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

Theorem

Let $\ln t$ denote the natural logarithm function for real $t > 0$.

Let $e^{-t}$ denote the real exponential.

Then:

$\ds \int_{0^+}^{\mathop \to +\infty} \ln t \, e^{-t} \rd t = - \gamma$

where the left hand side is an improper integral, and $\gamma$ is the Euler-Mascheroni Constant.

Proof

\(\ds \int_{0^+}^{ \mathop \to +\infty} \ln t \, e^{-t} \rd t\)

\(=\)

\(\ds \int_{0^+}^{ \mathop \to +\infty} t^{1 - 1} \ln t \, e^{-t} \rd t\)

\(\ds \)

\(=\)

\(\ds \map {\Gamma'} 1\)

Derivative of Gamma Function

\(\ds \)

\(=\)

\(\ds -\gamma\)

Derivative of Gamma Function at $1$

$\blacksquare$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Logarithmic Functions: $15.99$