I want to find a closed form of the following integral : $I_n =\int_0^1\frac{\ln(x)^n}{x-1}dx$ , for $n\geq 1$.
My attempt was to evaluate , using series : $I_1,I_2,I_3,I_4$ and noticed a pattern : $$I_1=\zeta(2)\\I_2=-2\zeta(3)\\I_3=2.3\zeta(4)\\I_4=-2.3.4\zeta(5)\\I_5=2.3.4.5\zeta(6)$$ Can we deduce that $I_n=(-1)^{n+1}n!\zeta(n+1)$ ? How to prove it ?
On
By enforcing the substitution $x=e^{-t}$ and using integration by parts, or just by Feynman's trick, we have that for any $k,n\in\mathbb{N}$ the identity $$ \int_{0}^{1}x^n \log(x)^k\,dx = \frac{d^k}{d\alpha^k}\left.\int_{0}^{1}x^{n+\alpha}\,dx\right|_{\alpha=0^+}=\frac{(-1)^kk!}{(n+1)^{k+1}} $$ holds. By expanding $\frac{1}{1-x}$ as $1+x+x^2+x^3+\ldots$ it follows that $$ \int_{0}^{1}\frac{\log(x)^k}{x-1}\,dx = (-1)^{k+1}k!\zeta(k+1) $$ as claimed.
On
I try to tackle the integral with differentiation by defining $$ I(a):=\int_{0}^{1} \frac{x^{a}}{x-1} d x $$ whose $n^{th}$ derivative $$ I^{(n)}(a)=\int_{0}^{1} \frac{x^{a} \ln ^{n} x}{x-1} d x $$
at $a=0$ gives the exact value of the integral $$ I^{(n)}(0)=\int_{0}^{1} \frac{\ln ^{n} x}{x-1} d x. $$ Back to $I(a)$, $$ I(a)=\int_{0}^{1} \frac{x^{a}}{x-1} d x=-\int_{0}^{b} x^{a} \sum_{k=0}^{\infty} x^{k} d x=-\sum_{k=0}^{\infty}\left[\frac{x^{k+a+1}}{k+a+1}\right]_{0}^{1} =-\sum_{k=0}^{\infty} \frac{1}{k+a+1} $$
By differentiating $I(a)$ by $n-1$ times yields $$ \begin{aligned} I^{(n)}(0) &=\frac{d^{n}}{d x^{n}}\left(-\left.\sum_{k=0}^{\infty} \frac{1}{k+a+1} \right)\right|_{a=0}\\ &=\sum_{k=0}^{\infty}-\left.\frac{(-1)(-2) \ldots(-n)}{(k+a+1)^{n+1}}\right|_{a=0} \\ &=(-1)^{n+1} n !\zeta(n+1) \end{aligned} $$ We can now conclude that$$\boxed{\int_{0}^{1} \frac{\ln ^{n} x}{x-1} d x= (-1)^{n+1} n !\zeta(n+1)}$$
The zeta function can be written as $$\zeta(n+1)=\frac1{\Gamma(n+1)}\int_0^\infty\frac{x^{n}}{e^x-1}\,dx$$ and your integral can be written as $$I(n)=n!\cdot\frac1{\Gamma(n+1)}\int_0^\infty\frac{\ln^n x}{e^{\ln x}-1}\,dx$$ since $\Gamma(n+1)=n!$