TITLE: Solving the Definite Integral $\int_0^{\infty} \frac{1}{t^{\frac{3}{2}}} e^{-\frac{a}{t}} \, \mathrm{erf}(\sqrt{t})\, \mathrm{d}t$
QUESTION [7 upvotes]: I would like to solve the following integral
$$\int_0^{\infty} \frac{1}{t^{\frac{3}{2}}} e^{-\frac{a}{t}} \, \mathrm{erf}(\sqrt{t})\, \mathrm{d}t$$
with Re$(a)>0$ and erf the error function.
Is it possible to given an closed form solution for this integral? Thank you.
Edit: Maybe this helps
$$\mathrm{L}(\mathrm{erf}(\sqrt{t}),s)=\frac{1}{s \, \sqrt{1+s}}$$
$$\mathrm{L}^{-1}(t^{-\frac{3}{2}} e^{-\frac{a}{t}})=\frac{1}{\sqrt{\pi \, a}}\mathrm{sin}(2 \sqrt{a \, s})$$
with L the Laplace transform.
Therefore it should be
$$\int_0^{\infty} \frac{1}{t^{\frac{3}{2}}} e^{-\frac{a}{t}} \, \mathrm{erf}(\sqrt{t})\, \mathrm{d}t = \int_0^{\infty} \frac{1}{s \, \sqrt{1+s} \, \sqrt{\pi \, a}} \, \mathrm{sin}(2 \sqrt{a \, s}) \mathrm{d}s$$

REPLY [5 votes]: You can also follow your Laplace approach. Define
$$
I(\alpha)=\int_0^{\infty}\frac{\sin(2\alpha\sqrt{s})}{s\sqrt{1+s}}ds
$$
now set $s=q^2$ 
$$
I(\alpha)=2\int_0^{\infty}\frac{\sin(2\alpha q)}{q\sqrt{1+q^2}}ds
$$
Now differentiate with respect to $\alpha$ 
$$
I'(\alpha)=4\int_0^{\infty}\frac{\cos(2\alpha q)}{\sqrt{1+q^2}}ds
$$
this integral now furnishs a representation of the modified Besselfunction $K_0(z)$ 
$$
I'(\alpha)=4 K_0(2\alpha )
$$
according to 10.43.2 backintegrating w.r.t. to $\alpha$ yields

$$
I(\alpha)=2\pi \alpha (K_0(2\alpha )L_{-1}(2\alpha )-K_1(2\alpha )L_{0}(2\alpha ))+C
$$

where $L_{\nu}(z)$ are modified Struve function. The constant of integration $C$ is fixed to be zero by the condition $I(0)=0$. Multiplying with $1/\sqrt{\pi}\alpha$ yields the result obtained by @Ron Gordon