1-dimensional Gaussian integral
\[ \begin{gathered} \int_{-\infty}^{\infty} e^{-x^2} d x=\sqrt{\pi} \\ \int_{-\infty}^{\infty} e^{-\frac{1}{2} a x^2+b x} d x=\sqrt{\frac{2 \pi}{a}} \exp \left(\frac{b^2}{2 a}\right) \\ \int_{-\infty}^{\infty} e^{-\frac{1}{2} a x^2+i b x} d x=\sqrt{\frac{2 \pi}{a}} \exp \left(-\frac{b^2}{2 a}\right) \\ I=\int_{-\infty}^{\infty} e^{-\frac{1}{2} a x^2+b x} d x=\sqrt{\frac{2 \pi}{a}} \exp \left(\frac{b^2}{2 a}\right) \\ \frac{\partial I}{\partial b}=\int_{-\infty}^{\infty} x e^{-\frac{1}{2} a x^2+b x} d x=\frac{b}{a} \sqrt{\frac{2 \pi}{a}} \exp \left(\frac{b^2}{2 a}\right) \\ \frac{\partial^2 I}{\partial b^2}=\int_{-\infty}^{\infty} x^2 e^{-\frac{1}{2} a x^2+b x} d x=\frac{1}{a}\left(1+\frac{b^2}{a}\right) \sqrt{\frac{2 \pi}{a}} \exp \left(\frac{b^2}{2 a}\right) \\ \int_{-\infty}^{\infty} x^2 e^{-\frac{1}{2} a x^2} d x=\frac{1}{a} \sqrt{\frac{2 \pi}{a}} \\ \int_{-\infty}^{\infty} x^4 e^{-\frac{1}{2} a x^2} d x=\frac{3}{a^2} \sqrt{\frac{2 \pi}{a}} \\ \int_{-\infty}^{\infty} x^{2 n} e^{-\frac{1}{2} a x^2} d x=\frac{(2 n-1)!!}{a^n} \sqrt{\frac{2 \pi}{a}} \\ \int_0^{\infty} x^{2 n+1} e^{-a x^2} d x=\frac{n!}{2 a^{n+1}} \\ \int_0^{\infty} x^n e^{-a x^2} d x=\frac{\Gamma\left(\frac{n+1}{2}\right)}{2 a^{\frac{n+1}{2}}} \end{gathered} \]
n-dimensional \({ }^{+}\)Gaussian integral (such as multivariate normal distribution)
\[ \begin{aligned} & \int_{-\infty}^{\infty} \exp \left(-\frac{1}{2} \sum_{i, j=1}^n A_{i j} x_i x_j\right) d \mathbf{x}=\int_{-\infty}^{\infty} \exp \left(-\frac{1}{2} x^T A x\right) d \mathbf{x}=\sqrt{\frac{(2 \pi)^n}{\operatorname{det} A}} \\ & \int_{-\infty}^{\infty} \exp \left(-\frac{1}{2} \sum_{i, j=1}^n A_{i j} x_i x_j+\sum_{i=1}^n B_i x_i\right) d^n x \\ & =\int_{-\infty}^{\infty} \exp \left(-\frac{1}{2} x^T A x+B^T x\right) d^n x=\sqrt{\frac{(2 \pi)^n}{\operatorname{det} A}} \exp \left(\frac{1}{2} B^T A^{-1} B\right) \\ & \int_{-\infty}^{\infty} \exp \left(-\frac{1}{2} x^T A x+i B^T x\right) d^n x=\sqrt{\frac{(2 \pi)^n}{\operatorname{det} A}} \exp \left(-\frac{1}{2} B^T A^{-1} B\right) \\ & \int_{-\infty}^{\infty} \exp \left(\frac{i}{2} x^T A x+i B^T x\right) d^n x=\sqrt{\frac{(2 \pi i)^n}{\operatorname{det} A}} \exp \left(-\frac{i}{2} B^T A^{-1} B\right) \end{aligned} \]
some useful integrals
\[ a \int_{-\infty}^{\infty} \frac{\cos m x d m}{a^2+m^2}= \pm \int_{-\infty}^{\infty} \frac{m \sin m x d m}{a^2+m^2}=\pi e^{\mp a x} \]
- if \(a\) be positive we have
\[ \int_{-\infty}^{\infty} \frac{e^{i m x} d m}{a+i m}=\left\{\begin{array}{cc} 2 \pi e^{-a x}, & {[x>0]} \\ 0, & {[x<0]} \end{array} \right. \]
whilst \[ \int_{-\infty}^{\infty} \frac{e^{i m x} d m}{a-i m}=\left\{\begin{array}{cc} 0, & {[x>0]} \\ 2 \pi e^{a x}, & {[x<0]} \end{array}\right. \]
- \[ f(\varpi)=\int^{\infty} J_0(k \varpi) k d k \int_0^{\infty} f(\alpha) J_0(k \alpha) \alpha d \alpha \]