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

Question

Saludos estoy tratando de evaluar $\int_0^1 \frac{\log^4(x)}{(1-x)^4}\,dx$, ahora este tipo de integral ya tiene dos respuestas aquí: forma Cerrada para $\int_0^1 \frac {\log^n(x)}{(1-x)^m} dx$. Sin embargo, yo deseo para evaluar con un método diferente, o a encontrar otra forma cerrada a saber: a Partir de $$I(a)=\int_0^1 \frac{\ln^n(x)}{a-x}\,dx$$ 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:$$I(a)=\sum_{j=0}^{\infty} \frac{(-1)^n n!}{a^{j+1} (1+j)^{n+1}}= \sum_{j=1}^{\infty} \frac{(-1)^n n!}{a^{j} j^{n+1}}=(-1)^n n!Li_{n+1}(\frac{1}{a})$$ Where $Li_n(z)=\sum_{k=1}^{\infty}\frac{z^k}{k^n}$ is the Polylogarithm. Now we have to derivate $I(a)$ four times with respect to $a$ using $\frac{d}{dz} \left(Li_n(z))= \frac{1}{z}Li_{n-1}(z)\right)$ then plug in $a=1$ and $n=4$ in order to get the desired integral, but I am struggling. $$4!\int_0^1 \frac{\log^4(x)}{(1-x)^4}\,dx=(-1)^nn!\frac{d^4}{da^4}\left(Li_{n+1}(\frac{1}{a})\right)$$ For the first derivate: $$\frac{d}{da} (Li_{n+1}(1/a)) = \frac{1}{\frac{1}{a}}Li_{n}(1/a)\frac{-1}{a^2} = - \frac{1}{a}Li_{n}(1/a)$$ Second time: $$\frac{d}{da} \left(-\frac{1}{a}Li_n(1/a)\right) = \frac{1}{a^2}\left(Li_{n}(1/a) + Li_{n-1}(1/a)\right)$$ Now when I get to the fourth one: $$\frac{1}{a^4}\left(6Li_{n}(\frac{1}{a})+12Li_{n-1}(\frac{1}{a})+6Li_{n-2}(\frac{1}{a})+Li_{n-3}(\frac{1}{a})\right)$$ But obviously this is wrong because if one would plug in $n=4$ and $a=1$ the last term will be $Li_1(1)$ which is not nice. Could you help me finish this or perhaps find another closed form for $\int_0^1 \frac {\log^n(x)}{(1-x)^m}\,dx$ usando este método?

Answer 1

En tu ejemplo, el denominador $(1-x)^4$, lo que significa que sólo se necesita tomar un tercero derivados de: $$ \frac{d^3}{da^3}\frac{1} {- x}=-6\frac{1}{(a-x)^4}. $$ Esto significa que la integral en el título es igual a $-I'''(1)/6$, en el caso de $n=4$. Para general $n$, calcular la tercera derivada de $I(a)$ el uso de la expresión y de enchufe $a=1$ para obtener $$ \int_0^1\frac{\log(x)^n}{(1-x)^4}\,dx=\frac{1}{3}\mathrm{Li}_n(1)+\frac{1}{2}\mathrm{Li}_{n-1}(1)+\frac{1}{6}\mathrm{Li}_{n-2}(1)=\frac{1}{3}\zeta(n)+\frac{1}{2}\zeta(n-1)+\frac{1}{6}\zeta(n-2). $$ Más generalmente, se puede comprobar que el $k$-ésima derivada de $\mathrm{Li}_n(a)$ implica la (firmado) los números de Stirling de primera especie: $$ \frac{d^k}{da^k}I(a)=(-1)^k\sum_{j=0}^k \genfrac{[}{]}{0pt}{}{k}{j}\mathrm{Li}_{n+1-j}(1/a), $$ y así $$ \int_0^1\frac{\log(x)^n}{(1-x)^m}\,dx=\frac{(-1)^{m-1}}{(m-1)!}\frac{d^{m-1}}{da^{m-1}}\bigg|_{a=1}\mathrm{Li}_{n+1}(1/a)=\frac{1}{(m-1)!}\sum_{j=0}^{m-1} \genfrac{[}{]}{0pt}{}{m-1}{j}\zeta(n+1-j). $$