İntegralimiz :
$$\int_0^\pi\,\frac{\ln(\sin{x})}{\sqrt{\sin{x}}}\,dx$$
Buradaki eşitlikte $n$ yerine $2$ verelim.Eşitlik :
$$\int_0^\pi\,\ln(\sin{x})\,\sqrt[n]{\csc{x}}\:dx=\frac{\Gamma^2\Big(\frac{1}{2}-\frac{1}{2n}\Big)}{\sqrt[n]{2}\:\Gamma\Big(1-\frac{1}{n}\Big)}\Bigg[\ln(2)+\psi\bigg(\frac{1}{2}-\frac{1}{2n}\bigg)-\psi\bigg(1-\frac{1}{n}\bigg)\Bigg]$$
$$\int_0^\pi\,\frac{\ln(\sin{x})}{\sqrt{\sin{x}}}\,dx=\frac{\Gamma^2\Big(\frac{1}{4}\Big)}{\sqrt{2}\:\Gamma\Big(\frac{1}{2}\Big)}\Bigg[\ln(2)+\psi\bigg(\frac{1}{4}\bigg)-\psi\bigg(\frac{1}{2}\bigg)\Bigg]$$
$\Gamma\big(\frac{1}{2}\big)=\sqrt{\pi}$ , $\psi\big(\frac{1}{2}\big)=-2\ln(2)-\gamma$ , $\psi\big(\frac{1}{4}\big)=-\frac{\pi}{2}-3\ln(2)-\gamma$ olduğunu biliyoruz.
$$\large\color{#A00000}{\boxed{\int_0^\pi\,\frac{\ln(\sin{x})}{\sqrt{\sin{x}}}\,dx=-\sqrt{\frac{\pi}{8}}\:\Gamma^2\bigg(\frac{1}{4}\bigg)}}$$