Explained: "How do I solve this:∫01log⁡(x2−2xcos⁡a+1)xdx"

Haylee Bowen

Answered question

2022-04-15

How do I solve this:

\int_{0}^{1} \frac{\log (x^{2} - 2 x \cos a + 1)}{x} d x

prangijahnot

Beginner2022-04-16Added 17 answers

Assume

0 < a < 2 π

. In this case

x^{2} - 2 x \cos a + 1 > 0

for

0 \leq x \leq 1

, so integral converges.
Denote

I (a)

the value of the integral. Consider

d_{a} I (a)

. You can exchange the order of integration and differentiation, getting

\partial_{a} I (a) = \int_{0}^{1} d x \frac{2 \sin a}{x^{2} + 1 - 2, x, \cos a} = π - 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{π}{2} = \frac{π^{2}}{24}

, we get

I (a) = \frac{π^{2}}{24} + \int_{\frac{π}{2}}^{a} d α (π - α) = π a - \frac{a^{2}}{2} - \frac{π^{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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