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.
How to approximate $\int_{0}^{\infty}x^{7/6}\ln(\frac{1}{1-e^{-x}})dx$
approximationdefinite integrals
Best Answer
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 $$\left( \begin{array}{cc} p & \text{approximation} \\ 1 & 1.305418432 \\ 2 & 1.364436687 \\ 3 & 1.364614603 \\ 4 & 1.366187225 \\ 5 & 1.366050056 \\ 6 & 1.366093078 \\ 7 & 1.366086316 \\ 8 & 1.366087690 \\ 9 & 1.366087444 \\ 10 & 1.366087490 \\ 11 & 1.366087481 \\ 12 & 1.366087483 \end{array} \right)$$