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As we known, the gamma integral $$ \Gamma (\alpha) = \int_0^\infty x^{\alpha-1} e^{-x} dx. $$ Now, I would like to know how to reduce the following integral to this gamma integral (or give a concerned expression which allows us to calculate its value), $$ I(\alpha) = \int_0^\infty x^{\alpha-1} e^{-x} \ln x dx. $$ Any comment?

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Since, when differentiating $x^{\alpha-1}$ with respect to $\alpha$, you get $x^{\alpha-1}\log x$, what you have got is the derivative of $\Gamma(\alpha)$. It is connected with the polygamma function, which is a special function. In your case $$ I(\alpha)=\Gamma'(\alpha)=\Gamma(\alpha)\psi^{(0)}(\alpha). $$

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  • $\begingroup$ Thanks for your answer, mickep. $\endgroup$ – Akai Shuichi Sep 15 '15 at 8:00
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The integral is the Laplace transform of $x^{\alpha-1}ln(x)$ which is : $$\Gamma(a)p^{-1}\left(\psi(a)-\ln(p)\right)$$ in the particular case $p=1$ $$I(\alpha)=\Gamma(a)\psi(a)$$ $\psi$ is the digamma function.

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