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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?

Olivier Oloa
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user263059
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  • Welcome to math.SE! This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – user37238 Aug 19 '15 at 11:24
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    The first identity can be rephrased in terms of third-order derivatives of the beta function. The second one, after a bit of rewriting, will require the substitution $u=\dfrac\pi2-t$. – Lucian Aug 19 '15 at 11:36

2 Answers2

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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.

Community
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Olivier Oloa
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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

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

tired
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