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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?

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Try eplacing (x,y) by (x,z) with z=y/x. –  Did Aug 23 '11 at 19:42
    
You might want to let $y/x=u$. This is natural, makes things look a little nicer. But soon you will want $u=v^2$. –  André Nicolas Aug 23 '11 at 19:47
    
With that replacement, it's necessary to evaluate $\int_1^2 \dfrac{e^z}{\sqrt z}\ dz$, but this brings about erfc again –  Daniel Aug 23 '11 at 19:50
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Yes, it is unavoidable in the long run. –  André Nicolas Aug 23 '11 at 19:56

2 Answers 2

up vote 8 down vote accepted

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.

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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

$$ \frac{3}{4} \sqrt{\pi } \left(\text{erfi}\left(\sqrt{2}\right)-\text{erfi}(1)\right) $$

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