Evaluation of $\int_{0}^1 \frac{1}{x} \log^3{(1-x)}dx =-\frac{\pi^4}{15}$ and $\int_{-\pi}^{\pi} \log(2\cos{\frac{x}{2}}) dx =0$

Question

In the following encyclopedia, http://m.encyclopedia-of-equation.webnode.jp/including-integral/

I found the relations below

\begin{eqnarray} \int_{0}^1 \frac{1}{x} \log^3{(1-x)}dx &=&-\frac{\pi^4}{15} \tag{1} \\ \int_{-\pi}^{\pi} \log(2\cos{\frac{x}{2}}) dx &=& 0 \tag{2} \end{eqnarray} I tried to prove these equation, but I didn't success to prove. How do you go about evaluating those integrals to obtain the repsective values?

Answer 1

For the first integral: substitute $1-x=q$, the integral becomes

$$ I_1=\int_0^1 \frac{\log^3(q)}{1-q} $$

The next substitution is obvious $e^{r}=q$, which yields after some some algebra

$$ I_1=\int_{0}^{\infty}\frac{r^3}{1-e^{r}} $$

This integral is now solvable by various means, for example by expanding the denominator into a geometric series, integrate termwise and using the special value $\zeta(4)=\frac{\pi^4}{90}$ to obtain

$$ I_1=-\frac{\pi^4}{15} $$

Answer 2

You may write

$$\begin{align} \int_{0}^1 \frac{1}{x} \log^3{(1-x)}dx&=\int_{0}^1 \frac{1}{1-x} \log^3xdx\\ &=\sum_{n=0}^{\infty}\int_0^1x^n\log^3x \:dx\\ &=-6\sum_{n=0}^{\infty}\frac{1}{(n+1)^4}\\ &=-\frac{\pi ^4}{15} \end{align}$$

where we have used $$ \begin{align} &\int_0^1x^\alpha\log^3x \:dx=-\frac{3!}{(\alpha+1)^4} \quad (\text{a direct integration by parts)}\\ &\zeta(4)=\frac{\pi ^4}{90} . \end{align} $$ (Riemann zeta function at even integers)

Your second integral rewrites, by parity of the integrand and a change of variable,

$$ \int_{-\pi}^{\pi} \log(2\cos{\frac{x}{2}}) dx=2\int_0^{\pi} \log(2\cos{\frac{x}{2}}) dx=4\int_0^{\pi/2} \log(2\cos x) dx=0 $$

a proof of the latter equality may be found here.