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I have this integral

$$\int_0^4\int_0^1\int_{2y}^2 \frac{4\cos (x^2)}{2\sqrt{z}}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$

I tried to change the order of integration in the following way

$$\int_0^1\int_{2y}^2\int_0^4 \frac{4\cos (x^2)}{2\sqrt{z}}\,\mathrm{d}z\,\mathrm{d}x\,\mathrm{d}y$$

After I integrated w.r.t $z$ I got

$$8\int_0^1\int_{2y}^2\cos(x^2)\,\mathrm{d}x\,\mathrm{d}y$$

And there is no integral from $\cos(x^2)$. Is there another another change of order of integration?

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    $\begingroup$ Do you know Fresnel Integrals and functions? $\endgroup$ – Martigan Jun 1 '15 at 11:00
  • $\begingroup$ No, I don't. I have this exercise for the Triple Integrals Topic $\endgroup$ – Tom Jun 1 '15 at 11:03
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The integral over $z$ is independent anyway, so your change of order was trivial. To change the $x$ and $y$ order, you will have to change the bounds of the integrals. You can sketch your domain on the $xy$ plane, you get a triangle. If you switch the variables, the same triangle is achieved by

$$\int_0^2\int_0^{x/2} \ldots {\rm d}y\,{\rm d}x$$

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