Integral to Infinity of e^-t by Error Function of Root t

Theorem

$\ds \int_0^\infty e^{-t} \erf \sqrt t \rd t = \dfrac {\sqrt 2} 2$

where $\erf$ denotes the error function.

Proof

Using the technique of Evaluation of Integral using Laplace Transform:

\(\ds \int_0^\infty e^{-s t} \erf \sqrt t \rd t\)

\(=\)

\(\ds \dfrac 1 {s \sqrt {s + 1} }\)

Laplace Transform of Error Function of Root

\(\ds \leadsto \ \ \)

\(\ds \int_0^\infty \erf \sqrt t \rd t\)

\(=\)

\(\ds \dfrac 1 {1 \sqrt {1 + 1} }\)

letting $s \to 0^+$

\(\ds \)

\(=\)

\(\ds \dfrac 1 {\sqrt 2}\)

simplifying

\(\ds \)

\(=\)

\(\ds \dfrac {\sqrt 2} 2\)

$\blacksquare$

Sources

• 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: Evaluation of Integrals: $46 \ \text{(b)}$