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

Question

I am trying to evaluate $$\int_0^1 \frac{\log^4(x)}{(1-x)^4}\,dx$$

This type of integral already has two answers here: Closed form for $\int_0^1 \frac {\log^n(x)}{(1-x)^m} dx$. However I desire to evaluate it with a different method. Starting with: $$I(a)=\int_0^1 \frac{\ln^n(x)}{a-x}\,dx$$ we have that $$\frac{1}{a-x}=\sum_{j=0}^{\infty} \frac{x^j}{a^{j+1}}$$ therefore $$I(a)=\sum_{j=0}^{\infty} \frac{1}{a^{j+1}} \int_0^1 x^j\ln^n(x) \,dx$$ using the substitution$$x=e^{-y}$$ $$I(a)=\sum_{j=0}^{\infty} \frac{(-1)^n}{a^{j+1}}\int_0^{\infty}y^ne^{-y(j+1)}\,dy $$now with $(j+1)y=t\,$ and using gamma function:$$I(a)=\sum_{j=0}^{\infty} \frac{(-1)^n n!}{a^{j+1} (1+j)^{n+1}}= \sum_{j=1}^{\infty} \frac{(-1)^n n!}{a^{j} j^{n+1}}=(-1)^n n!\operatorname{Li}_{n+1}\left(\frac{1}{a}\right)$$ Where $\operatorname{Li}_n(z)=\sum_{k=1}^{\infty}\frac{z^k}{k^n}$ is the Polylogarithm. Now we have to derivate $I(a)$ four times with respect to $a$ using $\frac{d}{dz} \left(\operatorname{Li}_n(z))= \frac{1}{z}\operatorname{Li}_{n-1}(z)\right)$ then plug in $a=1$ and $n=4$ in order to get the desired integral, but I am struggling here. $$4!\int_0^1 \frac{\log^4(x)}{(1-x)^4}\,dx=(-1)^nn!\frac{d^4}{da^4}\left(\operatorname{Li}_{n+1}\left(\frac{1}{a}\right)\right)$$ For the first derivate: $$\frac{d}{da} (\operatorname{Li}_{n+1}(1/a)) = \frac{1}{\frac{1}{a}}\operatorname{Li}_{n}(1/a)\frac{-1}{a^2} = - \frac{1}{a}\operatorname{Li}_{n}(1/a)$$ Second time: $$\frac{d}{da} \left(-\frac{1}{a}Li_n(1/a)\right) = \frac{1}{a^2}\left(\operatorname{Li}_{n}(1/a) + \operatorname{Li}_{n-1}(1/a)\right)$$ Now when I get to the fourth one: $$\frac{1}{a^4}\left(6\operatorname{Li}_{n}\left(\frac{1}{a}\right)+12\operatorname{Li}_{n-1}\left(\frac{1}{a}\right)+6\operatorname{Li}_{n-2}\left(\frac{1}{a}\right)+\operatorname{Li}_{n-3}\left(\frac{1}{a}\right)\right)$$ But obviously this is wrong because if one would plug in $n=4$ and $a=1$ the last term will be $\operatorname{Li}_1(1)$ which is not nice.

Could you help me finish this or perhaps find another closed form for $\displaystyle{\int_0^1 \frac {\log^n(x)}{(1-x)^m}\,dx}$ using this method?

Answer 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 $$ \int_0^1\frac{\log(x)^n}{(1-x)^m}\,dx=\frac{(-1)^{m-1}}{(m-1)!}\frac{d^{m-1}}{da^{m-1}}\bigg|_{a=1}\mathrm{Li}_{n+1}(1/a)=\frac{1}{(m-1)!}\sum_{j=0}^{m-1} \genfrac{[}{]}{0pt}{}{m-1}{j}\zeta(n+1-j). $$