# an integral of my own creation

#### Random Variable

##### Well-known member
MHB Math Helper
Show that $\displaystyle \int_{0}^{\infty} \left( \frac{1}{x^{2}} - \frac{1}{x \sinh x} \right) \ dx = \ln 2$.

I actually might have seen this one evaluated before but in a way that didn't make much sense to me.

Last edited:

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
I am interested to see the solution !

#### Random Variable

##### Well-known member
MHB Math Helper
Since $\displaystyle \frac{1}{\sinh x}$ behaves like $\displaystyle \frac{1}{x}$ near $0$, the singularity at $x=0$ is removable.

Let $f(z) = \displaystyle \frac{1}{z^{2}} - \frac{1}{z \sinh z}$ and integrate around a rectangle with vertices at $z=N, z= N + i \pi (N+\frac{1}{2}),$

$z = -N + i \pi (N+ \frac{1}{2}),$ and $z= - N,$ where $N$ is an positive integer.

Letting $N$ go to infinty, $\displaystyle \int \frac{dz}{z^{2}}$ and $\int \displaystyle \frac{dz}{z \sinh z}$ will evaluate to zero along the top and sides of the rectangle.

But it's not obvious that $\displaystyle \int \frac{dz}{z \sinh z}$ evaluates to $0$ along the top of the rectangle. So I'll show that.

$\displaystyle \Big| \int_{-N}^{N} \frac{dt}{[t+i \pi (N+\frac{1}{2})] \sinh[ t + i \pi(N+\frac{1}{2})]} \Big| \le \int_{-N}^{N} \frac{dt}{[\pi(N+\frac{1}{2})-t] \cosh t}$

$\displaystyle \le \int_{-N}^{N} \frac{1}{\pi(N+\frac{1}{2}) \cosh t} \ dt \le \frac{1}{\pi(N+ \frac{1}{2})} \int_{-\infty}^{\infty} \frac{1}{\cosh t} \ dt = \frac{1}{N+\frac{1}{2}} \to 0$ as $N \to \infty$

So $\displaystyle \int_{0}^{\infty} \Big( \frac{1}{x^{2}} - \frac{1}{x \sinh x} \Big) \ dx = \frac{1}{2} \int_{-\infty}^{\infty} \Big( \frac{1}{x^{2}} - \frac{1}{x \sinh x} \Big) \ dx= \pi i \sum_{n=1}^{\infty} \text{Res} [f(z), n \pi i]$

$\displaystyle \text{Res} [f,n \pi i] = \lim_{z \to n \pi i} \frac{\sinh z -z}{2z \sinh z + z^{2} \cosh x} = \frac{-n \pi i}{(n \pi i)^{2} (-1)^{n}} = (-1)^{n-1} \frac{1}{n \pi i}$

$\implies \displaystyle \int_{0}^{\infty} \Big( \frac{1}{x^{2}} - \frac{1}{x \sinh x} \Big) \ dx = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} = \ln 2$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
I was sure that it can be solved by contour integration but I am really surprized that this easy-looking value cannot be obtained by real methods , or can it be ?

#### Random Variable

##### Well-known member
MHB Math Helper
I was sure that it can be solved by contour integration but I am really surprized that this easy-looking value cannot be obtained by real methods , or can it be ?
I'll link to the evaluation I referred to in the my original post. It uses analytic continuation in a way that just doesn't make sense to me.

AoPS Forum

Last edited:

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Hey , I found the following answer .

#### Random Variable

##### Well-known member
MHB Math Helper
Hey , I found the following answer .
Incredibly I had that page bookmarked. Maybe I forget about it because it's more like an outline of a possible evaluation than an actual evaluation.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Incredibly I had that page bookmarked. Maybe I forget about it because it's more like an outline of a possible evaluation than an actual evaluation.
Maybe because it looks a little different