An integral of my own creation

In summary, the integral $\displaystyle \int_{0}^{\infty} \left( \frac{1}{x^{2}} - \frac{1}{x \sinh x} \right) \ dx$ can be evaluated using contour integration, by integrating around a rectangle and taking the limit as one of the vertices goes to infinity. This results in the sum of residues of the function $f(z) = \frac{1}{z^{2}} - \frac{1}{z \sinh z}$, which can be evaluated using the Residue Theorem. The final result is $\ln 2$. However, there is also an alternative method that uses analytic continuation, which may be confusing to some
  • #1
polygamma
229
0
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.
 
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  • #2
I am interested to see the solution !
 
  • #3
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$
 
  • #4
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 ?
 
  • #5
ZaidAlyafey said:
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.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=503261&p=2828173&hilit=dirichlet+lambda#p2828173
 
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  • #6
Hey , I found the following answer .
 
  • #7
ZaidAlyafey said:
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.
 
  • #8
Random Variable said:
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 ;)
 

FAQ: An integral of my own creation

What is an integral of my own creation?

An integral of my own creation is a mathematical concept created by me, the scientist. It is a unique integral that I have developed for a specific purpose or to solve a specific problem.

How is an integral of my own creation different from a regular integral?

An integral of my own creation is different from a regular integral in that it is not a well-known or commonly used integral in mathematics. It is a new integral that I have created and is not found in textbooks or other resources.

What is the purpose of creating my own integral?

The purpose of creating my own integral is to solve a specific problem or to make a contribution to the field of mathematics. It allows for new and unique solutions to be discovered and can lead to further advancements in the field.

How do I know if my integral is valid?

To determine if your integral is valid, it must follow the rules and principles of integration. It should also be able to solve the problem it was created for and produce accurate and consistent results.

Can I use an integral of my own creation in my research or studies?

Yes, you can use an integral of your own creation in your research or studies. However, it is important to explain and justify its use, as it may not be a common or well-established integral in the field of mathematics.

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