Inverse Laplace transform of $\frac1{\sqrt t}\operatorname{erf}\sqrt{t}$

Question

I'm trying to show that $$\mathcal{L}\left\{\frac{1}{\sqrt{t}}\operatorname{erf}\sqrt{t}\right\}= \frac{2}{\sqrt{\pi s}}\arctan\frac{1}{\sqrt {s}}$$ but I am still stuck in finding the Laplace transform of the error function of $\sqrt{t}$. Is there anyone who can help me?

Answer 1

I'm going to assume the version in your title is correct. Suppose we have

$$\mathcal{L}\left\{\frac{1}{\sqrt{t}}\text{erf}(\sqrt{t})\right\} = \int_0^\infty \frac{1}{\sqrt{t}}\text{erf}(\sqrt{t}) e^{-st}dt$$

Then substituting $t=x^2$

$$\int_0^\infty \frac{1}{\sqrt{t}}\text{erf}(\sqrt{t}) e^{-st}dt = 2\int_0^\infty \text{erf}(x)e^{-sx^2}dx$$

By definition of error function

$$2\int_0^\infty \text{erf}(x)e^{-sx^2}dx = \frac{4}{\sqrt{\pi}}\int_0^\infty \int_0^x e^{-y^2}e^{-sx^2}dydx$$

Converting to polar coordinates

$$\frac{4}{\sqrt{\pi}}\int_0^\infty \int_0^x e^{-y^2}e^{-sx^2}dydx = \frac{4}{\sqrt{\pi}}\int_0^{\frac{\pi}{4}} \int_0^\infty re^{-r^2(\sin^2\theta + s\cos^2\theta)}drd\theta$$

$$= \frac{2}{\sqrt{\pi}}\int_0^{\frac{\pi}{4}} \frac{1}{\sin^2\theta+s\cos^2\theta}d\theta = \frac{2}{\sqrt{\pi}}\int_0^{\frac{\pi}{4}} \frac{\sec^2\theta}{\tan^2\theta+s}d\theta$$

$$= \frac{2}{\sqrt{s\pi}} \tan^{-1}\left(\frac{\tan\theta}{\sqrt{s}}\right)\Biggr|_0^{\frac{\pi}{4}} = \boxed{\frac{2}{\sqrt{s\pi}} \tan^{-1}\left(\frac{1}{\sqrt{s}}\right)}$$

Answer 2

Thank you for your answer Minad but there are some points that made me confused in your solution: If we let $$x=r\cos\theta$$ and $$y=r\sin\theta$$ so $$dx=\cos\theta dr$$ $$dy=r\cos\theta d\theta$$ => $$dx dy=r\cos^2\theta dr d\theta$$ Converting to polar coordinates, it should be $$\frac{4}{\sqrt{\pi}}\int_0^\infty \int_0^x e^{-y^2}e^{-sx^2}dydx = \frac{4}{\sqrt{\pi}}\int_0^{\frac{\pi}{4}} \int_0^\infty r\cos^2\theta e^{-r^2(\sin^2\theta + s\cos^2\theta)}drd\theta$$