Log trigonométrico integral con radical

Question

Muestra esa:

$$\int_{0}^{\Large\frac{\pi}{2}}\frac{\log(\sin\theta)}{\sqrt{1+\sin^{2}\theta}}\ d\theta=-\frac{\Gamma^{2}\left(\dfrac14\right)\sqrt{\dfrac\pi2}}{16},$ $ o alguna otra forma equivalente.

Answer 1

Considere la integral \begin{align} I = \int_{0}^{\pi/2} \frac{\ln(\sin\theta)}{\sqrt{1+\sin^{2}\theta}} \ d\theta. \end {align} Al realizar la sustitución$t = \sin\theta$, la integral se reduce a \begin{align} I = \int_{0}^{1} \ln(t) \ (1-t^{4})^{-1/2} \ dt. \end {align} Realizar la sustitución$u = t^{4}$ de los rendimientos \begin{align} I &= \frac{1}{16} \int_{0}^{1} \ln(u) \ u^{-3/4} \ (1-u)^{-1/2} \ du \\ &= \frac{1}{16} \partial_{\mu} \left. \int_{0}^{1} u^{\mu-1} (1-u)^{-1/2} \ du \right|_{\mu=1/4} \\ &= \frac{1}{16} \ B(1/4, 1/2) \left[ \psi(1/4) - \psi(3/4) \right]. \end {align} El uso de \begin{align} \psi(1/4) &= - \gamma - \frac{\pi}{2} - 3 \ln(2) \\ \psi(3/4) &= - \gamma + \frac{\pi}{2} - 3 \ln(2) \\ \Gamma(3/4) &= \frac{\sqrt{2} \pi}{\Gamma(1/4)} \end {align} lleva a \begin{align} I = - \frac{\Gamma^{2}\left( \frac{1}{4} \right)}{16} \sqrt{\frac{\pi}{2}}. \end {align} Por lo tanto \begin{align} \int_{0}^{\pi/2} \frac{\ln(\sin\theta)}{\sqrt{1+\sin^{2}\theta}} \ d\theta = - \frac{\Gamma^{2}\left( \frac{1}{4} \right)}{16} \sqrt{\frac{\pi}{2}}. \end {align}