# closed form of the following integral :$\int_{0}^{\infty}- \sqrt{x}+ \sqrt{x}\coth (x) dx$?

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 $$(-1,1)$$ ?

• The definite integral that you linked on Wolfram Alpha has a $\coth\left(\sqrt{x}\right)$ rather than $\coth\left(x\right)$.
– user281997
Commented Mar 29, 2019 at 21:30
• Thanks , I edited the link for Wolfram alpha , the result is 1.63 Commented Mar 29, 2019 at 21:35

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).

• Thanks , I edited the link for Wolfram alpha , the result is 1.63 Commented Mar 29, 2019 at 21:34
• @zeraouliarafik Well, 1.63706 etc. Thanks for clarifying. FWIW the problem in your previous WA link would have also been solvable the same way, viz. $$\int_0^\infty\frac{2\sqrt{x}e^{-2\sqrt{x}}dx}{1-e^{-2\sqrt{x}}}=4\sum_{n\ge 1}\int_0^\infty t^2e^{-2nt}dt=\zeta(3)$$ (assuming I haven't made any mistakes in my arithmetic).
– J.G.
Commented Mar 29, 2019 at 21:38
• In fact, more generally I find $$\int_{0}^{\infty}x^{a}\left(\coth x^{b}-1\right)dx=\frac{2^{1-\frac{a+1}{b}}}{b}\Gamma\left(\frac{a+1}{b}\right)\zeta\left(\frac{a+1}{b}\right).$$
– J.G.
Commented Mar 29, 2019 at 21:52
• Nice closed form Commented Mar 29, 2019 at 21:57
• @zeraouliarafik Well, insofar as we consider special functions such as $\Gamma,\,\zeta$ closed forms. That's part of what we like about them.
– J.G.
Commented Mar 29, 2019 at 22:16