How can I approximate $\int_{0}^{\infty}x^{7/6}\ln(\frac{1}{1-e^{-x}})dx$? My approximation here needs to be valid for large values of $x$ here.
$$\int_{0}^{\infty}x^{7/6}\ln\Bigg(\frac{1}{1-e^{-x}}\Bigg)dx$$
I can't seem to get around this natural log no matter what I do. How can this integral be approximated without going to $0$? I know there is a solution in an integral table if the natural log wasn't there.
 A: If we use the Maclaurin series of the logarithm, we obtain that the integral is
$$
\sum\limits_{n = 1}^\infty  {\frac{1}{n}\int_0^{ + \infty } {x^{7/6} e^{ - nx} dx} }  = \Gamma \left( {2 + \tfrac{1}{6}} \right)\sum\limits_{n = 1}^\infty  {\frac{1}{{n^{3 + 1/6} }}}  = \Gamma \left( {2 + \tfrac{1}{6}} \right)\zeta \left( {3 + \tfrac{1}{6}} \right),
$$
where $\Gamma$ is the gamma function and $\zeta$ is the Riemann zeta function.
A: There is a very simple way to get rid of the logarithm: integration by parts. So we have
$$\int^\infty_0x^a\ln\left(\frac1{1-e^{-x}}\right)\,dx=\frac1{a+1}\int^\infty_0x^{a+1}\frac{e^{-x}}{1-e^{-x}}\,dx.$$ The fraction can be written as a geometric series:
$$\frac{e^{-x}}{1-e^{-x}}=\sum^\infty_{k=1}e^{-kx},$$ and since it's a positive series, we can exchange summation and integral:
$$\int^\infty_0x^{a+1}\frac{e^{-x}}{1-e^{-x}}\,dx=\sum^\infty_{k=1}\int^\infty_0x^{a+1}e^{-kx}\,dx.$$ Substituting $kx=t$, we see that
$$\int^\infty_0x^{a+1}e^{-kx}\,dx=\frac1{k^{a+2}}\int^\infty_0t^{a+1}e^{-t}\,dt=\frac{\Gamma(a+2)}{k^{a+2}},$$ using the well-known integral representation of the Gamma function. So our integral becomes
$$\int^\infty_0x^{a+1}\frac{e^{-x}}{1-e^{-x}}\,dx=\Gamma(a+2)\,\sum^\infty_{k=1}\frac1{k^{a+2}}=\Gamma(a+2)\,\zeta(a+2).$$ Putting all of that together, we obtain
$$\int^\infty_0x^a\ln\left(\frac1{1-e^{-x}}\right)\,dx=\frac{\Gamma(a+2)\,\zeta(a+2)}{a+1}=\Gamma(a+1)\,\zeta(a+2)$$ (many thanks to @Gary for the last simplification).
For $a=7/6$, that's approximately $1.2686090974543140685705195862758272411$.
