Stuck on evaluating this double integral

Question

I was going through some integral problems and I stumbled upon this.

$$\int_{0}^{1} \int_{y^{2}}^{1} \frac{\cosh(\pi x)}{ \sqrt{x} }\,dx\,dy$$

Initially I tried using the $u$ substitution $u$ = $\sqrt{x}$ when integrating wrt $x$, but I ended up with just $\int_{y = 0}^{y = 1} \int_{u=y}^{1} $$\ e^{\pi u^{2}} + e^{- \pi u^{2}} $$\,du\,dy$ which is even more difficult. I thought of using reverse product rule but found no luck with that either. Any thoughts?

Answer 1

If you reverse the order, you'll get$$\int_0^1\int_0^{\sqrt x}\frac{\cosh(\pi x)}{\sqrt x}\,\mathrm dy\,\mathrm dx.$$Can you take it from here?

Answer 2

It does help to change the order of integration. The region is bound by $x = y^2, x = 1$ and $y = 0$. So we can rewrite it as,

$\displaystyle \int_{0}^{1} \int_{0}^{\sqrt x} \frac{\cosh(\pi x)}{ \sqrt{x} }\,dy \,dx $

which is easier to evaluate.