MDoubt No 000001 - JEE

Question : For $m > 0, n > 0$, let $\displaystyle I_{m,n}=\int\limits_{0}^{1}x^m(\log x)^ndx$, then $I_{5,5}$ is given by

(a) $-\dfrac{5!}{6^5}$
(b) $-\dfrac{5!}{5^5}$
(c) $-\dfrac{5!}{6^6}$
(d) $\dfrac{5!}{6^6}$
Solution : Given $\displaystyle I_{m,n}=\int\limits_{0}^{1}x^m(\log x)^ndx$
put $\log x =-t$
We get $\displaystyle I_{m,n}=(-1)^n\int\limits_{0}^{\infty}t^ne^{-(m+1)t}dt$
now put $(m+1)t=u$
$\implies\displaystyle I_{m,n}=(-1)^n\int\limits_{0}^{\infty}\left(\dfrac{u}{m+1}\right)^ne^{-u}\dfrac{du}{m+1}$
$\implies\displaystyle I_{m,n}=\frac{(-1)^n}{(m+1)^{n+1}}\int\limits_{0}^{\infty}u^ne^{-u}du$
As we know Gamma Function is
$\displaystyle\left[\int\limits_{0}^{\infty}x^{(z-1)}e^{-x}dx=\Gamma(z)=(z-1)!\right]$
$\implies I_{m,n}=\dfrac{(-1)^n\Gamma(n+1)}{(m+1)^{n+1}}$
$\implies I_{5,5}=\dfrac{(-1)^5\Gamma(5+1)}{(5+1)^{5+1}}$
$\implies I_{5,5}=-\dfrac{5!}{6^6}$