Catalan 常数的引入

2020-02-02 本文已影响0人洛玖言

$\displaystyle\int_0^1\dfrac{\ln(1+x^2)}{1+x^2}\text{d}x$

Sol:
令 $x=\tan\theta$
$\begin{aligned} \int_0^1\dfrac{\ln(1+x^2)}{1+x^2}\text{d}x=&\int_0^{\frac\pi4}\dfrac{\ln(1+\tan^2\theta)}{1+\tan^2\theta}\sec^2\theta\text{d}\theta\\ =&\int_0^{\frac\pi4}\ln(1+\tan^2\theta)\text{d}\theta\\ =&\int_0^{\frac\pi4}\ln(\sec^2\theta)\text{d}\theta\\ =&-2\int_0^{\frac\pi4}\ln\cos\theta\text{d}\theta\\ =&-2\left(-\dfrac\pi4\ln2+\dfrac12\text{G}\right)\\ =&\frac\pi2\ln2-\text{G} \end{aligned}$

这里最后的

$\begin{aligned} &\int_0^{\frac\pi4}\ln\cos x\text{d}x \end{aligned}$
大家可能不是很熟

$\begin{aligned} &\int_0^{\frac\pi4}\ln\cos x\text{d}x\\ =&\dfrac12\int_0^{\frac\pi2}\ln\cos\frac{y}{2}\text{d}y\\ =&\dfrac12\int_0^{\frac\pi2}\ln\left(\dfrac{1+\cos y}{2}\right)^{\frac12}\text{d}y\\ =&\dfrac12\cdot\dfrac{1}{2}\left[\int_0^{\frac\pi2}\ln(1+\cos y)\text{d}y-\int_0^{\frac\pi2}\ln2\text{d}y\right]\\ =&\dfrac14\left(-\dfrac{\pi}{2}\ln2+2\text{G}-\dfrac{\pi}{2}\ln2\right)\\ =&-\dfrac{\pi}{4}\ln2+\dfrac12\text{G} \end{aligned}$

这边此时又出现了一个

$\displaystyle\int_0^{\frac\pi2}\ln(1+\cos y)\text{d}y$
可能也见的不多

因为
$\tan\dfrac{x}{2}=\dfrac{\sin x}{1+\cos x}$

所以
$1+\cos x=\dfrac{\sin x}{\tan\dfrac{x}{2}}$

$\begin{aligned} &\int_0^{\frac\pi2}\ln(1+\cos x)\text{d}x\\ =&\int_0^{\frac\pi2}\ln\dfrac{\sin x}{\tan\dfrac{x}{2}}\text{d}x\\ =&\int_0^{\frac\pi2}\ln\sin x\text{d}x-\int_0^{\frac\pi2}\ln\tan\dfrac{x}{2}\text{d}x\\ =&\int_0^{\frac\pi2}\ln\sin x\text{d}x-2\int_0^{\frac\pi4}\ln\tan y\text{d}y\\ =&-\dfrac\pi2-2(-\text{G})\\ =&2\text{G}-\dfrac\pi2\ln2 \end{aligned}$

这边出现了一个

$\displaystyle\int_0^{\frac\pi4}\ln\tan\theta\text{d}\theta=-\text{G}$
其中 G 表示Catalan常数

令 $\tan x=t$ ，则
$x=\arctan t ,\text{d}x=\dfrac{\text{d}t}{1+t^2}$

$\begin{aligned} &\int_0^{\frac\pi4}\ln\tan x\text{d}x\\ =&\int_0^1\ln t\cdot\dfrac{\text{d}t}{1+t^2}\\ =&\ln t\cdot\arctan t\bigg|_0^1-\int_0^1\arctan x\cdot\dfrac{\text{d}t}{t}\\ =&-\int_0^1\dfrac{\arctan t}{t}\text{d}t\\ =&-\text{G} \end{aligned}$

好啦，到最后一步了！

$\displaystyle\int_0^1\dfrac{\arctan t}{t}\text{d}t=\text{G}$ 怎么来的呢？

首先了解一下 Catalan 常数的定义

$\displaystyle\text{G}=\sum_{n=0}^{\infty}(-1)^n\dfrac{1}{(2n+1)^2}$

近似值
$\text{G}=0.915965594\cdots$

$\arctan x=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7}+\cdots+(-1)^n\dfrac{x^{2n+1}}{2n+1}+\cdots$

把 $\arctan x$ 的展开式带入，并逐项积分，得到

$\begin{aligned} &\int_0^1\dfrac{\arctan x}{x}\text{d}x\\ =&\int_0^1\dfrac{1}{x}\sum_{n=0}^{\infty}(-1)^n\dfrac{1}{2n+1}x^{2n+1}\text{d}x\\ =&\int_0^1\sum_{n=0}^{\infty}(-1)^n\dfrac{1}{2n+1}x^{2n}\text{d}x \end{aligned}$

根据积分号与求和号可以互换的原理，得到

$\begin{aligned} &\int_0^1\dfrac{\arctan x}{x}\text{d}x\\ =&\int_0^1\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n}}{2x+1}\text{d}x\\ =&\sum_{n=0}^{\infty}(-1)^n\dfrac{1}{2n+1}\int_0^1x^{2n}\text{d}x\\ =&\sum_{n=0}^{\infty}(-1)^n\dfrac{1}{2n+1}\cdot\dfrac{1}{2n+1}(x^{2n+1})\bigg|_0^1\\ =&\sum_{n=0}^{\infty}(-1)^n\dfrac{1}{(2n+1)^2}\\ =&\text{G} \end{aligned}$

Catalan 常数的引入

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