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

Answer

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.