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: enter image description here

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 $$\int_0^{\frac12}\int_0^{y^2}\dfrac{\cos\pi y}{y^2}\ dx \ dy$$

  • $\begingroup$ Thanks, will give this a go now $\endgroup$ – user606649 Oct 21 '18 at 13:38
  • $\begingroup$ OK, good luck!. $\endgroup$ – Nosrati Oct 21 '18 at 13:39
  • 1
    $\begingroup$ Just comes out straight away from that, thanks so much! $\endgroup$ – user606649 Oct 21 '18 at 13:40

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