# Evaluate $\int_{0}^{+\infty }{\left( \frac{x}{{{\text{e}}^{x}}-{{\text{e}}^{-x}}}-\frac{1}{2} \right)\frac{1}{{{x}^{2}}}\text{d}x}$

Evaluate : $$\int_{0}^{+\infty }{\left( \frac{x}{{{\text{e}}^{x}}-{{\text{e}}^{-x}}}-\frac{1}{2} \right)\frac{1}{{{x}^{2}}}\text{d}x}$$

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How many weird integrals and sums have you posted now with absolutely no context or thought of your own? –  mrf Feb 4 '13 at 9:34
@Ryan: if I understand mrf correctly, when you post one of these integrals, perhaps you may want to add a brief description of how you may have tried doing it, or how you think it may need to be done. That said, I do not want to discourage your postings, as we have been able to learn some techniques for these problems. –  Ron Gordon Feb 4 '13 at 9:53
It's hard for me to refrain myself from upvoting the question since I only focus on the problem itself (+1). On the other hand, I understand people that ask for more details because your question might be wrongly understood. –  Chris's sis Feb 4 '13 at 10:30
maple12 couldn't get closed form, numerically it's -0.34657... –  coffeemath Feb 4 '13 at 13:18
@coffeemath: this seems like $-\ln 2 / 2$ –  Chris's sis Feb 4 '13 at 13:23

Related technique. Here is a closed form solution of the integral

$$\int_{0}^{+\infty }{\left( \frac{x}{{{\text{e}}^{x}}-{{\text{e}}^{-x}}}-\frac{1}{2} \right)\frac{1}{{{x}^{2}}}\text{d}x} = -\frac{\ln(2)}{2}.$$

Here is the technique, consider the integral

$$F(s) = \int_{0}^{+\infty }{e^{-sx}\left( \frac{x}{{{\text{e}}^{x}}-{{\text{e}}^{-x}}}-\frac{1}{2} \right)\frac{1}{{{x}^{2}}}\text{d}x},$$

which implies

$$F''(s) = \int_{0}^{+\infty }{e^{-sx}\left( \frac{x}{{{\text{e}}^{x}}-{{\text{e}}^{-x}}}-\frac{1}{2} \right)\text{d}x}.$$

The last integral is the Laplace transform of the function

$$\frac{x}{{{\text{e}}^{x}}-{{\text{e}}^{-x}}}-\frac{1}{2}$$

and equals

$$F''(s) = \frac{1}{4}\,\psi' \left( \frac{1}{2}+\frac{1}{2}\,s \right) -\frac{1}{2s}.$$

Now, you need to integrate the last equation twice and determine the two constants of integrations, then take the limit as $s\to 0$ to get the result.

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Interesting method, I use trying to solve this yesterday, by $\psi$ do you mean $\psi(x)=\frac{\Gamma '(x)}{\Gamma(x)}$ –  user10444 Feb 5 '13 at 11:32
@MhenniBenghorbal :Thx for your nice method! I struggled for few days and find a ugly solution...:) –  Ryan Feb 5 '13 at 11:43
Can you show how to get that two constants plz? :) –  Ryan Feb 5 '13 at 12:11
@MhenniBenghorbal: brilliant! (+1) –  Chris's sis Feb 5 '13 at 14:19
@MhenniBenghorbal: I think that Laplace point is a tricky one :+) –  B. S. Jul 14 '13 at 6:37