Solving the Definite Integral $\int_0^{\infty} \frac{1}{t^{\frac{3}{2}}} e^{-\frac{a}{t}} \, \mathrm{erf}(\sqrt{t})\, \mathrm{d}t$ 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
$$\int_0^{\infty} \frac{1}{t^{\frac{3}{2}}} e^{-\frac{a}{t}} \, \mathrm{erf}(\sqrt{t})\, \mathrm{d}t = \int_0^{\infty} \frac{1}{s \, \sqrt{1+s} \, \sqrt{\pi \, a}} \, \mathrm{sin}(2 \sqrt{a \, s}) \mathrm{d}s$$
 A: Represent the erf as an integral and work a substitution.  To wit, the integral is
$$\frac{2}{\sqrt{\pi}} \int_0^1 dv \, \int_0^{\infty} \frac{dt}{t} e^{-(a/t+v^2 t)} $$
To evaluate the inner integral, we sub $y=a/t+v^2 t$.  Then the reader can show that
$$\int_0^{\infty} \frac{dt}{t} e^{-(a/t+v^2 t)} = 2 \int_{2 v \sqrt{a}}^{\infty} \frac{dy}{\sqrt{y^2-4 a v^2}} e^{-y}$$
The latter integral is easily evaluated using the sub $y=2 v \sqrt{a} \cosh{w} $ and is equal to
$$2 \int_{2 v \sqrt{a}}^{\infty} \frac{dy}{\sqrt{y^2-4 a v^2}} e^{-y} = 2 \int_0^{\infty} dw \, e^{-2 v \sqrt{a} \cosh{w}} = 2 K_0 \left ( 2 v \sqrt{a} \right )$$
where $K_0$ is the modified Bessel function of the second kind of zeroth order.  Now we integrate this expression with respect to $v$ and multiply by the factors outside the integral to get the final result:

$$\begin{align} \int_0^{\infty} dt \, t^{-3/2} e^{-a/t} \operatorname{erf}{\left ( \sqrt{t} \right  )} &= \frac{4}{\sqrt{\pi}} \int_0^1 dv \, K_0 \left ( 2 v \sqrt{a} \right ) \\ &= 2 \sqrt{\pi} \left [K_0 \left ( 2 \sqrt{a} \right ) \mathbf{L}_{-1}\left ( 2 \sqrt{a} \right ) + K_1 \left ( 2 \sqrt{a} \right ) \mathbf{L}_{0}\left ( 2 \sqrt{a} \right ) \right ] \end{align}$$

where $\mathbf{L}$ is a Struve function.
A: 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
A: Meh, interesting integral! I can give an heuristic approach but I believe someone else will do better. I'm on the bus and you know, it's not easy.
I would use Taylor Series for $e^{-a/t}$, hence
$$\int_0^{+\infty}\sum_{k = 0}^{+\infty} \frac{\left(-a/t\right)^k}{k!}t^{-3/2}\ \text{Erf}(\sqrt{t})\ \text{d}t$$
And we get
$$\sum_{k = 0}^{+\infty}\frac{(-a)^k}{k!}\int_0^{+\infty} t^{-k - 3/2}\ \text{Erf}(\sqrt{t})\ \text{d}t$$
Now if we call for simplicity $b = -k - 3/2$ we obtain a computable integral (I checked on Mathematica), which says:
$$\int_0^{+\infty} t^{b}\ \text{Erf}(\sqrt{t})\ \text{d}t = -\frac{\Gamma[3/2 + b]}{(1 + b)\sqrt{\pi}} ~~~ \to ~~~ -\frac{\Gamma[-k]}{(-k - 1/2)\sqrt{\pi}}$$
BUT there is condition over this result:
$$-\frac{3}{2} < \Re(b) < -1$$
This would give then
$$-\ \sum_{k = 0}^{+\infty}\frac{(-a)^k}{k!}\frac{\Gamma[-k]}{(-k - 1/2)\sqrt{\pi}}$$
And here I do stop because I cannot go on (mostly because I have no paper and pencils with me.. I'll check again when I'll be at home).
