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 |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Logarithmic Functions: $15.100$