# Computing a double integral: $\int _0^{\frac{1}{4}}\int _{\sqrt{x}}^{\frac{1}{2}}\:\frac{\cos\left(\pi y\right)}{y^2}~\mathrm dy~\mathrm dx$

I've been trying to solve this integral for the past few hours but to no avail:

$$\int _0^{\frac{1}{4}}\int _{\sqrt{x}}^{\frac{1}{2}}\:\frac{\cos\left(\pi y\right)}{y^2}~\mathrm dy~\mathrm dx$$

I have attempted integration by parts but it doesn't actually help me.

Wolfram alpha works it out to be $$\frac{1}{\pi}$$, but even knowing the answer isn't helping me at all.

Any hints/tips would be greatly appreciated.

consider the bounds of integration in the original integral you're trying to solve:

$$0 \le x \le \frac{1}{4}$$ $$\sqrt{x} \le y \le \frac{1}{2}$$

this is the following region:

the region is bounded by $$y=\sqrt{x}$$ or $$x=y^2$$ and $$y= 1/2$$ with $$x$$ ranging from $$0$$ to $$\frac{1}{4}$$. We can rewrite this as the region with $$x$$ ranging from $$0$$ to $$y^2$$ with the y-values ranging from $$0$$ to $$\frac{1}{2}$$. So

$$0\le y \le \frac{1}{2}$$ $$0 \le x \le y^2$$

So the integral becomes: $$\int_{y=0}^{y=\frac{1}{2}}\int_{x=0}^{x=y^2}\frac{\cos(\pi y)}{y^2} \mathrm dx \mathrm d y$$

Hint: Change the arrangement of variables and write it as $$\int_0^{\frac12}\int_0^{y^2}\dfrac{\cos\pi y}{y^2}\ dx \ dy$$

• Thanks, will give this a go now – user606649 Oct 21 '18 at 13:38
• OK, good luck!. – Nosrati Oct 21 '18 at 13:39
• Just comes out straight away from that, thanks so much! – user606649 Oct 21 '18 at 13:40