s+2n}\right)\,ds\\&= \frac{1}{2}\int_0^{\infty} \sum\limits_{n=0}^{\infty} \left(\frac{2}{1+s+2n} - \frac{2(2n+1)}{(1+s+2n)^2}+\frac{2n}{s+2n} - \frac{2n+2}{s+2n+2}\right)\,ds\\&= -\lim\limits_{N\to \infty}\sum\limits_{n=0}^{N}\left(\log (1+2n) + \frac{(2n+1)}{2n+1}+n\log (2n) - (n+1)\log (2n+2)\right)\\&= -\lim\limits_{N \to \infty} \log \frac{(2N+1)!e^{N+1}}{2^{2N+1}N!(N+1)^{N+1}}\\&= -\frac{1}{2}\log 2 \qquad(\text{by Stirling's approximation})\end{align*}

Nice use of the partial sum/stirling thing there at the end RD. I like.

Yet another solution (that is not mine) and I found in my notes.

We note that the integrand function is even , hence:

\begin{align*}
\int_{0}^{\infty}\left ( \frac{x}{e^x-e^{-x}} - \frac{1}{2} \right )\frac{{\rm d}x}{x^2} &=\frac{1}{2}\int_{-\infty}^{\infty}\left ( \frac{xe^x}{e^{2x}-1} - \frac{1}{2} \right )\frac{{\rm d}x}{x^2} \\
&=\frac{1}{2}\int_{-\infty}^{\infty}\frac{2xe^x -e^{2x}+1}{2\left ( e^{2x}-1 \right )x^2}\, {\rm d}x
\end{align*}

Consider the function $\displaystyle f(z)= \frac{2ze^z -e^{2z}+1}{2\left ( e^{2z}-1 \right )z^2}$ and $f(0)= -\frac{1}{12}$. Since $\lim \limits_{z \rightarrow 0+} f(z)= -\frac{1}{12}$ then the function will be analytic in the upper half plane and will still have simple poles at $z_k = i \pi \kappa, \; \kappa =1, 2,3, \dots$.

We can easily establish the residue:

$$\mathfrak{Res}\left ( f(z); i \pi \kappa \right )=\frac{(-1)^{\kappa+1}i}{2\pi \kappa}$$

Now we are integrating $f$ on a semicircle located at the UHP. We can easily see that the contour integral over the arc vanishes as the radius $R \rightarrow +\infty$, since:

\begin{align*}
\left | f(z) \right | &= \frac{1}{2R^2}\cdot \frac{\left | 2ze^z-e^{2z}+1 \right |}{\left | e^{2z}-1 \right |}\\
& \! \! \! \! \! \! \! \! \! \! \! \overset{z=Re^{i\theta}, \; \theta \in [0, \pi]}{\leq } \frac{1}{2R^2}\frac{2Re^R+e^{2R}+1}{e^{2R}-1} \xrightarrow{R \rightarrow +\infty}0
\end{align*}

Hence , if $\gamma$ denotes the circle then:

\begin{align*}
\oint_{\gamma}f(z)\, {\rm d}z &=2\pi i \sum_{\kappa=1}^{\infty}\frac{(-1)^{\kappa+1}i}{2\pi \kappa} \\
&=- \sum_{\kappa=1}^{\infty}\frac{(-1)^{\kappa+1}}{\kappa} \\
&= - \ln 2
\end{align*}

Hence if $\mathcal{J}$ denotes the initial integral then:

$$\mathcal{J}=\frac{1}{2}\int_{-\infty}^{\infty}\frac{2xe^x -e^{2x}+1}{2\left ( e^{2x}-1 \right )x^2}\, {\rm d}x = -\frac{\ln 2}{2}$$

as proven in the previous solutions.

Cool, T

Well yesterday I encountered a problem where I had to use the Mellin Transform of $\frac{1}{\sinh ax}$. I was remembering this post , but I had to make a little adjustment to the output.

What I discovered is the following formula:

$$\int_{0}^{\infty}\frac{x^{s-1}}{\sinh ax} \, {\rm d}x = \frac{2}{a^2}\Gamma(s) \lambda(s) , \;\; a>0, \;\; s>1$$

I don't think that it is too difficult to prove. For example , one might begin using the series expansion of $\frac{1}{\sinh ax}$ which is quite easy to get. Then interchange integral and sum and I believe the result falls immediately.

I'll leave the derivation to the reader. I also guess that in the domain $0<s<1$ we'll have a similar formula. C, what do you think?

Yeah, T, I would say that is probably the case.

There are some fun integrals associated with this one.

One may derive the integral you mention by using Zeta, so it is more than likely safe to assume we can continue it to $0<s<1$.


Like you said, showing the one you mention can be done using geometric series and the 'e' series for sinh.

There is a similar series for cosh:

$$\int_{0}^{\infty}\frac{x^{s-1}}{\cosh(ax)}dx=\frac{2\Gamma(s)}{(2a)^{s}}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)^{s}}$$

There's that Dirichlet Lambda thing again.