積分公式
不定積分
\[\begin{equation} \int \frac{1}{a^2 + x^2} \dd x = \frac{1}{a}\arctan\frac{x}{a} \end{equation}\] \[\begin{equation} \int \frac{1}{a^2 - x^2} \dd x = \frac{1}{a}\artanh\frac{x}{a} \end{equation}\] \[\begin{equation} \int \frac{1}{\sqrt{a^2 - x^2}} \dd x = \arcsin\frac{x}{a} \end{equation}\] \[\begin{equation} \int \frac{1}{\sqrt{x^2 + a}} \dd x = \log\left| x + \sqrt{x^2 + a}\right| \end{equation}\]$\tan{\frac{x}{2}} = t$ の置換積分
$x$ から $t$ へ $\tan{\frac{x}{2}} = t$ の置換。次のように変換される：
\[\begin{align} \dd x = \frac{2}{1 + t^2} \dd t \end{align}\] \[\begin{align} \sin x = \frac{2t}{1 + t^2}, \quad\cos x = \frac{1 - t^2}{1 + t^2}, \quad\tan x = \frac{2t}{1 - t^2} \end{align}\]定積分
Gauss関数
\[\begin{gather} \int_{-\infty}^{\infty} e^{-ax^2} \dd x = \sqrt{\frac{\pi}{a}} \\ \int_{-\infty}^{\infty} x^2 e^{-ax^2} \dd x = \frac{1}{2}\sqrt{\frac{\pi}{a^3}} \\ \int_{-\infty}^{\infty} x^4 e^{-ax^2} \dd x = \frac{3}{4}\sqrt{\frac{\pi}{a^5}} \end{gather}\]複素平面上で積分経路を虚軸方向にずらした場合：
\[\begin{align} \int_{-\infty}^{\infty} e^{-a(x + ib)^2} \dd x = \sqrt{\frac{\pi}{a}} \end{align}\]