Evaluate $\int_{0}^{\pi}x\ln[b^{2a}+2b^a\cos(ax)+1]dx$

Question

How to evaluate this integral?

Let $(a,b)\ge2$

$$I=\int_{0}^{\pi}x\ln[b^{2a}+2b^a\cos(ax)+1]\mathrm dx$$

$u=\ln[b^{2a}+2b^a\cos(ax)+1]$

$u^{'}=\frac{2ab^a\sin(ax)}{b^{2a}+2b^a\cos(ax)+1}$

$v=\int x=\frac{x^2}{2}$

$$I=\frac{x^2}{2}\cdot \ln[b^{2a}+2b^a\cos(ax)+1]-\int\frac{ab^ax^2\sin(ax)}{b^{2a}+2b^a\cos(ax)+1}\mathrm dx$$

$$I=\pi^2\cdot \ln(b^a+1)-\int_{0}^{\pi}\frac{ab^ax^2\sin(ax)}{b^{2a}+2b^a\cos(ax)+1}\mathrm dx$$

$w=b^a\cos(ax)$

$$J=\int_{0}^{\pi}\frac{ab^ax^2\sin(ax)}{b^{2a}+2b^a\cos(ax)+1}\mathrm dx$$

$$J=-\int \frac{x^2}{b^{2a}+2w+1}\mathrm dw$$

$$J=-\frac{1}{a}\int \arccos^2\left(\frac{w}{b^a}\right)\frac{1}{b^{2a}+2w+1}\mathrm dw$$

The case when $a$ is even is quite trivial ($\pi^2 a\ln b$) whereas the other case looks quite hard. — Zacky, Jul 08 '19 at 19:20
I am very interested if you can evaluate for the other case. — , Jul 08 '19 at 19:34
So, you just need a special case of $$j(s;p,q)=\int_0^\pi x\ln\left(p+q\cos(sx)\right)\ dx$$ — clathratus, Jul 09 '19 at 05:29

score 1 · Accepted Answer · answered Jul 08 '19 at 22:13

Let $w=b^a.$ Then there exists a closed form in terms of polylogarithms for the integral

$$ I=\int_0^\pi x \log\big(w^2+2\,w\,\cos{(a\,x)}+1 \big) dx = \pi^2\,\log{w} + \frac{i\,\pi}{a} \Big( \text{Li}_2\big(-\frac{e^{i\,a\,\pi}}{w} \big) - \text{Li}_2\big(-\frac{e^{-i\,a\,\pi}}{w} \big) \Big)+$$ $$+\frac{1}{a^2}\Big(2\, \text{Li}_3(-1/w) - \Big( \text{Li}_3\big(-\frac{e^{i\,a\,\pi}}{w} \big) + \text{Li}_3\big(-\frac{e^{-i\,a\,\pi}}{w} \big) \Big) \Big). $$ The proof consists of factoring the argument of the logarithm as $(w+\exp{(i\,\pi\,a)})(w+\exp{(-i\,\pi\,a)}),$ using the additive property of the log to get a sum of two integrals, and let Mathematica do the integral. In the comments it was said that it can be assumed that $a$ can be an integer greater than two. For this case we have

$$ I = \pi^2\,\log{w} + (1 - (-1)^a)/a^2 \Big( \text{Li}_3\big(-\frac{1}{w} \big) - \text{Li}_3\big(\frac{1}{w} \big) \Big) .$$

The special case for even $a,$ as mentioned by Zachy, is easily recovered.

Evaluate $\int_{0}^{\pi}x\ln[b^{2a}+2b^a\cos(ax)+1]dx$

1 Answers1