Solutions for $ \int_0^{\infty} \frac{1}{\sqrt{t^3\left(t+\tau\right)^3}}\exp\left[- \frac{2t+\tau}{t\left(t+\tau\right)}\right] \, dt $ I am trying to find an analytical solution to the above integral. The context is as follows:
I am interested in obtaining an expression for the autocovariance function of the change in groundwater level in an idealised aquifer following a unit pulse input. The governing equation for fluid motion in this idealised aquifer is given by:
\begin{equation}
\frac{\partial h}{\partial t} = \frac{T}{S}\frac{\partial^2 h}{\partial x^2}
\end{equation}
where $T$ and $S$ are characteristics of the aquifer system. The unit impulse-response function is given by:
\begin{equation}
h^{\delta}\left(x,\, t\right) = \frac{Sx^2}{4T}\left(t^*\right)^{-\frac{3}{2}}\mathrm{e}^{-1/t^*} \quad \mbox{where} \quad t^* = \frac{4Tt}{Sx^2}
\end{equation}
The autocovariance function, $\gamma\left(x,\,\tau\right)$ for the causal system is given by:
\begin{equation}
\gamma\left(x,\, \tau\right) = \int_0^{\infty} h^{\delta}\left(x,\,t\right)h^{\delta}\left(x,\,t+\tau\right) dt = \left(\frac{Sx^2}{4T}\right)^2 \int_0^{\infty} \left(t^*\right)^{-\frac{3}{2}}\mathrm{e}^{-1/t^*}\left(t^*+\tau\right)^{-\frac{3}{2}}\mathrm{e}^{-1/(t^*+\tau)}\,  dt^*
\end{equation}
I have noted that:
\begin{equation}
\int_0^{\infty} \left(t^*\right)^{-\frac{3}{2}}\mathrm{e}^{-1/t^*} \, dt^* = \sqrt{\pi} \quad \mbox{and} \quad \int_0^{\infty} \left(t^*+\tau\right)^{-\frac{3}{2}}\mathrm{e}^{-1/(t^*+\tau)} \, dt^* = \sqrt{\pi} \,\mathrm{erf}\left[\frac{1}{\tau}\right]
\end{equation}
but can get no further. Any assistance or thoughts would be much appreciated.
 A: Long Comment and Conjectured Result
With a little bit of work integral transforms to
$$I=\int_0^{\infty } \frac{1}{v^2 \sqrt{\left(\frac{\tau }{2}\right)^2+v^2}} \, \exp \left(-\frac{2 \sqrt{\left(\frac{\tau }{2}\right)^2+v^2}}{v^2}\right)dv \tag{1}$$
which using the trigonometric substitution $v=\frac{\tau}{2}\cot \theta$ remarkably becomes
$$I=\frac{4 }{\tau ^2}\int_0^{\frac{\pi }{2}} \tan (\theta ) \sec (\theta )\, \exp{\left(-\frac{4 \tan (\theta ) \sec (\theta )}{\tau }\right)} \, d\theta \tag{2}$$
Since $\tan (\theta ) \sec (\theta )\approx \theta$ to a very weak first order approximation we have
$$I\approx \frac{4 }{\tau ^2} \int_0^{\frac{\pi }{2}} \theta\,  \exp{\left(-\frac{4 \theta }{\tau }\right)} \, d\theta=\frac{\tau -e^{-\frac{2 \pi }{\tau }} (\tau +2 \pi )}{4 \tau }$$
which looks surprisingly good at first because function starts and ends in the right places, however unfortunately it is not a very accurate approximation in between. 
This at least gives you some ideas to play with.
Update 1
Using the substitution $x=\sin \theta$ and $\sqrt{1-x^2}=\cos \theta$ in (2) we get an integral between the limits of $0$ and $1$.
$$I=\frac{4 }{\tau ^2}\int_0^1 \frac{x \exp\left({-\frac{4 x}{\tau  \left(1-x^2\right)}}\right) }{\left(1-x^2\right)^{3/2}} \, dx \tag{3}$$
Update 2
Using Mathematica I can integrate another equivalent integral
$$I=\frac{4 }{\tau ^2}\int_0^{\infty } \left(\frac{i}{\sqrt{1+2 i w}}-\frac{i}{\sqrt{1-2 i w}}\right) e^{-\frac{4 w}{\tau }} \, dw\tag{4}$$
with the result
$$I=\frac{\left(\frac{1}{2}+\frac{i}{2}\right) \sqrt{\pi } e^{-\frac{2 i}{\tau }} \left(i\, e^{\frac{4 i}{\tau }} \left(-1+\text{erf}\left(\frac{1+i}{\sqrt{\tau }}\right)\right)+i \,\text{erfi}\left(\frac{1+i}{\sqrt{\tau }}\right)+1\right)}{\tau ^{3/2}}\tag{5}$$
for $\tau > 0$ and where $\text{erf}$ is the error function and $\text{erfi}$ is the imaginary error function,
and where according to Mathematica the differentiated inverse trigonometric function for $w=\tan (\theta ) \sec (\theta )$ is
$$\frac{\sin \left(\frac{1}{2} \tan ^{-1}(2 w )\right)}{w\sqrt[4]{4 w ^2+1}}= \frac{i}{w\sqrt{1+2 i w}}-\frac{i}{w\sqrt{1-2 i w}}$$
