Inverse Laplace transform of $\frac1{\sqrt t}\operatorname{erf}\sqrt{t}$ 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?
 A: 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
$$\frac{4}{\sqrt{\pi}}\int_0^\infty \int_0^x e^{-y^2}e^{-sx^2}dydx = \frac{4}{\sqrt{\pi}}\int_0^{\frac{\pi}{4}} \int_0^\infty re^{-r^2(\sin^2\theta + s\cos^2\theta)}drd\theta$$
$$= \frac{2}{\sqrt{\pi}}\int_0^{\frac{\pi}{4}} \frac{1}{\sin^2\theta+s\cos^2\theta}d\theta = \frac{2}{\sqrt{\pi}}\int_0^{\frac{\pi}{4}} \frac{\sec^2\theta}{\tan^2\theta+s}d\theta$$
$$= \frac{2}{\sqrt{s\pi}} \tan^{-1}\left(\frac{\tan\theta}{\sqrt{s}}\right)\Biggr|_0^{\frac{\pi}{4}} = \boxed{\frac{2}{\sqrt{s\pi}} \tan^{-1}\left(\frac{1}{\sqrt{s}}\right)}$$
A: 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 
$$\frac{4}{\sqrt{\pi}}\int_0^\infty \int_0^x e^{-y^2}e^{-sx^2}dydx = \frac{4}{\sqrt{\pi}}\int_0^{\frac{\pi}{4}} \int_0^\infty r\cos^2\theta e^{-r^2(\sin^2\theta + s\cos^2\theta)}drd\theta$$
