How can I approximate $\int_{0}^{\infty}x^{7/6}\ln(\frac{1}{1-e^{-x}})dx$?

Question

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.

Answer 1

This is not a full answer.

Considering $$I(a)=\int_{0}^{\infty}x^{(1+a)}\log\Bigg(\frac{1}{1-e^{-x}}\Bigg)\,dx $$ by Taylor $$I(a)=\zeta (3)+\sum_{n=1}^\infty \frac {a^n}{n!}\int_{0}^{\infty} x \log ^n(x)\log\Bigg(\frac{1}{1-e^{-x}}\Bigg)\,dx $$

The remaining integrals

$$J_n=\int_{0}^{\infty} x \log ^n(x)\log\Bigg(\frac{1}{1-e^{-x}}\Bigg)\,dx$$ do not show closed form but they can be tabulated and, as long as $a$ is small, the summation seems to converge quite fast. Summing up to $p$, for $a=\frac 16$ the results would be $$\left( \begin{array}{cc} p & \text{approximation} \\ 1 & 1.253737667 \\ 2 & 1.268492231 \\ 3 & 1.268514471 \\ 4 & 1.268612760 \\ 5 & 1.268608473 \\ 6 & 1.268609145 \\ 7 & 1.268609093 \\ 8 & 1.268609098 \\ 9 & 1.268609097 \end{array} \right)$$

Tested for $a=\frac 13$, we have the same kind of convergence $$\left( \begin{array}{cc} p & \text{approximation} \\ 1 & 1.305418432 \\ 2 & 1.364436687 \\ 3 & 1.364614603 \\ 4 & 1.366187225 \\ 5 & 1.366050056 \\ 6 & 1.366093078 \\ 7 & 1.366086316 \\ 8 & 1.366087690 \\ 9 & 1.366087444 \\ 10 & 1.366087490 \\ 11 & 1.366087481 \\ 12 & 1.366087483 \end{array} \right)$$

Answer 2

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.

Answer 3

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 $1.2686090974543140685705195862758272411$.