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Logarithmic Gaussian integral

Let \gamma denote the Euler’s constant. Prove that:

    \[\int_0^\infty \log x e^{-x^2} \, \mathrm{d}x = -\frac{\sqrt{\pi}}{4} \left ( 2 \log 2 + \gamma \right )\]

Solution

First of all by making the substitution x \mapsto \sqrt{x} we get that

    \[\int_0^\infty \log x e^{-x^2} \, \mathrm{d}x = \frac{1}{4} \int_{0}^{\infty} \frac{\log x}{\sqrt{x}} \; e^{-x} \, \mathrm{d}x\]

For t>-1 let us consider the function

    \[F(t) = \int_{0}^{\infty} x^t e^{-x} \, \mathrm{d}x = \Gamma\left ( t+1 \right )\]

where \Gamma is the Euler’s Gamma function. Differentiating once we get:

    \begin{align*} \Gamma'(t+1) &=\frac{\mathrm{d} }{\mathrm{d} t} \int_{0}^{\infty} x^t e^{-x} \, \mathrm{d}t \\ &=\int_{0}^{\infty} \frac{\partial }{\partial t} x^t e^{-x}\; \mathrm{d}t \\ &=\int_{0}^{\infty} x^t \log x e^{-x} \; \mathrm{d}x \end{align*}

Thus, the desired integral is obtained by setting t=-\frac{1}{2}. Hence,

    \begin{align*} \frac{1}{4}\int_{0}^{\infty} \frac{\log x}{\sqrt{x}} \; e^{-x} \, \mathrm{d}x &= \frac{1}{4} \cdot \Gamma' \left ( 1-\frac{1}{2} \right )\\ &= \frac{1}{4} \cdot \Gamma' \left ( \frac{1}{2} \right )\\ &=\frac{1}{4} \cdot \psi^{(0)} \left ( \frac{1}{2} \right ) \Gamma \left ( \frac{1}{2} \right ) \\ &= \frac{1}{4} \cdot \left ( -2\ln 2 -\gamma \right ) \cdot \sqrt{\pi} \\ &= -\frac{\sqrt{\pi}}{4} \left ( 2\ln 2+\gamma \right ) \end{align*}

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