Evaluate the integral from 0 to 1 of (ln x)^4 * ln(1 - x) / (x(1 - x)) dx.

Step-by-step solution

1. 1

   Split the kernel into two standard pieces. Partial fractions give $\frac{1}{x(1-x)}=\frac1x+\frac1{1-x}$, so $I=I_1+I_2$ with $I_1=\int_0^1\frac{\ln^{4}x\,\ln(1-x)}{x}\,dx,\qquad I_2=\int_0^1\frac{\ln^{4}x\,\ln(1-x)}{1-x}\,dx .$ The two halves need different series, which is exactly why the naive symmetry argument that sends $x\mapsto 1-x$ fails here.

2. 2

   Evaluate the first half with the logarithm series. Using $\ln(1-x)=-\sum_{n\ge1}\frac{x^{n}}{n}$ and the standard moment $\int_0^1 x^{n-1}\ln^{4}x\,dx=\frac{4!}{n^{5}}=\frac{24}{n^{5}}$, $I_1=-\sum_{n\ge1}\frac{1}{n}\cdot\frac{24}{n^{5}}=-24\zeta(6).$

3. 3

   Evaluate the second half with the harmonic generating function. Here $\frac{\ln(1-x)}{1-x}=-\sum_{n\ge1}H_n x^{n}$, and $\int_0^1 x^{n}\ln^{4}x\,dx=\frac{24}{(n+1)^{5}}$, so $I_2=-24\sum_{n\ge1}\frac{H_n}{(n+1)^{5}} .$ Shifting the index with $m=n+1$ and $H_{m-1}=H_m-\frac1m$ turns this into $I_2=-24\left(\sum_{m\ge1}\frac{H_m}{m^{5}}-\zeta(6)\right).$

4. 4

   Add the halves. The $\zeta(6)$ terms cancel: $I=I_1+I_2=-24\zeta(6)-24\sum_{m\ge1}\frac{H_m}{m^{5}}+24\zeta(6)=-24\sum_{m\ge1}\frac{H_m}{m^{5}} .$ Everything now hinges on one linear Euler sum.

5. 5

   Apply Euler's formula for the linear sum. With $\sum_{m\ge1}\frac{H_m}{m^{s}}=\left(1+\frac{s}{2}\right)\zeta(s+1)-\frac12\sum_{k=1}^{s-2}\zeta(k+1)\zeta(s-k)$ at $s=5$, $\sum_{m\ge1}\frac{H_m}{m^{5}}=\frac72\zeta(6)-\zeta(2)\zeta(4)-\frac12\zeta(3)^{2}.$ Hence $I=-84\zeta(6)+24\zeta(2)\zeta(4)+12\zeta(3)^{2}$.

6. 6

   Reduce to a closed form and check numerically. Using $\zeta(6)=\frac{\pi^{6}}{945}$ and $\zeta(2)\zeta(4)=\frac{\pi^{2}}{6}\cdot\frac{\pi^{4}}{90}=\frac{\pi^{6}}{540}$, the $\pi^{6}$ terms combine as $-\frac{4}{45}\pi^{6}+\frac{2}{45}\pi^{6}=-\frac{2}{45}\pi^{6}$, giving $I=12\zeta(3)^{2}-\frac{2\pi^{6}}{45}$. Numerically this is $-25.389119\ldots$, and double-exponential quadrature of the original integral returns $-25.389119\ldots$ as well.