Laplace transform of the error function

Example

Let $\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-u^2}\, du$ be the error function. The Laplace transform of $f(t) = \operatorname{erf}(\sqrt{t})$ is $$\mathcal{L}\{\operatorname{erf}(\sqrt{t})\}(s) = \frac{1}{s \sqrt{s + 1}} \quad (s > 0).$$ Check: $f(0) = 0$ and, for $t > 0$, $f'(t) = \frac{2}{\sqrt{\pi}} e^{-t} \cdot \frac{1}{2\sqrt{t}} = \frac{e^{-t}}{\sqrt{\pi t}}$. By the transform of a power with $\alpha = -1/2$ and the shift $s \mapsto s + 1$ caused by the factor $e^{-t}$, $$\mathcal{L}\{f'\}(s) = \frac{1}{\sqrt{\pi}} \cdot \frac{\Gamma(1/2)}{\sqrt{s + 1}} = \frac{1}{\sqrt{s + 1}},$$ using $\Gamma(1/2) = \sqrt{\pi}$ for the gamma function. Since $f(t) = \int_0^t f'(u)\, du$, its transform is that of $f'$ divided by $s$, which gives $1/(s\sqrt{s + 1})$. The formula extends to $\operatorname{Re} s > 0$ with the principal square root.