Random Problems

$\displaystyle \text{Lemma : } \int_{0}^{1} x^a(lnx)^bdx=\frac{(-1)^b b!}{(a+1)^{b+1}}$

Proof : Let $\displaystyle I(a,b)=\int_{0}^{1} x^a(lnx)^bdx$ , Put $lnx=y$ and the integral turns,

$\displaystyle I(a,b)=\int_{\infty}^{0} e^{y(a+1)}y^bdy$ , Put $y(a+1)=-u$ and the integral turns,

$\displaystyle I(a,b)=\frac{(-1)^b}{(a+1)^{b+1}}\int_{0}^{\infty}e^{-u}u^{(b+1)-b}du=\frac{(-1)^b\Gamma(b+1)}{(a+1)^{b+1}}=\frac{(-1)^bb!}{(a+1)^{b+1}}$

We have $b=3$ , so $\displaystyle \sum_{n=0}^{\infty} I_n=\sum_{n=0}^{\infty} \frac{-6}{(n+1)^4}$

Therefore, $\displaystyle \sum_{n=0}^{\infty} I_n=-6\zeta(4)=-6\frac{\pi^4}{90}=-\frac{\pi^4}{15}\approx -6.493$