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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 $(-1,1)$ ?

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

1 Answer 1

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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).

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  • $\begingroup$ Thanks , I edited the link for Wolfram alpha , the result is 1.63 $\endgroup$
    – zeraoulia rafik
    Commented Mar 29, 2019 at 21:34
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    $\begingroup$ @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). $\endgroup$
    – J.G.
    Commented Mar 29, 2019 at 21:38
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    $\begingroup$ In fact, more generally I find $$\int_{0}^{\infty}x^{a}\left(\coth x^{b}-1\right)dx=\frac{2^{1-\frac{a+1}{b}}}{b}\Gamma\left(\frac{a+1}{b}\right)\zeta\left(\frac{a+1}{b}\right).$$ $\endgroup$
    – J.G.
    Commented Mar 29, 2019 at 21:52
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    $\begingroup$ Nice closed form $\endgroup$
    – zeraoulia rafik
    Commented Mar 29, 2019 at 21:57
  • $\begingroup$ @zeraouliarafik Well, insofar as we consider special functions such as $\Gamma,\,\zeta$ closed forms. That's part of what we like about them. $\endgroup$
    – J.G.
    Commented Mar 29, 2019 at 22:16

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