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How do I solve this: $$\int\limits_{0}^{1} \frac{\log(x^{2}-2\cdot x \cdot \cos{a}+1)}{x} \ dx$$

Integration by parts is the only thing which I could think of, clearly that seems cumbersome. Substitution also doesn't work.

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The integral is improper and fails to converge from a quick check in Mathematica. – Adam Saltz Aug 15 '11 at 15:37
@IAmBrianDawkins It does converge. Near the origin integrand behaves as $-2\cos a - x \cos(2 a) + O(x^2)$. – Sasha Aug 15 '11 at 16:10
Yikes! How embarrassing. Thank you for the correction, Sasha. – Adam Saltz Aug 15 '11 at 16:15
up vote 12 down vote accepted

Assume $0<a<2 \pi$. In this case $x^2-2 \, x \, \cos a+1 > 0$ for $0\le x \le 1$, so integral converges.

Denote $I(a)$ the value of the integral. Consider $\partial_a I(a)$. You can exchange the order of integration and differentiation, getting

$$ \partial_a I(a) = \int_0^1 \mathrm{d} x \frac{2 \sin a}{x^2 + 1 - 2 \, x \, \cos a} = \pi - a $$

The integral above is computed after a substitution $x = \cos a + \sin a \cdot \tan \frac{t}{2}$ and carefully considering integration region.

From here, knowing that for $I(\frac{\pi}{2}) = \frac{\pi^2}{24}$, we get

$$I(a) = \frac{\pi^2}{24} + \int_{\frac{\pi}{2}}^a \mathrm{d} \alpha ( \pi - \alpha) = \pi a - \frac{a^2}{2} - \frac{\pi^2}{3}.$$

The formula also extends to $a=0$ because the $I(0)$ converges.

The answer is extended to real $a$ by periodicity.

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This is very nice and useful. +1. I have used this approach when I was teaching math at the polytechnic school. Too bad that in our first year course my teachers forgot to mention this. :) – Beni Bogosel Aug 15 '11 at 16:42

Rewriting the integral as $$ \int\limits_{0}^{1} \frac{\log|x-e^{i a}|^2}{x} \ dx= \int\limits_{0}^{1} \frac{\log|1-xe^{-i a}|^2}{x} \ dx= $$ $$ \int\limits_{0}^{1} \frac{\log(1-xe^{i a })}{x} \ dx+ \int\limits_{0}^{1} \frac{\log(1-xe^{-i a })}{x} \ dx, $$ expanding and integrating termwise leads to an answer in the form of series.

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You mean $e^{i a x}$ to be $e^{i a}$, right ? – Sasha Aug 15 '11 at 16:12
@Sasha Yes, thanks. Fixed it. – Andrew Aug 15 '11 at 16:17
Notice the so obtained series can be summed in terms of dilogarithms as $-\mathrm{Li}_2(\mathrm{e}^{i a}) - \mathrm{Li}_2(\mathrm{e}^{-i a})$. For $0\le a<2 \pi$ by Jonquière's inversion formula this would give $ I(a) = - \frac{(2 \pi i)^2}{2!} B_2( \frac{a}{2\pi})$, where $B_2$ is a second Bernoulli polynomial, this coincides with the answer I gave. – Sasha Aug 15 '11 at 16:29
Also one can note that this series is the Fourier series of the answer. – Andrew Aug 15 '11 at 16:35

Please see $\textbf{Problem 4.30}$ in the book: ($\text{solutions are also given at the end}$)

  • $\text{The Math Problems Notebook,}$ by Valentin Boju and Louis Funar.
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