0-pi-2-ln-sinx-ln-cosx-dx-

Question Number 160358 by Ar Brandon last updated on 28/Nov/21

\int_{0}^{\frac{π}{2}} \ln (\sin x) \ln (\cos x) d x

Answered by amin96 last updated on 28/Nov/21

\boldsymbol \sin (\boldsymbol x) = \boldsymbol t \frac{\boldsymbol dt}{\boldsymbol dx} = \boldsymbol \cos (\boldsymbol x) = \sqrt{1 - \boldsymbol t^{2}}

\boldsymbol Ω = \int_{0}^{1} \frac{\boldsymbol \ln (\boldsymbol t) \boldsymbol \ln (\sqrt{1 - \boldsymbol t^{2}})}{\sqrt{1 - \boldsymbol t^{2}}} \boldsymbol dt = \frac{1}{2} \int_{0}^{1} \frac{\boldsymbol \ln (\boldsymbol t) \boldsymbol \ln (1 - \boldsymbol t^{2})}{\sqrt{1 - \boldsymbol t^{2}}} \boldsymbol dt

\boldsymbol t^{2} = \boldsymbol x \frac{\boldsymbol dx}{\boldsymbol dt} = 2 \boldsymbol t = 2 \sqrt{\boldsymbol x}

\boldsymbol Ω = \frac{1}{4} \int_{0}^{1} \frac{\boldsymbol \ln (\sqrt{\boldsymbol x}) \boldsymbol \ln (1 - \boldsymbol x)}{\sqrt{\boldsymbol x} \sqrt{1 - \boldsymbol x}} \boldsymbol dx = \frac{1}{8} \int_{0}^{1} \frac{(\sqrt{1 - \boldsymbol x}) \boldsymbol \ln (\boldsymbol x) \boldsymbol \ln (1 - \boldsymbol x)}{\sqrt{\boldsymbol x} \times (1 - \boldsymbol x)} \boldsymbol dx =

= - \frac{1}{8} \underset{\boldsymbol n = 1}{\sum^{\infty}} \boldsymbol H_{\boldsymbol n} \int_{0}^{1} \boldsymbol x^{\boldsymbol n - \frac{1}{2}} {(1 - \boldsymbol x)}^{\frac{1}{2}} \boldsymbol \ln (\boldsymbol x) \boldsymbol dx =

= - \frac{1}{8} Σ \boldsymbol H_{\boldsymbol n} (Σ \frac{{(- 1)}^{\boldsymbol n + 1}}{\boldsymbol n}) \int_{0}^{1} \boldsymbol x^{\boldsymbol n - \frac{1}{2}} {(1 - \boldsymbol x)}^{\frac{1}{2}} {(\boldsymbol x - 1)}^{\boldsymbol n} \boldsymbol dx =

= \frac{1}{8} \underset{\boldsymbol n = 1}{\sum^{\infty}} \boldsymbol H_{\boldsymbol n} {\underset{\boldsymbol n = 1}{\sum^{\infty}} \frac{1}{\boldsymbol n} \int_{0}^{1} \boldsymbol x^{\boldsymbol n - \frac{1}{2}} {(1 - \boldsymbol x)}^{\boldsymbol n + \frac{1}{2}} \boldsymbol dx} =

= \frac{1}{8} \underset{\boldsymbol n = 1}{\sum^{\infty}} \boldsymbol H_{\boldsymbol n} {\underset{\boldsymbol n = 1}{\sum^{\infty}} \frac{1}{\boldsymbol n} \int_{0}^{1} \boldsymbol x^{(\boldsymbol n + \frac{1}{2}) - 1} {(1 - \boldsymbol x)}^{(\boldsymbol n + \frac{3}{2}) - 1} \boldsymbol dx} =

= \frac{1}{8} \underset{\boldsymbol n = 1}{\sum^{\infty}} \boldsymbol H_{\boldsymbol n} {\underset{\boldsymbol n = 1}{\sum^{\infty}} \frac{1}{\boldsymbol n} \boldsymbol β (\boldsymbol n + \frac{1}{2}; \boldsymbol n + \frac{3}{2})}

Commented by Ar Brandon last updated on 28/Nov/21

I wish it is further simplified .

Commented by amin96 last updated on 28/Nov/21

do you know the answer?

Commented by Ar Brandon last updated on 29/Nov/21

Yeah

Commented by Ar Brandon last updated on 29/Nov/21

\frac{π}{2} \ln^{2} 2 - \frac{π^{3}}{48}

Commented by amin96 last updated on 29/Nov/21

\int_{0}^{\frac{π}{2}} \boldsymbol xln (\boldsymbol sinx) \boldsymbol \ln (\boldsymbol cosx) \boldsymbol dx

The answer to this integral

Commented by Ar Brandon last updated on 29/Nov/21

\int_{0}^{\frac{π}{2}} x \ln (\sin x) \ln (\cos x) d x = \frac{π^{2}}{8} \ln^{2} 2 - \frac{π^{4}}{192}

\int_{0}^{\frac{π}{2}} x \ln (\sin x) \ln (\cos x) d x = \frac{π}{4} \int_{0}^{\frac{π}{2}} \ln (\sin x) \ln (\cos x) d x