I have tried to evaluate this:$\int_{0}^{\infty}- \sqrt{x}+ \sqrt{x}\coth (x)$ using the the following formula
$$2 \Gamma(a) \zeta(a) \left(1-\frac{1}{2^{a}} \right) = \int_{0}^{\infty}\Big( \frac{x^{a-1}}{\sinh x} – x^{a-2}\Big) \, dx \ , \ {\color{bleu}{-1}} <\text{Re}(a) <1. \tag{1}$$
in order to present the preceding integral in closed form but I failed, This
$\int_{0}^{\infty}- \sqrt{x}+ \sqrt{x}\coth (x)$ close to $1.63$ using wolfram alpha and I really believe that integral could be represented in a closed form value using $(1)$ since Coth function has a relationship with Sinh function, Now Is there a general formula for:$$\int_{0}^{\infty}\Big( \frac{x^{a-1}}{\tanh x} – x^{a-1}\Big) \, dx \ , \ {\color{bleu}{-1}} <\text{Re}(a) <1. \tag{2}$$ if we assume $a$ as a complex varible with real part lie in $(-1,1)$ ?
Best Answer
An easier approach is to rewrite the integral as $$\int_0^\infty\frac{2\sqrt{x}e^{-2x}dx}{1-e^{-2x}}=2\sum_{n\ge 1}\int_0^\infty\sqrt{x}e^{-2nx}dx=2\Gamma\left(\frac{3}{2}\right)\sum_{n=1}(2n)^{-3/2}=\sqrt{\frac{\pi}{8}}\zeta\left(\frac{3}{2}\right).$$This zeta constant isn't known, it seems, to have a nice closed form (but see also its discussion here, which up to $x=t^2$ reports the same integral representation).