Is it possible to resolve this integral using integral contour? What should be the contour? \begin{equation} \int_0^\infty \log(x) e^{-x} dx = -\gamma \approx -0.577216 \end{equation} where $\gamma$ is the Euler‐Mascheroni constant.
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Why are you asking for a complex-analytic proof? By differentiation under the integral sign, your integral is trivially: $$\Gamma'(1) = \Gamma(1)\psi(1) = \psi(1) = -\gamma.$$ – Jack D'Aurizio Jun 03 '15 at 11:50
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@JackD'Aurizio and why is $\psi(1) = -\gamma$? – Ant Jun 03 '15 at 11:55
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@Ant: because for any natural number $n$, $\psi(n) = H_{n-1}-\gamma$. – Jack D'Aurizio Jun 03 '15 at 11:58
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@JackD'Aurizio Ahah okay! But it seems to me like begging the question, how do you prove your last equality? If you're going to assume that, you may just as well assume that we know the value of the integral :-) – Ant Jun 03 '15 at 12:00
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A related question. – Lucian Jun 03 '15 at 15:12
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I am studing integral contour. My question is to learn how to think when we have logarithm and exponential. How to threat the poles/branches/singularities and what should be the contour. – Mauro Jun 04 '15 at 07:42
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We have: $$ \int_{0}^{+\infty} x^{\alpha}e^{-x}\,dx=\Gamma(\alpha+1) \tag{1}$$ and: $$\begin{eqnarray*}\frac{d}{d\alpha}\int_{0}^{+\infty} x^{\alpha}e^{-x}\,dx &=& \int_{0}^{+\infty}x^{\alpha} \log(x)\, e^{-x}\,dx = \Gamma'(\alpha+1) \\[0.2cm]&=& \Gamma(\alpha+1)\cdot\frac{d}{d\alpha}\log\Gamma(\alpha+1)\\[0.2cm] &=& \Gamma(\alpha+1)\cdot\psi(\alpha+1)\tag{2}\end{eqnarray*}$$ by the definition of $\Gamma$ and $\psi$, so $\int_{0}^{+\infty}\log(x)\,e^{-x}\,dx$ equals $\psi(1)$.
On the other hand, the Weierstrass product: $$ \Gamma(z+1) = e^{-\gamma z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{\frac{z}{n}}\tag{3}$$ gives, through logarithmic differentiation, $$ \psi(z+1) = -\gamma+\sum_{n\geq 1}\frac{z}{n(n+z)}\tag{4}$$ so $\color{red}{\psi(1)=-\gamma}$ follows.
Jack D'Aurizio
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