How to compute the following integral? ∫log⁡(sin⁡x)dx

Question

How to compute the following integral?  ∫log⁡(sin⁡x)dx

zurilomk4 · Accepted Answer

You can calculate  ∫0πlog⁡(sin⁡x)dx=−πlog⁡2  and integrating up to π2 would give half of this.  Note that integrating log⁡(sin⁡x) from 0 to π2 is the same as integrating log⁡(cos⁡x) so that  ∫0π2log⁡(sin⁡x)dx=12∫0π2log⁡(sin⁡xcos⁡x)dx  =12∫0π2log⁡(sin⁡2x)dx−π4log⁡2  After a change of variables, this can be rearranged to get the result.

Thomas Nickerson · Accepted Answer

Series expansion can be used for this integral too.
We use the following identity;

\log (\sin x) = - \log 2 - \sum_{k \geq 1} \frac{\sin (2 k b) - \sin (2 k a)}{2 k^{2}}

(a, b < π)

for example,

\int_{0}^{\frac{π}{4}} \log (\sin x) d x = - \frac{π}{4} \log 2 - \sum_{k \geq 1} \frac{\sin (π \frac{k}{2})}{2 k^{2}}

- \frac{π}{4} \log (\sin x) d x = - \frac{π}{2} \log 2

\int_{0}^{\frac{π}{2}} \log (\sin x) d x = - \frac{π}{2} \log 2

\int_{0}^{π} \log (\sin x) d x = - π \log 2

Vasquez · Accepted Answer

I=∫0π/2log⁡(sin⁡x)dx=∫0π/2log⁡(cos⁡x)dx⇒2I=∫0π/2log⁡(sin⁡xcos⁡x)dx=∫0π/2log⁡(12⋅2sin⁡xcos⁡x)dx=∫0π/2log⁡(1/2)dx+∫0π/2log⁡(sin⁡2x)dxPut t=2x⇒dt=2dx2I=π2log⁡(12)+12∫0πlog⁡(sin⁡t)dt=−π2log⁡2+12=−π2log⁡2+12∫0π/2log⁡(sin⁡x)dx+12∫0π/2log⁡(cos⁡x)dx=−π2log⁡2+II=−π2log⁡2

How to compute the following integral? \int\log(\sin x)dx

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