积分题目4

这篇博客用于记录一些比较有趣的积分问题。

求\(I=\displaystyle \int_{0}^{+\infty}\sin(x^2)dx\)

题目为著名的 Fersnel 积分，下面分步进行求解。

\[I\xlongequal{x^2=t}\frac{1}{2}\int_{0}^{+\infty}\frac{\sin t}{\sqrt{t}}dt\]

考虑积分\(\displaystyle \int_{0}^{+\infty}e^{-ax^2}dx\)，其中\(a>0\)：

\[\int_{0}^{\infty}e^{-ax^2}dx = \frac{1}{2} \sqrt{\frac{\pi}{a}} \int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}\cdot\frac{1}{\sqrt{2a}}}e^{-\frac{x^2}{2(\frac{1}{\sqrt{2a}})^2}}dx = \frac{1}{2} \sqrt{\frac{\pi}{a}} \Rightarrow\sqrt{\frac{1}{a}}=\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}e^{-ax^2}dx\]

则有：

\[I\xlongequal{x^2=t}\frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}\sin t (\int_{0}^{+\infty} e^{-tx^2}dx)dt=\frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}dx\int_{0}^{+\infty}e^{-x^2t}\sin t dt=\frac{1}{\sqrt{\pi}}J\]

上式最后一步直接使用了\(\mathcal{L}(\sin at)\)的结果，其中\(J=\displaystyle \int_{0}^{+\infty}\frac{1}{x^4+1}dx\)，下面求解\(J\)。

\[J \xlongequal{ x = \frac{1}{t} }\int_{0}^{+\infty}\frac{t^2}{1+t^4}dt=\int_{0}^{+\infty}\frac{x^2}{1+x^4}dx\]

\[J = \frac{1}{2}\int_{0}^{+\infty}\frac{1+x^2}{1+x^4}dx =\frac{1}{2} \int_{0}^{+\infty}\frac{d(x-\frac{1}{x})}{(x-\frac{1}{x})^2+2}=[\frac{1}{2\sqrt{2}}\arctan(\frac{x-\frac{1}{x}}{\sqrt{2}})]_{0}^{+\infty}=\frac{\pi}{2\sqrt{2}}\]

最终得到\(\displaystyle\int_{0}^{+\infty} \sin(x^2) dx = \displaystyle \frac{1}{2} \sqrt{\frac{\pi}{2}}\)。此外，\(\displaystyle\int_{0}^{+\infty} \sin(x^2) dx = \displaystyle\int_{0}^{+\infty} \cos(x^2) dx\)，这一性质根据\(\mathcal{L}(\sin at)\)和\(\mathcal{L}(\cos at)\)的关系显而易见。

求\(I = \displaystyle \int_{0}^{\frac{\pi}{2}}\ln \sin xdx\)

显然\(\displaystyle \int_{0}^{\frac{\pi}{2}}\ln \sin xdx =\int_{0}^{\frac{\pi}{2}}\ln \cos xdx\)，那么有：

\[I = \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\ln (\sin x \cos x)dx=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\ln \sin 2x dx - \frac{\pi}{4}\ln 2\]

\[\int_{0}^{\frac{\pi}{2}}\ln \sin 2x dx \xlongequal{2x = u} \frac{1}{2}\int_{0}^{\pi}\ln \sin udu = \frac{1}{2}(\int_{0}^{\frac{\pi}{2}}\ln \sin xdx + \int_{\frac{\pi}{2}}^{\pi}\ln \sin xdx)\]

\[\int_{\frac{\pi}{2}}^{\pi}\ln \sin xdx \xlongequal{u=\pi-x}\int_{0}^{\frac{\pi}{2}}\ln \sin udu\]

最终得到\(I = \displaystyle-\frac{\pi}{2}\ln 2\)

求\(I = \displaystyle\int_{0}^{\frac{\pi}{2}}\sin x \ln \sin x dx\)

首先求解被积函数的原函数：

\[\int\sin x \ln \sin x dx=\int\ln \sin xd(-\cos x)=-\cos x\ln \sin x + \int \frac{1-\sin^2x}{\sin x} \]

\[= (1-\cos x)\ln \sin x - \ln(1+\cos x)+\cos x+C\]

再代入积分限有：

\[I = \ln 2-1-[\lim_\limits{x\to 0}(1-\cos x)\ln \sin x]=\ln2 - 1\]

在此可以复习一下自适应 Simpson 积分的 C++ 实现。

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#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;

const double eps = 1e-9;

double f(double x) {
    return sin(x) * log(sin(x));
}

double simpson(double l, double r) {
    double mid = (l + r) / 2;
    return (r - l) * (f(l) + f(r) + 4 * f(mid)) / 6;
}

double asr(double l, double r, double area) {
    double mid = (l + r) / 2;
    double area_l = simpson(l, mid), area_r = simpson(mid, r);
    return fabs(area - area_l - area_r) < eps? (area_l + area_r):
           asr(l, mid, area_l) + asr(mid, r, area_r);
}

int main() {
    // l should approach 0 close enough but not equal 0
    double l = 1e-9, r = acos(-1) / 2;
    cout << fixed << setprecision(4) << asr(l, r, simpson(l, r)) << endl;
    return 0;   
}

求\(I = \displaystyle \int_{0}^{1}\frac{\ln x}{1-x^2}dx\)

\[I = \displaystyle \int_{0}^{1}\sum_{n=0}^{\infty}x^{2n}\ln xdx=\sum_{n=0}^{\infty}\int_{0}^{1}x^{2n}\ln xdx \xlongequal{\ln x = t}-\sum_{n=0}^{\infty}\int_{0}^{+\infty}te^{-(2n+1)t}dt=-\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}\]

由 Basel 级数易得\(\displaystyle\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{3}{4}\cdot\frac{\pi^2}{6}=\frac{\pi^2}{8}\)，那么\(I=\displaystyle -\frac{\pi^2}{8}\)。

求\(I = \displaystyle \int_{0}^\pi \frac{dx}{a+b\cos x}, a^2>b^2\)

\(\displaystyle \int_{0}^\pi \frac{dx}{a+b\cos x}=\int_{0}^{+\infty}\frac{\frac{2}{1+t^2}dt}{a+b\frac{1-t^2}{1+t^2}}=\frac{2}{a-b}\int_{0}^{+\infty}\frac{dt}{t^2+\frac{a+b}{a-b}}=\frac{2}{a-b}[\sqrt{\frac{a-b}{a+b}}\arctan \frac{x}{\sqrt{\frac{a-b}{a+b}}}]_{0}^{+\infty}=\frac{\pi}{\sqrt{a^2-b^2}}\)

\(\displaystyle \int_{0}^{+\infty}\frac{dx}{1+x^n}=\frac{\pi}{n\sin\frac{\pi}{n}}\)

求\(I = \displaystyle \int_{-\infty}^{+\infty}\frac{\cos x}{1+x^2}dx\)

费曼积分法

令\(y(a) = \displaystyle \int_{0}^{+\infty} \frac{\cos ax}{1+x^2}dx,a>0\)，则\(I = 2y(1)\)，并且有\(y(0) = \displaystyle\frac{\pi}{2}\)。

\[y'(a) = \int_{0}^{+\infty}(\frac{\partial}{\partial a}\frac{\cos ax}{1+x^2})dx = -\frac{\pi}{2}+\int_{0}^{+\infty}\frac{\sin ax}{x(1+x^2)}dx \Rightarrow y'(0) = -\frac{\pi}{2}\]

\[y''(a) = \int_{0}^{+\infty}[\frac{\partial}{\partial a}\frac{\sin ax}{x(1+x^2)}]dx = y(a) \Rightarrow y(a) = C_1e^{a}+C_2e^{-a}\]

求解常数\(C_1\)和\(C_2\)，得到\(y(a) = \displaystyle \frac{\pi}{2}e^{-a}\)，那么\(I = \displaystyle \frac{\pi}{e}\)。

留数法

注意到\(I = \displaystyle \int_{-\infty}^{+\infty}\frac{\text{Re}(e^{ix})}{1+x^2}dx = \text{Re}(\int_{-\infty}^{+\infty}\frac{e^{ix}}{1+x^2}dx)\)，那么有：

\[\int_{-\infty}^{+\infty}\frac{e^{ix}}{1+x^2}dx = 2\pi i \text{Res}[\frac{e^{ix}}{1+x^2},i]=\frac{\pi}{e} \Rightarrow I = \frac{\pi}{e}\]

Frullani积分

假设\(f(+\infty)\)存在且有限，那么：

\[\int_{0}^{+\infty}\frac{f(ax)-f(bx)}{x}dx=(f(0)-f(+\infty))\ln\frac{b}{a}\]