Integrate $\int t^{-\frac{1}{2}} e^{-\frac{c^2}{4t}} \, \mathrm{d}t$ I am trying to solve the following integral by hand, where $c$ is a constant:
$$\int t^{-\frac{1}{2}} e^{-\frac{c^2}{4t}} \, \mathrm{d}t$$
I have found a solution using Mathematica, but I was interested in how one might solve it manually.
At first I tried the substitution $t=v^2$, which almost yields the Error function, but I wasn't sure how to proceed further.
According to Mathematica, the correct solution is
$$2e^{\frac{-c^2}{4t}}\sqrt{t}+c\sqrt{\pi} \,\mathrm{erf}\!\left(\frac{c}{2\sqrt{t}} \right) $$ (plus an arbitrary constant).

Using the substitution suggested by Robert Israel quickly yields a solution:
Let $t=\cfrac{c^2}{4v^2}$.
Thus $\mathrm{d}t = -\cfrac{c^2}{2v^3}$ and $v=\cfrac{c}{2\sqrt{t}}$.
So, 
\begin{align}
\int t^{-\frac{1}{2}} e^{-\frac{c^2}{4t}} \, \mathrm{d}t &= \int \left(\frac{c^2}{4}v^{-2}\right)^{-\frac{1}{2}} e^{-\frac{c^2}{4}\left(\frac{c^2}{4}v^{-2}\right)^{-1}} \cdot \left(-\frac{c^2}{2}v^{-3}\right)\, \mathrm{d}v \\
&= -\int \left(\frac{2}{c}v\right) e^{-v^2} \cdot \left(\frac{c^2}{2}v^{-3}\right)\, \mathrm{d}v \\
&= -c \int v^{-2} e^{-v^2} \, \mathrm{d}v \\
\textrm{Now, integrate by parts:} \\
&= -c\left(-v^{-1}e^{-v^2} -\int -v^{-1} \left(-2ve^{-v^2} \right) \, \mathrm{d}v \right) + k \\
&= -c \left(-v^{-1}e^{-v^2} -2 \int e^{-v^2} \mathrm{d}v\right) + k\\
\textrm{Substitute } v=\cfrac{c}{2\sqrt{t}} \mathrm{:}\\
&= 2 \sqrt{t} e^{-\frac{c^2}{4t}} + c \sqrt{\pi} \; \mathrm{erf}\!\left(\frac{c}{2\sqrt{t}} \right) + k
\end{align}
Thus we find the desired solution.
