泊松（Poisson）积分表达式：
\[J_\nu\left(z\right)=\dfrac{\left(z/2\right)^\nu}{\sqrt{\pi}\Gamma\left(\nu+1/2\right)}\int_0^{\pi}\cos\left(z\cos\theta\right)\sin^{2\nu}\theta\mathrm{d}\theta\left(\mathrm{Re}\left(\nu\right)\gt-\dfrac{1}{2}\right) \]证明：
\[\begin{aligned} \int_0^{\pi/2}\cos\left(z\cos \theta\right)\sin^{2\nu}\theta\mathrm{d}\theta &= \sum\limits_{m=0}^{\infty}\dfrac{\left(-1\right)^m}{\left(2m\right)!}z^{2m}\int_0^{\pi/2}\cos^{2m}u\sin^{2\nu}\theta\mathrm{d}\theta\\ &= \dfrac{1}{2}\sum\limits_{m=0}^{\infty}\dfrac{\left(-1\right)^m}{\left(2m\right)!}\dfrac{\Gamma\left(\nu+1/2\right)\Gamma\left(m+1/2\right)}{\Gamma\left(\nu+m+1\right)}z^{2m}\\ &= \dfrac{\sqrt{\pi}2^{n-1}\Gamma\left(\nu+1/2\right)}{z^{\nu}}J_\nu\left(z\right) \end{aligned} \]移项即可得到原式。
由这个式子也可以得到
\[\begin{aligned} \int_0^{\pi/2}\sin\left(z\sin\theta\right)\sin\theta\cos^{2\nu}\theta\mathrm{d}\theta &= -\dfrac{1}{z}\int_0^{\pi/2}\cos^{2\nu-1}\theta\sin\theta\mathrm{d}\left(\cos\left(z\sin\theta\right)\right)\\ &= \dfrac{1}{z}\int_0^{\pi/2}\cos\left(z\sin\theta\right)\left(2\nu\cos^{2\nu}\theta-\left(2\nu-1\right)\cos^{2\nu-2}\theta\right)\mathrm{d}\theta\\ &= \dfrac{\sqrt{\pi}}{z}\left(2\nu\dfrac{2^{\nu-1}\Gamma\left(n+1/2\right)}{z^\nu}J_\nu\left(z\right)-\left(2\nu-1\right)\dfrac{\sqrt{\pi}2^{\nu-1}\Gamma\left(\nu-1\right)}{z^{\nu-1}}J_{\nu-1}\left(z\right)\right)\\ &= \dfrac{\sqrt{\pi}2^{\nu-1}\Gamma\left(n+1/2\right)}{z^\nu}\left(\dfrac{2\nu}{z}J_\nu\left(z\right)-J_{\nu-1}\left(z\right)\right)\\ &= \dfrac{\sqrt{\pi}2^{\nu-1}\Gamma\left(n+1/2\right)}{z^\nu}J_{\nu+1}\left(z\right) \end{aligned} \]移项可以得到
\[J_\nu\left(z\right)=\dfrac{a^{\nu-1}}{\sqrt{\pi}2^{\nu-2}\Gamma\left(\nu-1/2\right)}\int_0^{\pi/2}\cos^{2\nu-2}\theta\sin\left(a\sin\theta\right)\sin\theta\mathrm{d}\theta \]第一索宁（Sonine）有限积分公式：
\[\int_0^{\pi/2}J_{\nu}\left(z\sin\theta\right)\sin^{\mu+1}\theta\cos^{2\nu-1}\theta\mathrm{d}\theta = \dfrac{2^\nu}{z^{\nu+1}}\Gamma\left(\nu+1\right)J_{\mu+\nu+1}\left(z\right) \]证明：
\[\begin{aligned} \int_0^{\pi/2}J_{\nu}\left(z\sin\theta\right)\sin^{\mu+1}\theta\cos^{2\nu-1}\theta\mathrm{d}\theta &= \sum\limits_{k=0}^{\infty}\dfrac{\left(-1\right)^k}{k!\Gamma\left(k+\mu+1\right)}\left(\dfrac{z}{2}\right)^{2k+\mu}\int_0^{\pi/2}\sin^{2\mu+2k+1}\theta\cos^{2\nu+1}\theta\mathrm{d}\theta\\ &= \sum\limits_{k=0}^{\infty}\dfrac{\left(-1\right)^k}{2\Gamma\left(\mu+\nu+k+2\right)}\Gamma\left(\nu+1\right)\left(\dfrac{z}{2}\right)^{2k+\mu}\\ &= \dfrac{2^\nu}{z^{\nu+1}}\Gamma\left(\nu+1\right)J_{\mu+\nu+1}\left(z\right) \end{aligned} \]第二索宁（Sonine）有限积分公式：
\[\begin{aligned} \int_0^{\pi/2}J_{\mu}\left(x\sin\theta\right)J_\nu\left(y\cos\theta\right)\sin^{\mu+1}\theta\cos^{\nu+1}\theta\mathrm{d}\theta = \dfrac{x^\mu y^\nu J_{\mu+\nu+1}\left(\sqrt{x^2+y^2}\right)}{\left(x^2+y^2\right)^{\left(\mu+\nu+1\right)/2}} \end{aligned} \]证明：，
\[\begin{aligned} \text{LHS} &= \left(\dfrac{x}{2}\right)^\mu\left(\dfrac{y}{2}\right)^\nu\sum\limits_{m=0}^{\infty}\sum\limits_{n=0}^{\infty}\dfrac{x^{2m}y^{2n}}{\Gamma\left(m+\mu+1\right)\Gamma\left(n+\nu+1\right)}\dfrac{\left(-1\right)^{m+n}}{m!n!2^{2m+2n}}\int_0^{\pi/2}\sin^{2m+2\nu+1}\theta\cos^{2n+2\nu+1}\theta\mathrm{d}\theta\\ &= \left(\dfrac{x}{2}\right)^\mu\left(\dfrac{y}{2}\right)^\nu\sum\limits_{m=0}^{\infty}\sum\limits_{n=0}^{\infty}\dfrac{\left(-1\right)^{m+n}}{m!n!2^{2m+2n}}\dfrac{x^{2m}y^{2n}}{2\Gamma\left(m+n+\mu+\nu+2\right)}\\ &= \left(\dfrac{x}{2}\right)^\mu\left(\dfrac{y}{2}\right)^\nu\sum\limits_{t=0}^{\infty}\dfrac{\left(-1\right)^t}{2^{2t+1}t!\Gamma\left(t+\mu+\nu+2\right)}\sum\limits_{m=0}^t\binom{t}{m}x^{2m}y^{2\left(t-m\right)}\\ &= \left(\dfrac{x}{2}\right)^\mu\left(\dfrac{y}{2}\right)^\nu\dfrac{2^{\mu+\nu}}{\left(x^2+y^2\right)^{\left(\mu+\nu+1\right)/2}}\sum\limits_{t=0}^{\infty}\dfrac{\left(-1\right)^t}{t!\Gamma\left(t+\mu+\nu+2\right)}\left(\dfrac{\sqrt{x^2+y^2}}{2}\right)^{2t+\mu+\nu+1}\\ &= \dfrac{x^\mu y^\nu J_{\mu+\nu+1}\left(\sqrt{x^2+y^2}\right)}{\left(x^2+y^2\right)^{\left(\mu+\nu+1\right)/2}} \end{aligned} \]斯图鲁弗（Struve）积分公式：
\[\int_0^{\infty}\dfrac{J_\mu\left(t\right)J_\nu\left(t\right)}{t^{\mu+\nu}}\mathrm{d}t = \dfrac{\Gamma\left(1/2\right)\Gamma\left(\mu+\nu\right)}{2^{\mu+\nu}\Gamma\left(\mu+\nu+1/2\right)\Gamma\left(\mu+1/2\right)\Gamma\left(\nu+1/2\right)} \]证明：
将 \(J_\nu\) 的积分表示
\[J_\nu\left(a\right)=\dfrac{a^{\nu-1}}{\sqrt{\pi}2^{\nu-2}\Gamma\left(\nu-1/2\right)}\int_0^{\pi/2}\cos^{2\nu-2}u\sin\left(a\sin u\right)\sin u\mathrm{d}u \]代入得
\[\int_0^{\infty}\dfrac{J_\mu\left(t\right)J_\nu\left(t\right)}{t^{\mu+\nu}}\mathrm{d}t =\dfrac{\int_0^{\infty}\int_0^{\pi/2}\int_0^{\pi/2}\dfrac{\sin\left(t\sin u\right)\sin\left(t\sin v\right)}{t^2}\cos^{2\mu-2}u\cos^{2\nu-2}v\sin u\sin v\mathrm{d}t\mathrm{d}u\mathrm{d}v}{\pi2^{\mu+\nu-4}\Gamma\left(\mu-1/2\right)\Gamma\left(\nu-1/2\right)} \]令 \(\alpha=\sin u,\beta=\sin v\)，则 \(0\le \alpha,\beta\le 1\)，
\[\begin{aligned} \int_0^{\infty}\dfrac{\sin\alpha t\sin\beta t}{t^2}\mathrm{d}t &= \dfrac{1}{2}\int_0^{\infty}\dfrac{\cos\left(\alpha-\beta\right)t-\cos\left(\alpha+\beta\right)t}{t^2}\mathrm{d}t\\ &=\dfrac{1}{2}\left[\int_0^{\infty}\dfrac{1-\cos 2\beta t}{t^2}\mathrm{d}t+\int_{\beta}^{\alpha}\int_0^{\infty}\dfrac{-\sin\left(\alpha'-\beta\right)t+\sin\left(\alpha'+\beta\right)t}{t}\mathrm{d}\alpha'\mathrm{d}t\right] \end{aligned} \]对于第一个积分，
\[\int_0^{\infty}\dfrac{1-\cos 2\beta t}{t^2}\mathrm{d}t = 2\int_0^{\infty}\dfrac{\sin^2\beta t}{t^2}\mathrm{d}t = \pi\beta \]而
\[\int_0^{\infty}\dfrac{\sin \alpha x}{x}\mathrm{d}x=\begin{cases}\dfrac{\pi}{2}, & \alpha\gt 0\\-\dfrac{\pi}{2}, & \alpha\lt 0\end{cases} \]因此
\[\int_0^{\infty}\dfrac{-\sin\left(\alpha'-\beta\right)t+\sin\left(\alpha'+\beta\right)t}{t}\mathrm{d}\alpha' = \begin{cases}\pi,&\alpha'\lt \beta\\0, & \alpha'\gt \beta\end{cases} \]即
\[\int_0^{\infty}\dfrac{\sin\alpha t\sin\beta t}{t^2}\mathrm{d}t=\pi\min\left(\alpha,\beta\right) \]因此
\[\begin{aligned} \int_0^{\infty}\dfrac{J_\mu\left(t\right)J_\nu\left(t\right)}{t^{\mu+\nu}}\mathrm{d}t &= \dfrac{\int_0^{\infty}\int_0^{\pi/2}\int_0^{\pi/2}\dfrac{\sin\left(t\sin u\right)\sin\left(t\sin v\right)}{t^2}\cos^{2\mu-2}u\cos^{2\nu-2}v\sin u\sin v\mathrm{d}t\mathrm{d}u\mathrm{d}v}{\pi2^{\mu+\nu-3}\Gamma\left(\mu-1/2\right)\Gamma\left(\nu-1/2\right)}\\ &= \dfrac{\int_0^{\pi/2}\int_0^{u}\cos^{2\mu-2}u\cos^{2\nu-2}v\sin u\sin^2 v\mathrm{d}u\mathrm{d}v+\int_0^{\pi/2}\int_0^{v}\cos^{2\mu-2}u\cos^{2\nu-2}v\sin^2 u\sin v\mathrm{d}v\mathrm{d}u}{2^{\mu+\nu-3}\Gamma\left(\mu-1/2\right)\Gamma\left(\nu-1/2\right)}\\ &= \dfrac{\dfrac{1}{2\mu-1}\int_0^{\pi/2}\cos^{2\mu+2\nu-3}v\sin^2 v\mathrm{d}v+\dfrac{1}{2\nu-1}\int_0^{\pi/2}\cos^{2\mu+2\nu-3}u\sin^2 u\mathrm{d}u}{2^{\mu+\nu-3}\Gamma\left(\mu-1/2\right)\Gamma\left(\nu-1/2\right)}\\ &= \dfrac{\Gamma\left(3/2\right)\Gamma\left(\mu+\nu-1\right)\left(\mu+\nu-1\right)}{\left(2\mu-1\right)\left(2\nu-1\right)2^{\mu+\nu-3}\Gamma\left(\mu-1/2\right)\Gamma\left(\nu-1/2\right)\Gamma\left(\mu+\nu+1/2\right)}\\ &= \dfrac{\Gamma\left(1/2\right)\Gamma\left(\mu+\nu\right)}{2^{\mu+\nu}\Gamma\left(\mu+\nu+1/2\right)\Gamma\left(\mu+1/2\right)\Gamma\left(\nu+1/2\right)} \end{aligned} \]得证。
题外话：标签：right,函数,dfrac,sin,mu,性质,Bessel,nu,left From： https://www.cnblogs.com/bestlxm/p/17971191
这篇文章写于期末出分后，交上去用来防止绩点挂得太惨。主要内容都是从书上抄的。
学了一个学期数理方程，每个星期都会花接近一天时间，最后仍然得到了很烂的成绩。
能力确实不行。以后这博客也不一定更得动了。