常用定积分
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Dirichlet
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$$\int_{0}^{\infty}\frac{\sin x}{x}\text{d}x=\frac{\pi}{2}$$
$\nabla$证
$$ 引入狄利克雷核D_n=\frac{\sin\frac{(2n+1)x}{2}}{2\sin{\frac{x}{2}}}=\frac{1}{2}+\sum\limits_{i=1}^{n}\cos{nx}\\\\ \quad\\\\ \int_{0}^{\pi}D_n=\frac{\pi}{2}\\\\ \quad\\\\ 令f(x)=\frac{1}{x}-\frac{1}{2\sin\frac{x}{2}}=O(x)(x\to 0)，f(x)在[0,\pi]上常义可积\\\\ \quad\\\\ 由\text{Riemann}定理有\\\\ \quad\\\\ \lim\limits_{n\to\infty}\int_{0}^{\pi}f(x)\sin\frac{(2n+1)x}{2}\text{d}x=0\\\\ \quad\\\\ 即\lim\limits_{n\to\infty}\int_{0}^{\pi}\frac{\sin\frac{(2n+1)x}{2}}{x}\text{d}x=\int_{0}^{\infty}\frac{\sin x}{x}\text{d}x=\lim\limits_{n\to\infty}\int_{0}^{\pi}D_n\text{d}x=\frac{\pi}{2} $$
$\nabla$衍生积分
$$\int_{0}^{\infty}\frac{\sin^2x}{x^2}\text{d}x=\frac{\pi}{2}\qquad \int_{0}^{\infty}\frac{\sin^2x\cos ax}{x^2}\text{d}x=\left\lbrace\begin{array}&\frac{(2-a)\pi}{4}&a\in [0,2)\\\\ 0&a\in [2,\infty)\end{array}\right.$$
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Fresnel
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$$\int_{0}^{\infty}\cos\frac{\pi x^2}{2}\text{d}x=\frac{1}{2}\qquad \int_{0}^{\infty}\sin\frac{\pi x^2}{2}\text{d}x=\frac{1}{2}$$
$\nabla$证
$$ 先用\Gamma函数得出一个代换\\\\ \quad\\\\ \Gamma(x)=\int_{0}^{\infty}t^{x-1}e^{-t}\text{d}t\\\\ \quad\\\\ \int_{0}^{\infty}t^{x-1}e^{-yt}\text{d}t=y^{-x}\Gamma(x)\quad y^{-x}=\frac{1}{\Gamma(x)}\int_{0}^{\infty}t^{x-1}e^{-yt}\text{d}t\\\\ \quad\\\\ \begin{aligned}\int_{0}^{\infty}\cos\frac{\pi x^2}{2}\text{d}x&=\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}\frac{\cos t}{\sqrt{t}}\text{d}t\\\\ &=\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}\frac{\cos t}{\Gamma(\frac{1}{2})}\int_{0}^{\infty}y^{-\frac{1}{2}}e^{-ty}\text{d}y\text{d}t\\\\ &=\frac{\sqrt{2}}{2\pi}\int_{0}^{\infty}y^{-\frac{1}{2}}dy\int_{0}^{\infty}\cos t \;e^{-yt}\text{d}t\\\\ &=\frac{\sqrt{2}}{2\pi}\int_{0}^{\infty}\frac{y^{\frac{1}{2}}}{1+y^2}\text{d}y\\\\ &=\frac{\sqrt{2}}{4\pi}\int_{0}^{\infty}\frac{z^{-\frac{1}{4}}}{1+z}\text{d}z\\\\ &=\frac{\sqrt{2}}{4\pi}\text{B}(\frac{3}{4},\frac{1}{4})\\\\ &=\frac{1}{2} \end{aligned}\\\\ \quad\\\\ 同理可得\int_{0}^{\infty}\sin\frac{\pi x^2}{2}\text{d}x=\frac{1}{2} $$
$\nabla$衍生积分
$$\int_{0}^{\infty}\cos x^p=\frac{1}{p}\Gamma(\frac{1}{p})\cos\frac{\pi}{2p}\quad p>1$$
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Euler-Poisson
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$$\int_{0}^{\infty}e^{-x^2}\text{d}x=\frac{\sqrt{\pi}}{2}$$
$\nabla$证
$$ \begin{aligned}\int_{0}^{\infty}e^{-x^2}\text{d}x&=\sqrt{\int_{0}^{\infty}e^{-x^2}\text{d}x\int_{0}^{\infty}e^{-y^2}\text{d}y}\\\\ &=\sqrt{\int_{0}^{\frac{\pi}{2}}\int_{0}^{\infty}re^{-r^2}\text{d}r\text{d}\theta}\\\\ &=\frac{\sqrt{\pi}}{2}\end{aligned} $$
$\nabla$衍生积分
$$\int_{0}^{\infty}e^{\displaystyle-(a^2x^2+\frac{b^2}{x^2})}\text{d}x=\frac{e^{-2ab}\sqrt{\pi}}{2a}$$