A question on laplace transform

Question

I try to solve the following question on Laplace transform

$$L(\{ \int_0^{t}e^{-x^2}\})$$

I solved as following: $$L(\{ \int_0^{t}e^{-x^2}\})=\frac{1}{s}L(\{e^{-x^2}\})=\frac{1}{s}L(\{\sum_0^{\infty}\frac{(-t)^{2n}}{n!}\})$$ So $$\sum_0^{\infty}\frac{(-1)^{n}(2n)!}{n!s^{2n+2}}.$$

Can you suggest another way to solve this Laplace transform?

Answer 1

Let us start with the Laplace transform of $e^{-x^2}$:

$$ \mathcal{L}(e^{-x^2})(t) = \int_{0}^{+\infty} e^{-x^2-xt}\,dx =e^{t^2/4}\int_{0}^{+\infty}e^{-(x+t/2)^2}\,dx = e^{t^2/4}\int_{t/2}^{+\infty}e^{-x^2}\,dx.$$ The Laplace transform of $\int_{0}^{x}e^{-u^2}\,du$ is easily related to the previous one:

$$ \mathcal{L}\left(\int_{0}^{x}e^{-u^2}\,du\right)(t) = \frac{e^{t^2/4}}{\color{red}{t}}\int_{t/2}^{+\infty} e^{-x^2}\,dx. $$