求证一道极限

dim

\[
\lim_{n \to \infty} \left( n \int_0^1 (1-x)^n (\ln x)^2 \,dx - (\ln n)^2 - 2\gamma \ln n \right) = \gamma^2 + \frac{\pi^2}{6}
\]

战巡

回复 1# dim


首先把那个积分搞掉
考虑这个东西
\[B(m,n+1)=\int_0^1x^{m-1}(1-x)^ndx\]
\[\frac{\partial^2}{\partial m^2}B(m,n+1)=\int_0^1x^{m-1}(1-x)^n\ln^2(x)dx\]
\[\int_0^1(1-x)^n\ln^2(x)dx=\frac{\partial^2}{\partial m^2}B(m,n+1)|_{m=1}\]
另一方面
\[\frac{\partial^2}{\partial m^2}B(m,n+1)=\frac{\partial^2}{\partial m^2}\frac{\Gamma(m)\Gamma(n+1)}{\Gamma(m+n+1)}\]
\[=\frac{\Gamma(m)\Gamma(n+1)}{\Gamma(m+n+1)}·[(\psi(m)-\psi(m+n+1))^2+(\psi'(m)-\psi'(m+n+1))]\]
代入$m=1$可得
\[\int_0^1(1-x)^n\ln^2(x)dx=\frac{1}{n+1}[(\sum_{i=1}^{n+1}\frac{1}{i})^2+\frac{\pi^2}{6}-\psi'(n+2)]\]

\[\lim_{n\to \infty}[n\int_0^1(1-x)^n\ln^2(x)dx-\ln^2(n)-2\ln(n)\gamma]=\lim_{n\to\infty}[(\sum_{i=1}^{n+1}\frac{1}{i})^2+\frac{\pi^2}{6}-\psi'(n+2)-\ln^2(n)-2\ln(n)\gamma]\]
注意到$\lim_{n\to\infty}\psi'(n+2)=0$
\[=\lim_{n\to\infty}[(\ln(n)+\gamma)^2+\frac{\pi^2}{6}-\psi'(n+2)-\ln^2(n)-2\ln(n)\gamma]=\frac{\pi^2}{6}+\gamma^2\]

dim

回复 2# 战巡

懂了，非常感谢！