γとπが出てくる定積分
γとπが出てくる定積分
\[ \int_{0}^{\infty}e^{-x}\log^{2}\left(x\right)dx=? \]
\[ \int_{0}^{\infty}e^{-x}\log^{2}\left(x\right)dx=? \]
\begin{align*}
\int_{0}^{\infty}e^{-x}\log^{2}\left(x\right)dx & =\left[\frac{d^{2}}{dt^{2}}\int_{0}^{\infty}e^{-x}x^{t}dx\right]_{t=0}\\
& =\left[\frac{d^{2}}{dt^{2}}\Gamma\left(t+1\right)\right]_{t=0}\\
& =\left[\frac{d}{dt}\Gamma\left(t+1\right)\frac{d}{dt}\log\left(\Gamma\left(t+1\right)\right)\right]_{t=0}\\
& =\left[\frac{d}{dt}\Gamma\left(t+1\right)\psi\left(t+1\right)\right]_{t=0}\\
& =\left[\psi\left(t+1\right)\frac{d}{dt}\Gamma\left(t+1\right)+\Gamma\left(t+1\right)\frac{d}{dt}\psi\left(t+1\right)\right]_{t=0}\\
& =\left[\psi\left(t+1\right)\Gamma\left(t+1\right)\psi\left(t+1\right)+\Gamma\left(t+1\right)\psi^{\left(1\right)}\left(t+1\right)\right]_{t=0}\\
& =\left[\Gamma\left(t+1\right)\left(\psi^{2}\left(t+1\right)+\psi^{\left(1\right)}\left(t+1\right)\right)\right]_{t=0}\\
& =\psi^{2}\left(1\right)+\psi^{\left(1\right)}\left(1\right)\\
& =\left(-\gamma\right)^{2}-\left(-1\right)^{1}1!\zeta\left(1+1\right)\\
& =\gamma^{2}+\zeta\left(2\right)\\
& =\gamma^{2}+\frac{\pi^{2}}{6}
\end{align*}
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