\int_0^\frac{\pi}{2}\sin^n{x}\,dx=\int_0^\frac{\pi}{2}\cos^n{x}\,dx=\frac{2 \cdot 4 \cdot 6 \cdot \cdots \cdot (n-1)}{3 \cdot 5 \cdot 7 \cdot \cdots \cdot n}
( \scriptstyle{n}  عدد صحیح فرد و   \scriptstyle{n \ge 3} )
\int_0^\infty\frac{\sin^2{x}}{x^2}\,dx=\frac{\pi}{2}
\int_0^\infty x^{z-1}\,e^{-x}\,dx = \Gamma(z)
\int_{-\infty}^\infty e^{-(ax^2+bx+c)}\,dx=\sqrt{\frac{\pi}{a}}\exp\left[\frac{b^2-4ac}{4a}\right]
\int_{0}^{2 \pi} e^{x \cos \theta} d \theta = 2 \pi I_{0}(x)
\int_{0}^{2 \pi} e^{x \cos \theta + y \sin \theta} d \theta = 2 \pi I_{0} \left(\sqrt{x^2 + y^2}\right)
\int_{-\infty}^{\infty}{(1 + x^2/\nu)^{-(\nu + 1)/2}dx} = \frac { \sqrt{\nu \pi} \ \Gamma(\nu/2)} {\Gamma((\nu + 1)/2))}\,
(\nu > 0\,)