Expert Answer to "Find: ∫0π4xln⁡(sin⁡x)dx"

Madilyn Fitzgerald

Answered question

2022-01-26

Find:

\int_{0}^{\frac{π}{4}} x \ln (\sin x) d x

Dominique Green

Beginner2022-01-27Added 11 answers

\ln (\sin x) = - \ln 2 - \sum_{n = 1}^{\infty} \frac{\cos (2 n x)}{n}

We can use the Fourier series and integrate by parts to obtain

I \equiv \int_{0}^{\frac{π}{4}} x [- \ln (\sin (x))] d x = \frac{π^{2}}{32} \ln (2) + {\frac{14}{\sum}}_{n = 1}^{\infty} \frac{1}{n^{2}} [\frac{π}{2} \sin (\frac{π}{2} n) - \frac{1}{n} (1 - \cos (\frac{π}{2} n))]

\sin (\frac{π}{2} n)

is non-zero and alternating for odd n, while

\cos (\frac{π}{2} n)

is non-zero and alternating for even n. Therefore,

I = \frac{π^{2}}{32} \ln (2) + \frac{π}{8} \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{{(2 k + 1)}^{2}} - \frac{1}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{3}} - \frac{1}{32} \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k - 1}}{k^{3}}

The first series is Catalan's constant G, the second one is

ζ (3)

and the third one is

η (3) = \frac{34}{ζ} (3)

(with the Riemann zeta function

ζ

and the Dirichlet eta function

η

), so we obtain

I = \frac{π^{2}}{32} \ln (2) + \frac{π}{8} G - \frac{35}{128} ζ (3)

and your integral is −I.

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