级数求定积分一例

2020-02-06 本文已影响0人洛玖言

$\displaystyle\int_0^1\ln(1+x)\ln x\text{d}x=2-2\ln2-\dfrac{\pi^2}{12}$

群里有小伙伴问的，这个算是比较明显用级数解

首先

$\displaystyle\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{x^n}{n}$

其次

$\displaystyle\int_0^1x^k\ln x\text{d}x=\dfrac{1}{1+k}x^{k+1}\ln x\bigg|_0^1-\dfrac{1}{1+k}\int_0^1x^k\text{d}x=-\dfrac{1}{(1+k)^2}$

$\begin{aligned} &\int_0^1\ln(1+x)\ln x\text{d}x\\ =&\int_0^1\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{x^n}{n}\ln x\text{d}x\\ =&\sum_{n=1}^{\infty}(-1)^{n-1}\int_0^1\dfrac{x^n}{n}\ln x\text{d}x\\ =&\sum_{n=1}^{\infty}(-1)^{n}\dfrac{1}{n(n+1)^2}\\ =&\sum_{n=1}^{\infty}(-1)^{n}\left(\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)^2}\right)\\ =&\sum_{n=1}^{\infty}(-1)^{n}\dfrac{1}{n}+\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{1}{n+1}+1-\dfrac{\pi^2}{12}\\ =&\sum_{n=1}^{\infty}(-1)^{n}\dfrac{1}{n}+1+\sum_{n=1}^{\infty}(-1)^{n}\dfrac{1}{n}+1-\dfrac{\pi^2}{12} \\ =&2-2\ln2-\dfrac{\pi^2}{12} \end{aligned}$

感觉简书的markdown总是有莫名其妙的bug

级数求定积分一例

猜你喜欢

热点阅读