Laplace Transform of Reciprocal of Square Root

Theorem

$\forall t \in \R_{\ne 0}: \laptrans {\dfrac 1 {\sqrt t} } = \sqrt {\dfrac \pi s}$

where $\laptrans f$ denotes the Laplace transform of the real function $f$.

Proof

Let $\map f t = \dfrac 1 {\sqrt t}$.

By definition of the Laplace transform with a discontinuity at zero, $\laptrans f$ is the improper integral:

$\ds \laptrans f := \lim_{\epsilon \mathop \to 0^+} \int_\epsilon^{\to +\infty} \dfrac {e^{-s t} } {\sqrt t} \rd t$

if it exists.

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It follows that this improper integral converges.

So the Laplace transform $\laptrans f$ exists.

Hence:

$\blacksquare$

Sources

• 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: The Gamma Function: $32$