Evaluating the integral $\displaystyle\int_1^2 \int_x^{2x} \sqrt{\dfrac xy} e^{\tfrac yx}\; dy\; dx$ I'm trying to evaluate the integral $\displaystyle\int_1^2 \int_x^{2x} \sqrt{\dfrac xy} e^{\tfrac yx}\ dy\ dx$. Direct integration involves a non-elementary function (erfc), so a change of variables is necessary. However, I can't figure out any useful one. Any suggestions?
 A: Let $u = y/x$ in the inner integral. You get
$$ \int_1^2 \int_1^2 u^{-1/2} e^u \: x \: du \: dx $$
and this factors nicely, giving
$$ \int_1^2 x \: dx \int_1^2 u^{-1/2} e^u \: du.$$
The first factor is $3/2$. Now let $u = v^2$ in the second integral; then the second integral is
$$ \int_1^{\sqrt{2}} e^{v^2} \: dv $$
which can be written in terms of the imaginary error function.
A: First perform integration with respect to $y$ variable. To this end, perform the change of variabels $ \sqrt{2 y/x} = t$, i.e. $\frac{2/x}{\sqrt{2 (y/x)}} \mathrm{d} y = \mathrm{d}t$, which gives $\mathrm{d} y = \frac{x t}{2} \mathrm{d} t$. Now
$$
  \int_x^{2 x} \sqrt{\frac{x}{y}} \mathrm{e}^{\frac{y}{x}} \mathrm{d} y
  =
  \int_\sqrt{2}^{2} \frac{\sqrt{2}}{t} \cdot \mathrm{e}^{\frac{t^2}{2}} \cdot \frac{x t}{2} \, \mathrm{d} t = \frac{x}{\sqrt{2}} 
   \int_\sqrt{2}^{2}  \mathrm{e}^{\frac{t^2}{2}} \mathrm{d} t.
$$
Integration with respect to $x$ is now trivial: 
$$ \int_1^2 \frac{x}{\sqrt{2}} 
   \int_\sqrt{2}^{2}  \mathrm{e}^{\frac{t^2}{2}} \mathrm{d} t \mathrm{d} x = 
\left(\int_1^2 \frac{x}{\sqrt{2}} 
    \mathrm{d} x \right) \int_\sqrt{2}^{2}  \mathrm{e}^{\frac{t^2}{2}} \mathrm{d} t = 
 \frac{3}{2 \sqrt{2}} \int_\sqrt{2}^{2}  \mathrm{e}^{\frac{t^2}{2}} \mathrm{d} t.
$$
Now $ \int \mathrm{e}^{\frac{t^2}{2}} \mathrm{d} t = \sqrt{\frac{\pi}{2}}  \operatorname{erfi}(\frac{t}{\sqrt{2}})$. Hence the answer is 
$$
\frac{3}{4} \sqrt{\pi } \left(\text{erfi}\left(\sqrt{2}\right)-\text{erfi}(1)\right)
$$
