这道反常积分怎么求。

yigo

xiaoshuchong

挺经典的一个积分，跟$\Gamma$函数有很大关系。知乎和MSE上都能找到解答。

A. 来自知乎 https://zhuanlan.zhihu.com/p/114320236
\begin{eqnarray*}
\int_{0}^{\infty}e^{-t}\log tdt&=&\int_{0}^{1}e^{-t}\log tdt+\int_{1}^{\infty}e^{-t}\log tdt\\&=&I_{1}+I_{2}\\&=&\int_{0}^{1}\frac{e^{-x}-1}{x}dx+\int_{1}^{\infty}\frac{e^{-x}}{x}dx\\&=&\lim_{n\rightarrow\infty}\left[\int_{0}^{1}\frac{\left(1-\frac{x}{n}\right)^{n}-1}{x}dx+\int_{1}^{n}\frac{\left(1-\frac{x}{n}\right)^{n}}{x}dx\right]\\&=&\lim_{n\rightarrow\infty}\left[\int_{0}^{n}\frac{\left(1-\frac{x}{n}\right)^{n}-1}{x}dx+\int_{1}^{n}\frac{1}{x}dx\right]\\&=&\lim_{n\rightarrow\infty}\left[M+\int_{1}^{n}\frac{1}{x}dx\right]\\&=&\lim_{n\rightarrow\infty}\left[-\sum_{k=1}^{n}\frac{1}{k}+\log\left(n\right)\right]\\&=&-\gamma\\t&=&1-\frac{x}{n},dx=-ndt\\M&=&\int_{0}^{n}\frac{\left(1-\frac{x}{n}\right)^{n}-1}{x}dx\\&=&-\int_{0}^{1}\frac{1-t^{n}}{1-t}dt\\&=&-\int_{0}^{1}\sum_{k=1}^{n}t^{k-1}dt\\&=&-\sum_{k=1}^{n}\frac{1}{k}
\end{eqnarray*}

B. 来自MSE https://math.stackexchange.com/q ... -the-gamma-function
1. 与$\Gamma$函数的关联
\begin{eqnarray*}
\Gamma\left(x\right)&=&\int_{0}^{\infty}e^{-t}t^{x-1}dt\\\Gamma^{'}\left(x\right)&=&\int_{0}^{\infty}e^{-t}t^{x-1}\log tdt\\\Gamma^{'}\left(1\right)&=&\int_{0}^{\infty}e^{-t}\log tdt=-\gamma
\end{eqnarray*}
2.  $\Gamma^{'}(1)$的求解
\begin{eqnarray*}
\Gamma(z+1)&=&e^{-\gamma z}\cdot\prod_{n=1}^{\infty}\left(1+\frac{z}{n}\right)^{-1}e^{\frac{z}{n}}\\\log\left(\Gamma(z+1)\right)&=&-\gamma z+\sum_{n=1}^{\infty}\left[\frac{z}{n}-\log\left(1+\frac{z}{n}\right)\right]\\\frac{d}{dz}\log\left(\Gamma(z+1)\right)&=&\frac{\Gamma^{'}(z+1)}{\Gamma(z+1)}\\&=&-\gamma+\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{z+n}\right)\\\Gamma^{'}(1)&=&-\gamma
\end{eqnarray*}