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题:试证\(\,\displaystyle\int_0^{\infty}e^{-x}\ln x\,\text{d}x=-\gamma\)
提示:令\(\,\Psi(x)=\large\frac{\Gamma\ '(x)}{\Gamma(x)},\) 则
\(\begin{align}
{\int_{0}^{\infty}e^{-x}\ln x \,d x}&=
\lim_{\mu \to 0^+}\frac{d}{d\mu}\int_{0}^{\infty}x^{\mu} e^{-x}\,d x=\lim_{\mu \to 0^+}\frac{d}{d\mu}{\Gamma(\mu + 1)}\\
&=\lim_{\mu \to 0^+}\Gamma(\mu + 1)\Psi(\mu + 1)=\Gamma(1)\Psi(1)=-
\color{red}{\large \gamma}\end{align}\) |
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