Integral $\int_0^1 \frac{\log^4(x)}{(1-x)^4}dx$

Solution 1:

In your example, the denominator $(1-x)^4$, which means you need only take a third derivative: $$ \frac{d^3}{da^3}\frac{1}{a-x}=-6\frac{1}{(a-x)^4}. $$ This means the integral in the title is equal to $-I'''(1)/6$, in the case $n=4$. For general $n$, compute the third derivative of $I(a)$ using your expression and plug in $a=1$ to get $$ \int_0^1\frac{\log(x)^n}{(1-x)^4}\,dx=\frac{1}{3}\mathrm{Li}_n(1)+\frac{1}{2}\mathrm{Li}_{n-1}(1)+\frac{1}{6}\mathrm{Li}_{n-2}(1)=\frac{1}{3}\zeta(n)+\frac{1}{2}\zeta(n-1)+\frac{1}{6}\zeta(n-2). $$ More generally, one can check that the $k$-th derivative of $\mathrm{Li}_n(a)$ involves the (signed) Stirling numbers of the first kind: $$ \frac{d^k}{da^k}I(a)=(-1)^k\sum_{j=0}^k \genfrac{[}{]}{0pt}{}{k}{j}\mathrm{Li}_{n+1-j}(1/a), $$ and so $$ \int_0^1\frac{\log(x)^n}{(1-x)^m}\,dx=\frac{(-1)^{m-1}}{(m-1)!}\frac{d^{m-1}}{da^{m-1}}\bigg|_{a=1}\mathrm{Li}_{n+1}(1/a)=\frac{1}{(m-1)!}\sum_{j=0}^{m-1} \genfrac{[}{]}{0pt}{}{m-1}{j}\zeta(n+1-j). $$