常见定积分的计算
一些数学杂技而已。用作收集。
过于简单的(如不定积分都能算的,不包括那些不定积分只能给出机械化步骤无法或很难给出解的表达式的)就不记录了。
收敛性:\(\displaystyle \ln\left(\sin x\right)\mathrm{d}x = \ln x+\ln\left(\dfrac{\sin x}{x}\right)\),前者收敛,后者为常义积分,所以收敛。
\( \begin{aligned} \int_{0}^{\pi/2}\ln\left(\sin x\right)\mathrm{d}x &= \dfrac{1}{2}\int_{0}^{\pi}\ln\left(\sin x\right)\mathrm{d}x\\ &= \int_{0}^{\pi/2}\ln\left(\sin 2x\right)\mathrm{d}x\\ &= \dfrac{\pi\ln 2}{2}+2\int_{0}^{\pi/2}\ln\left(\sin x\right)\mathrm{d}x\\ &= -\dfrac{\pi\ln 2}{2} \end{aligned} \)
推广:
\(
\begin{aligned}
\int_0^{\pi/2}\ln\left(\cos x\right)\mathrm{d}x = -\dfrac{\pi\ln 2}{2}
\end{aligned}
\)
\(
\begin{aligned}
\int_0^{\pi/2}\dfrac{x}{\tan x}\mathrm{d}x &= x\ln\left(\sin x\right)\Bigg\vert_0^{\pi/2} - \int_0^{\pi/2}\ln\left(\sin x\right)\mathrm{d}x = \dfrac{\pi\ln 2}{2}
\end{aligned}
\)
\(a=b\) 时 \(\displaystyle \int_0^{\pi/2}\ln\left(a^2\sin^2 x+b^2\cos^2 x\right)\mathrm{d}x=\pi\ln a\)
\(a\neq b\) 时令 \(\displaystyle I\left(a\right)=\int_0^{\pi/2}\ln\left(a^2\sin^2 x+b^2\cos^2 x\right)\mathrm{d}x\),则
\(
\begin{aligned}
I'\left(a\right) &= \int_0^{\pi/2}\dfrac{2a\sin^2 x}{a^2\sin^2 x+b^2\cos^2 x}\mathrm{d}x\\
&= \dfrac{1}{a^2-b^2}\left(-2b\arctan\left(\dfrac{a\tan x}{b}\right)+2ax\right)\Bigg\vert_{0}^{\pi/2}\\
&= \dfrac{\pi}{a+b}
\end{aligned}
\)
因此 \(\displaystyle I\left(a\right)=\pi\ln b+\int_b^a\dfrac{\pi}{u+b}\mathrm{d}u=\pi\ln\dfrac{a+b}{2}\)
收敛性:\(x\lt 1\) 时为常义积分,\(x\gt 1\) 时 \(\mathrm{e}^{-x^2}\lt\mathrm{e}^{-x}\),而 \(\displaystyle\int_1^{+\infty}\mathrm{e}^{-x}\mathrm{d}x\) 收敛。
\( \begin{aligned} \int_0^{+\infty}\mathrm{e}^{-x^2}\mathrm{d}x &= \sqrt{\int_0^{+\infty}\int_0^{+\infty}\mathrm{e}^{-x^2-y^2}\mathrm{d}x\mathrm{d}y}\\ &= \sqrt{\int_0^{+\infty}\int_0^{\pi/2}\mathrm{e}^{-r^2}r\mathrm{d}r\mathrm{d}\theta}\\ &= \dfrac{\sqrt{\pi}}{2} \end{aligned} \)
令 \(\displaystyle I\left(u\right)=\int_0^{2\pi}\mathrm{e}^{u\cos x}\cos\left(u\sin x\right)\mathrm{d}x\),则 \(I\left(0\right)=2\pi\)
\(
\begin{aligned}
I'\left(u\right) &= \int_0^{2\pi}\mathrm{e}^{u\cos x}\left(\cos x\cos\left(u\sin x\right)-\sin x\sin\left(u\sin x\right)\right)\mathrm{d}x\\
&= u\int_0^{2\pi}\left(\mathrm{e}^{u\cos x}\mathrm{d}\left(\sin\left(u\sin x\right)\right)+\sin\left(u\sin x\right)\mathrm{d}\left(\mathrm{e}^{u\cos x}\right)\right)\\
&= 0
\end{aligned}
\)
因此 \(I\left(u\right)=2\pi\)
\( \begin{aligned} \int_0^1\dfrac{x^b-x^a}{\ln x}\mathrm{d}x &= \int_0^1\mathrm{d}x\int_a^b x^u\mathrm{d}u\\ &= \int_a^b\mathrm{d}u\int_0^1 x^u\mathrm{d}x\\ &= \int_a^b\dfrac{\mathrm{d}u}{u+1}\\ &= \ln\dfrac{b+1}{a+1} \end{aligned} \)
收敛性:\(x\gt 1\) 时 \(\dfrac{x^{p-1}}{1+x}\sim\dfrac{1}{x^{2-p}}\),\(x\lt 1\) 时 \(\dfrac{x^{p-1}}{1+x}\sim\dfrac{1}{x^{1-p}}\),所以收敛。
\( \begin{aligned} \int_0^1\dfrac{x^{p-1}}{1+x}\mathrm{d}x &= \int_0^1\mathrm{d}x\sum\limits_{n=0}^{+\infty}\left(-1\right)^n x^{n+p-1} \end{aligned} \)
\(x=0\) 为瑕点,\(x=1\) 时级数不收敛。
级数在 \(\left(0,1\right)\) 的任何闭区间一致收敛,且 \(\displaystyle \int_0^1\mathrm{d}x\sum\limits_{k=0}^{n-1}\left(-1\right)^k x^{k+p-1}\le\int_0^1 x^{p-1}\mathrm{d}x\) 对 \(n\in\mathbb{N}^{\star}\) 一致收敛,所以
\(
\begin{aligned}
\int_0^1\mathrm{d}x\sum\limits_{n=0}^{+\infty}\left(-1\right)^n x^{n+p-1} &= \sum\limits_{n=0}^{+\infty}\dfrac{\left(-1\right)^n}{n+p}
\end{aligned}
\)
\(
\begin{aligned}
\int_1^{+\infty}\dfrac{x^{p-1}}{1+x}\mathrm{d}x = \int_0^1 \dfrac{x^{-p}}{1+x}\mathrm{d}x = \sum\limits_{n=1}^{+\infty}\dfrac{\left(-1\right)^n}{p-n}
\end{aligned}
\)
\(
\begin{aligned}
\int_0^{+\infty}\dfrac{x^{p-1}}{1+x}\mathrm{d}x = \dfrac{1}{p}+2p\sum\limits_{n=0}^{+\infty}\dfrac{\left(-1\right)^n}{p^2-n^2}
\end{aligned}
\)
由 \(a\in\mathbb{R}\setminus\mathbb{Z}\) 时 \(\cos ax\) 在 \(\left[-\pi,\pi\right]\) 上的 Fourier 展开式 \(\displaystyle \cos ax=\dfrac{\sin a\pi}{\pi}\left(\dfrac{1}{a}+2a\sum\limits_{n=1}^{+\infty}\dfrac{\left(-1\right)^n}{a^2-n^2}\cos nx\right)\)
带入 \(x=0\) 可知 \(\displaystyle \dfrac{\pi}{\sin a\pi}=\dfrac{1}{a}+2a\sum\limits_{n=1}^{+\infty}\dfrac{\left(-1\right)^n}{a^2-n^2}\)
所以 \(\displaystyle\int_0^{+\infty}\dfrac{x^{p-1}}{1+x}=\dfrac{\pi}{\sin p\pi}\)
设 \(\displaystyle I\left(a\right)=\int_0^{+\infty}\mathrm{e}^{-ax}\dfrac{\sin x}{x}\mathrm{d}x\),则由于 \(\displaystyle\int_0^{+\infty}\dfrac{\sin x}{x}\mathrm{d}x\) 收敛、\(\mathrm{e}^{-ax}\) 关于 \(x\) 单调、关于 \(a\) 一致有界,所以 \(I\left(a\right)\) 在 \(\left[0,+\infty\right)\) 上一致收敛。
因为 \(\mathrm{e}^{-ax}\dfrac{\sin x}{x}\) 在 \(\left(a,x\right)\in\left[0,+\infty\right)\times\left[0,+\infty\right)\) 上连续,所以 \(I\left(a\right)\) 在 \(\left[0,+\infty\right)\) 上连续,\(I\left(0\right)=\lim\limits_{a\to 0^+}I\left(a\right)\)
因为 \(\displaystyle \int_0^{+\infty}\mathrm{e}^{-ax}\sin x\mathrm{d}x\) 在 \(\left[\dfrac{a}{2},+\infty\right)\) 一致收敛,所以 \(a\gt 0\) 时
\(
\begin{aligned}
I'\left(a\right) &= -\int_0^{+\infty}\mathrm{e}^{-ax}\sin x\mathrm{d}x\\
&= -\dfrac{\mathrm{e}^{-ax}\left(a\sin x+\cos x\right)}{a^2+1}\Bigg\vert_0^{+\infty}\\
&= -\dfrac{1}{1+a^2}
\end{aligned}
\)
即 \(\displaystyle I\left(a\right)=-\arctan a+C\),由 \(a\to +\infty\) 时 \(I\left(a\right)\to 0\) 知 \(C=\dfrac{\pi}{2}\)
所以 \(\displaystyle I\left(0\right)=\lim_{a\to 0^+}\left(\dfrac{\pi}{2}-\arctan a\right)=\dfrac{\pi}{2}\)
乘上 \(\mathrm{e}^{-ax}\) 再令 \(a\to 0^+\) 的方法等价于 Laplace 变换中的结论:
若 $F\left(p\right)$ 为 $f\left(t\right)$ 的 Laplace 变换,则在等号两边积分均存在的情况下有 $$ \int_0^{+\infty}F\left(p\right)\mathrm{d}p = \int_0^{+\infty}\dfrac{f\left(t\right)}{t}\mathrm{d}t $$
\(\forall a\ge 0\),令 \(\displaystyle I\left(a\right)=\int_0^{+\infty}\left(\dfrac{\sin ax}{x}\right)^2\mathrm{d}x\),则 \(I\left(0\right)=0\)。由 \(\displaystyle I\left(a\right)\le\int_0^1 a^2\mathrm{d}x + \int_1^{+\infty}\dfrac{\mathrm{d}x}{x^2}\) 知 \(I\left(a\right)\) 关于 \(a\) 一致收敛,由 \(\left(\dfrac{\sin ax}{x}\right)^2\) 在 \(\left(a,x\right)\in\left[0,+\infty\right)\times\left(0,+\infty\right)\) 连续知 \(I\left(a\right)\) 连续,即 \(I\left(0\right)=\lim\limits_{a\to 0^+}I\left(a\right)\)。
\(\forall \delta\gt 0\),\(A\to +\infty\) 时 \(\displaystyle \int_0^A\sin 2ax\mathrm{d}x\) 对任意 \(a\ge\delta\) 一致有界,\(\dfrac{1}{x}\) 关于 \(x\) 单调,所以 \(\displaystyle \int_0^{+\infty}\dfrac{\sin 2ax}{x}\mathrm{d}x\) 在 \(\left[\delta, +\infty\right)\) 上一致收敛。
由 \(\dfrac{\sin 2ax}{x}\) 在 \(\left(a,x\right)\in\left[0,+\infty\right)\times\left(0,+\infty\right)\) 连续知 \(\displaystyle \forall a\gt 0,I'\left(a\right)=\int_0^{+\infty}\dfrac{\sin 2ax}{x}\mathrm{d}x=\dfrac{\pi}{2}\)。
因此 \(I\left(1\right)=I\left(\delta\right)+\dfrac{\pi}{2}\left(1-\delta\right)\),令 \(\delta\to 0^+\) 得 \(I\left(1\right)=\dfrac{\pi}{2}\),即 \(\displaystyle \int_0^{+\infty}\left(\dfrac{\sin x}{x}\right)^2\mathrm{d}x = \dfrac{\pi}{2}\)
取以 \(R\) 为半径、\(0\) 为圆心的上半圆围道,从上方绕过原点,则函数 \(\dfrac{3\mathrm{e}^{\mathrm{i}z}-\mathrm{e}^{3\mathrm{i}z}-2}{z^3}\) 围道内无奇点。令 \(R\to +\infty\) 得
\( \begin{aligned} \int_0^{+\infty}\dfrac{\sin^3 x}{x^3}\mathrm{d}x &= \dfrac{1}{8}\mathrm{Im}\int_{-\infty}^{+\infty}\dfrac{3\mathrm{e}^{\mathrm{i}x}-\mathrm{e}^{3\mathrm{i}x}-2}{x^3}\mathrm{d}x\\ &= \dfrac{\pi}{8}\mathrm{Res}\dfrac{3\mathrm{e}^{\mathrm{i}z}-\mathrm{e}^{3\mathrm{i}z}}{z^3}\Bigg\vert_{z=0}\\ &= \dfrac{3}{8}\pi \end{aligned} \)
简单换元可得 \(\displaystyle \int_0^{+\infty}\sin\left(x^2\right)\mathrm{d}x = \dfrac{1}{2}\int_0^{+\infty}\dfrac{\sin x}{\sqrt{x}}\mathrm{d}x\),\(\forall a\ge 0\),设 \(\displaystyle I\left(a\right) = \int_0^{+\infty}\mathrm{e}^{-ax}\dfrac{\sin x}{\sqrt{x}}\mathrm{d}x\)。
\(a\gt 0\) 时,由 \(\displaystyle \int_0^{+\infty}\mathrm{e}^{-x^2}\mathrm{d}x = \dfrac{\sqrt{\pi}}{2}\) 可知 \(\displaystyle \forall t\gt 0, \int_0^{+\infty}\mathrm{e}^{-tx^2}\mathrm{d}x = \dfrac{\sqrt{\pi}}{2\sqrt{t}}\),则 \(\displaystyle I\left(a\right) = \dfrac{2}{\sqrt{\pi}}\int_0^{+\infty}\mathrm{d}x\int_0^{+\infty}\mathrm{e}^{-x\left(a+t^2\right)}\sin x\mathrm{d}t\)。
由 \(\displaystyle \int_0^{+\infty}\left\vert\mathrm{e}^{-x\left(a+t^2\right)}\sin x\right\vert\mathrm{d}x \le \int_0^{+\infty}\mathrm{e}^{-ax}\mathrm{d}x\) 知 \(\displaystyle \int_0^{+\infty}\mathrm{e}^{-x\left(a+t^2\right)}\sin x\mathrm{d}x\) 对 \(t\in\left[0,+\infty\right)\) 一致收敛;
由 \(\lim\limits_{x\to 0^+}\dfrac{\mathrm{e}^{-ax}\sin x}{\sqrt{x}} = 0\) 且 \(\displaystyle\int_0^{+\infty}\mathrm{e}^{-x\left(a+t^2\right)}\sin x\mathrm{d}t\) 对任意 \(\delta\gt 0\) 在 \(\left[\delta, +\infty\right)\) 上一致收敛可知 \(\displaystyle\int_0^{+\infty}\mathrm{e}^{-x\left(a+t^2\right)}\sin x\mathrm{d}t\) 在 \(\left[0, +\infty\right)\) 上一致收敛;
由 \(\displaystyle \int_0^{+\infty}\mathrm{e}^{-x\left(a+t^2\right)}\sin x\mathrm{d}x = \dfrac{1}{1+\left(a+t^2\right)^2}\) 关于 \(t\) 连续、\(\dfrac{\mathrm{e}^{-ax}\sin x}{\sqrt{x}}\) 关于 \(x\) 连续、\(\displaystyle \int_0^{+\infty}\left\vert\mathrm{e}^{-ax}\dfrac{\sin x}{\sqrt{x}}\right\vert\mathrm{d}x\) 收敛可知积分次序可交换。
由 \(I\left(a\right)\) 关于 \(a\) 一致收敛、\(\dfrac{1}{1+\left(a+t^2\right)^2}\) 在定义域内连续可得
\(
\begin{aligned}
\int_0^{+\infty}\sin\left(t^2\right)\mathrm{d}x &= \dfrac{1}{\sqrt{\pi}}\lim\limits_{a\to 0^+}\int_0^{+\infty}\mathrm{d}x\int_0^{+\infty}\mathrm{e}^{-x\left(a+t^2\right)}\sin x\mathrm{d}t\\
&= \dfrac{1}{\sqrt{\pi}}\lim\limits_{a\to 0^+}\int_0^{+\infty}\mathrm{d}t\int_0^{+\infty}\mathrm{e}^{-x\left(a+t^2\right)}\sin x\mathrm{d}x\\
&= \dfrac{1}{\sqrt{\pi}}\lim\limits_{a\to 0^+}\int_0^{+\infty}\dfrac{\mathrm{d}t}{1+\left(a+t^2\right)^2}\\
&= \dfrac{1}{\sqrt{\pi}}\int_0^{+\infty}\dfrac{\mathrm{d}t}{1+t^4}\\
&= \sqrt{\dfrac{\pi}{8}}
\end{aligned}
\)
将被积式解析延拓至复平面,构造幅角为 \(0\) 的实轴上 \(\left[0, R\right]\) 部分逆时针旋转 \(\dfrac{2\pi}{n}\) 构成的扇形围道,则围道内包含唯一奇点 \(\mathrm{e}^{\frac{\pi}{n}\mathrm{i}}\)。
设 \(\displaystyle I=\int_0^{+\infty}\dfrac{x^m}{1+x^{n}}\mathrm{d}x\),则 \(R\to +\infty\) 时由留数定理有 \(\displaystyle I-\int_0^{+\infty}\dfrac{r^m\mathrm{e}^{\frac{2\pi}{n}m\mathrm{i}}}{1+r^{n}}\mathrm{e}^{\frac{2\pi}{n}\mathrm{i}}\mathrm{d}r = 2\pi\mathrm{i}\dfrac{\mathrm{e}^{\frac{m}{n}\mathrm{i}\pi}}{n\mathrm{e}^{\frac{n-1}{n}\mathrm{i}\pi}}\)
化简得 \(\displaystyle \left(1-\mathrm{e}^{\frac{2\left(m+1\right)}{n}\mathrm{i}\pi}\right)I = -\dfrac{2\pi\mathrm{i}}{n}\mathrm{e}^{\frac{m+1}{n}\mathrm{i}\pi}\)
因此有 \(\displaystyle \int_0^{+\infty}\dfrac{x^m}{1+x^{n}}\mathrm{d}x = \dfrac{\pi}{n\sin\left(\frac{m+1}{n}\pi\right)}\)
题外话:
你可能会说这个结果是之前对 $\dfrac{x^{p-1}}{1+x}$ 积分的扩展,实际上通过代换 $u=x^n$ 可以发现两者是等价的。
虽然结果是一样的,但两种方法本身也有值得学习之处。
将被积式解析延拓至上半复平面,显然 \(\mathrm{i}\) 是二阶奇点,\(\left(2k+1\right)\mathrm{i}\left(k\in\mathbb{N}^{\star}\right)\) 为一阶奇点。
设 \(R=2r\left(r\in\mathrm{N}^{\star}\right)\) 并令 \(R\to +\infty\),取以 \(R\) 为半径、\(0\) 为圆心的上半圆围道,则由留数定理和大圆弧引理
\(
\begin{aligned}
\int_0^{+\infty}\dfrac{\mathrm{d}x}{\left(1+x^2\right)\cosh\dfrac{\pi x}{2}} &= \dfrac{1}{2}\cdot 2\pi\mathrm{i}\left(\mathrm{Res}\dfrac{1}{\left(1+z^2\right)\cosh\dfrac{\pi z}{2}}\Bigg\vert_{z=\mathrm{i}} + \sum\limits_{k=1}^{+\infty}\mathrm{Res}\dfrac{1}{\left(1+z^2\right)\cosh\dfrac{\pi z}{2}}\Bigg\vert_{z=\left(2k+1\right)\mathrm{i}}\right)\\
&= \pi\mathrm{i}\left(\lim\limits_{z\to\mathrm{i}}\dfrac{\mathrm{d}}{\mathrm{d}z}\dfrac{z-\mathrm{i}}{\left(z+\mathrm{i}\right)\cosh\dfrac{\pi z}{2}}+\sum\limits_{k=1}^{+\infty}\lim\limits_{z\to \left(2k+1\right)\mathrm{i}}\dfrac{z-\left(2k+1\right)\mathrm{i}}{\left(1-\left(2k+1\right)^2\right)\cosh\dfrac{\pi z}{2}}\right)\\
&= \dfrac{1}{2}\left(1+\sum\limits_{k=1}^{+\infty}\dfrac{\left(-1\right)^{k+1}}{k\left(k+1\right)}\right)\\
&= \ln 2
\end{aligned}
\)
设被积函数为 \(\dfrac{z^3\mathrm{e}^z}{\left(\mathrm{e}^z+1\right)^2}\),围道取 \(\left[-R,R\right]\times\left[0,2\pi\mathrm{i}\right]\) 的边界,并令 \(R\to +\infty\)。
则由留数定理有 \(\displaystyle \int_{-\infty}^{+\infty}\dfrac{x^3\mathrm{e}^x-\left(x+2\pi\mathrm{i}\right)^3\mathrm{e}^x}{\left(\mathrm{e}^x+1\right)^2}\mathrm{d}x = 2\pi\mathrm{i}\mathrm{Res}\dfrac{z^3\mathrm{e}^z}{\left(\mathrm{e}^z+1\right)^2}\Bigg\vert_{z=\pi\mathrm{i}}\)
化简得 \(-6\pi\mathrm{i}\displaystyle \int_{-\infty}^{+\infty}\dfrac{x^2\mathrm{e}^x}{\left(\mathrm{e}^x+1\right)^2}\mathrm{d}x + 8\pi^3\mathrm{i}\int_{-\infty}^{+\infty}\dfrac{\mathrm{e}^x}{\left(\mathrm{e}^x+1\right)^2}\mathrm{d}x = 6\pi^3\mathrm{i}\)
因此有 \(\displaystyle \int_{-\infty}^{+\infty}\dfrac{x^2\mathrm{e}^x}{\left(\mathrm{e}^x+1\right)^2}\mathrm{d}x = \dfrac{\pi^2}{3}\)
设 \(t\in\left(0,1\right)\)。对于主值积分 \(\displaystyle \mathrm{PV} \int_{-\infty}^{+\infty}\dfrac{\mathrm{e}^{tx}}{1-\mathrm{e}^x}\mathrm{d}x\),构造 \(\left[-R,+R\right]\times\left[0, 2\pi\mathrm{i}\right]\) 的围道并令 \(R\to +\infty\),从上方绕过 \(0\),从下方绕过 \(2\pi\mathrm{i}\),则围道内无奇点,可得 \(\displaystyle \mathrm{PV} \int_{-\infty}^{+\infty}\dfrac{\mathrm{e}^{tx}}{1-\mathrm{e}^x}\mathrm{d}x = -\pi\mathrm{i}\dfrac{1+\mathrm{e}^{2\pi t\mathrm{i}}}{1-\mathrm{e}^{2\pi t\mathrm{i}}} = -\pi\cot\pi t\)
对于积分 \(\displaystyle \int_{-\infty}^{+\infty}\dfrac{x\mathrm{e}^{tx}}{1-\mathrm{e}^x}\mathrm{d}x\),构造 \(\left[-R,+R\right]\times\left[0, 2\pi\mathrm{i}\right]\) 的围道并令 \(R\to +\infty\),从下方绕过 \(2\pi\mathrm{i}\),则围道内无奇点,在 \(2\pi\mathrm{i}\) 处留数为 \(-2\pi^2\mathrm{e}^{2\pi t\mathrm{i}}\),可得 \(\displaystyle \int_{-\infty}^{+\infty}\dfrac{x\mathrm{e}^{tx}}{1-\mathrm{e}^x}\mathrm{d}x = \dfrac{2\pi^2\mathrm{i}\cot\left(\pi t\right)\mathrm{e}^{2\pi t\mathrm{i}} + 2\pi^2\mathrm{e}^{2\pi t\mathrm{i}}}{1-\mathrm{e}^{2\pi t\mathrm{i}}} = -\dfrac{\pi^2}{\sin^2\pi t}\)
由 \(\displaystyle \int_{-\infty}^{+\infty}\dfrac{x\mathrm{e}^{tx}}{1-\mathrm{e}^x}\mathrm{d}x\) 在 \(t\in\left[p,q\right]\) 上一致收敛得
\(
\begin{aligned}
\int_{-\infty}^{+\infty}\dfrac{\mathrm{e}^{px}-\mathrm{e}^{qx}}{1-\mathrm{e}^x}\mathrm{d}x &= \int_{-\infty}^{+\infty}\int_p^q\dfrac{x\mathrm{e}^{tx}}{1-\mathrm{e}^x}\mathrm{d}t\mathrm{d}x\\
&= \int_p^q\int_{-\infty}^{+\infty}\dfrac{x\mathrm{e}^{tx}}{1-\mathrm{e}^x}\mathrm{d}x\mathrm{d}t\\
&= \pi\left(\cot\pi p-\cot\pi q\right)
\end{aligned}
\)
\( \begin{aligned} \int_0^1 x^x\mathrm{d}x &= \int_0^1\sum\limits_{n=0}^{+\infty}\dfrac{\left(x\ln x\right)^n}{n!}\mathrm{d}x\\ &= \sum\limits_{n=0}^{+\infty}\dfrac{1}{n!}\int_0^1\left(x\ln x\right)^n\mathrm{d}x\\ &= \sum\limits_{n=1}^{+\infty}\dfrac{\left(-1\right)^{n+1}}{n^n} \end{aligned} \)
\( \begin{aligned} \int_0^{\pi}\dfrac{\cos nx}{a-\mathrm{i}b\cos x}\mathrm{d}x &= \dfrac{1}{2}\int_{-\pi}^{\pi}\dfrac{\cos nx+\mathrm{i}\sin nx}{a-\mathrm{i}b\cos x}\mathrm{d}x\\ &= \dfrac{1}{2}\int_{\lvert z\rvert = 1}\dfrac{z^n}{a-\mathrm{i}b\dfrac{z+\frac{1}{z}}{2}}\dfrac{\mathrm{d}z}{\mathrm{i}z}\\ &= 2\pi\mathrm{i}\mathrm{Res}\dfrac{z^n}{bz^2+2\mathrm{i}az+b}\Bigg\vert_{z=\frac{\sqrt{a^2+b^2}-a}{b}\mathrm{i}}\\ &= \dfrac{\pi}{\sqrt{a^2+b^2}}\left(\dfrac{\sqrt{a^2+b^2}-a}{b}\right)^n\mathrm{i}^n \end{aligned} \)
换元,令 \(x=\sin^2 u\)
\( \begin{aligned} \int_0^{1}\left(1-x\right)^{n-1/2}\sin\left(a\sqrt{x}\right)\mathrm{d}x &= 2\int_0^{\pi/2}\cos^{2n}u\sin\left(a\sin u\right)\sin u\mathrm{d}u\\ &= -\dfrac{2}{a}\int_0^{\pi/2}\cos^{2n-1}u\sin u\mathrm{d}\left(\cos\left(a\sin u\right)\right)\\ &= \dfrac{2}{a}\int_0^{\pi/2}\cos\left(a\sin u\right)\left(2n\cos^{2n}u-\left(2n-1\right)\cos^{2n-2}u\right)\mathrm{d}u \end{aligned} \)
贝塞尔函数
满足递推关系
由
\(
\begin{aligned}
\int_0^{\pi/2}\cos\left(a\sin u\right)\cos^{2n}u\mathrm{d}u &= \sum\limits_{m=0}^{\infty}\dfrac{\left(-1\right)^m}{\left(2m\right)!}a^{2m}\int_0^{\pi/2}\sin^{2m}u\cos^{2n}u\mathrm{d}u\\
&= \dfrac{1}{2}\sum\limits_{m=0}^{\infty}\dfrac{\left(-1\right)^m}{\left(2m\right)!}\dfrac{\Gamma\left(n+1/2\right)\Gamma\left(m+1/2\right)}{\Gamma\left(n+m+1\right)}a^{2m}\\
&= \dfrac{\sqrt{\pi}2^{n-1}\Gamma\left(n+1/2\right)}{a^{n}}J_n\left(a\right)
\end{aligned}
\)
代入得
\( \begin{aligned} \int_0^{1}\left(1-x\right)^{n-1/2}\sin\left(a\sqrt{x}\right)\mathrm{d}x &= \dfrac{1}{a}\left(2n\dfrac{\sqrt{\pi}2^n\Gamma\left(n+\dfrac{1}{2}\right)}{a^n}J_n\left(a\right)-\left(2n-1\right)\dfrac{\sqrt{\pi}2^{n-1}\Gamma\left(n-\dfrac{1}{2}\right)}{a^{n-1}}J_{n-1}\left(a\right)\right)\\ &= \dfrac{1}{a}\dfrac{\sqrt{\pi}2^n\Gamma\left(n+\dfrac{1}{2}\right)}{a^{n-1}}\left(\dfrac{2n}{a}J_n\left(a\right)-J_{n-1}\left(a\right)\right)\\ &= \dfrac{\sqrt{\pi}2^n\Gamma\left(n+\dfrac{1}{2}\right)}{a^n}J_{n+1}\left(a\right) \end{aligned} \)
\(n\) 取整数时,结果可化为更简单的形式(假设 \(\left(-1\right)!!=1\))