Definite Integral to Infinity of Power of x by Exponential of -a x^2

Theorem

$\ds \int_0^\infty x^m e^{-a x^2} \rd x = \frac {\map \Gamma {\paren {m + 1}/2} } {2 a^{\paren {m + 1}/2} }$

where $m$ and $a$ are real numbers with $m > -1$ and $a > 0$.

Proof

\(\ds \int_0^\infty x^m e^{-a x^2} \rd x\)

\(=\)

\(\ds \int_0^\infty \frac 1 {2 \sqrt t} \paren {\sqrt t}^m e^{-a t} \rd t\)

substituting $t = x^2$

\(\ds \)

\(=\)

\(\ds \frac 1 2 \int_0^\infty t^{\paren {\paren {m + 1}/2} - 1} e^{-a t} \rd t\)

\(\ds \)

\(=\)

\(\ds \frac {\map \Gamma {\paren {m + 1}/2} } {2 a^{\paren {m + 1}/2} }\)

Laplace Transform of Real Power

$\blacksquare$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Exponential Functions: $15.77$