قرار دهید:
$$u:=Arcsinx \Rightarrow du=\frac{dx}{ \sqrt{1-x^2}}= \frac{dx}{ \sqrt{1-sin^2u}}= \frac{dx}{cosu},u(0)=0,u(1)= \frac{\pi}{2}$$
$$ \Rightarrow I= \int_{-1}^1 \frac{Arcsinx}{x}dx=2\int_0^1 \frac{Arcsinx}{x}dx=2\int_.^ \frac{\pi}{2} \frac{u}{sinu}.cosu.du$$
$$=2 \int_0^\frac{\pi}{2}ud(Ln(sinu))=2uLn(sinu)|_0^ \frac{\pi}{2}-2 \int_0^ \frac{\pi}{2}Ln(sinu)du$$
$$=0-2 \int_0^ \frac{\pi}{2}Ln(sinu)du=-2 \int_0^ \frac{\pi}{2}Ln(sinu)du$$
حالا انتگرال اخیر را جدا محاسبه میکنیم(در این مرحله غیر محسوس از تغییر متغیر $x:=2u$ و $x:=u-\frac{\pi}{2}$ استفاده میکنیم):
$$J:=\int_0^ \frac{\pi}{2}Ln(sinu)du=2 \int_0^ \frac{\pi}{2}Ln(sin(0+ \frac{\pi}{2}-u)du=\int_0^ \frac{\pi}{2}Ln(cosu)du$$
$$2J=J+J=2 \int_0^ \frac{\pi}{2}Ln(sinu)du+\int_0^ \frac{\pi}{2}Ln(cosu)du=2\int_0^ \frac{\pi}{2}Ln(sinucos)du$$
$$=\int_0^ \frac{\pi}{2}Ln( \frac{1}{2}sin2u)du=\int_0^ \frac{\pi}{2}Ln(sinu)du-Ln2\int_0^ \frac{\pi}{2}du$$
$$= \int_0^ \frac{\pi}{2}Ln(sin2u)du- \frac{\pi}{2}Ln2=2\int_0^\pi Lnsinxdx-\frac{\pi}{2}Ln2$$
$$=2 \int_0^ \frac{\pi}{2}Lnsinxdx+2 \int_\frac{\pi}{2}^\pi Lnsinudu-\frac{\pi}{2}Ln2$$
$$=2 \int_0^ \frac{\pi}{2}Lnsinxdx-2 \int_0^ \frac{\pi}{2}Lnsinxdx-\frac{\pi}{2}Ln=2J-2J-\frac{\pi}{2}Ln2=-\frac{\pi}{2}Ln2$$
$$ \Rightarrow I=-2J=-2.(-\frac{\pi}{2}Ln2)=\pi Ln2$$
$\Box$
