Definite Integral to Infinity of Exponential of -x^2 by Logarithm of x

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Theorem

$\ds \int_0^\infty e^{-x^2} \ln x \rd x = -\frac {\sqrt \pi} 4 \paren {\gamma + 2 \ln 2}$

where $\gamma$ denotes the Euler-Mascheroni constant.


Proof

Consider the integral:

$\ds \int_0^\infty x^t e^{-x^2} \rd x$

for positive real parameter $t$.

Using Definite Integral to Infinity of $x^m e^{-a x^2}$, we have:

$\ds \int_0^\infty x^t e^{-x^2} \rd x = \frac 1 2 \map \Gamma {\frac {1 + t} 2}$

Differentiating the left hand side with respect to $t$ we have:

\(\ds \frac \d {\d t} \int_0^\infty t^x e^{-x^2} \rd t\) \(=\) \(\ds \int_0^\infty \frac \partial {\partial t} \paren {x^t e^{-x^2} } \rd x\) Definite Integral of Partial Derivative
\(\ds \) \(=\) \(\ds \int_0^\infty x^t e^{-x^2} \ln x \rd x\) Derivative of Power of Constant

Differentiating the right hand side using the Chain Rule for Derivatives gives:

$\ds \int_0^\infty x^t e^{-x^2} \ln x \rd x = \frac 1 4 \map {\Gamma'} {\frac {1 + t} 2}$

We therefore have:

\(\ds \int_0^\infty e^{-x^2} \ln x \rd x\) \(=\) \(\ds \int_0^\infty x^0 e^{-x^2} \ln x \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 4 \map {\Gamma'} {\frac 1 2}\)
\(\ds \) \(=\) \(\ds \frac 1 4 \map \Gamma {\frac 1 2} \map \psi {\frac 1 2}\) Definition of Digamma Function
\(\ds \) \(=\) \(\ds -\frac {\sqrt \pi} 4 \paren {\gamma + 2 \ln 2}\) Gamma Function of One Half, Digamma Function of One Half

$\blacksquare$


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