$$ \int _0^{\sqrt{\pi}} \int_y^{\sqrt{\pi}} \sin (y^2 )\; dydx$$ Even if I change the order of integration I don't see how to get rid of this $\sin (x^2)$ which doesn't have antiderivate. It is possible to evaluate it without using any aproximation method?
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Is the integrand $\sin^2{x}$? – Ron Gordon Nov 18 '14 at 13:23
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In the notation $dy, dx$ we integrate w.r.t. y first, right? Some books say the opposite. So just clarifying. Here it seems you'd integrate w.r.t. x first. – Swapnil Tripathi Nov 18 '14 at 13:25
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Sorry, I've made a correction. – Nov 18 '14 at 13:30
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@michaelaba: what do you have to say about what I asked above? Am i right? – Swapnil Tripathi Nov 18 '14 at 13:31
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Yes, it is y first – Nov 18 '14 at 13:33
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1@michaelaba It can't be so IMO. If you integrate w.r.t $y$ first and put $y$ as a limit again, you're never going to get rid of $y$ in your final answer. I think we should integrate w.r.t. $x$ first. Check the notation. It differs from book to book. – Swapnil Tripathi Nov 18 '14 at 13:35
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The integral in the question makes no sense because the ordering of $\mathrm dy$ and $\mathrm dx$ is wrong. Instead it probably ought to read $$ I=\int _0^{\sqrt{\pi}} \int_x^{\sqrt{\pi}} \sin (y^2 )\; \mathrm dy\mathrm dx,$$ that is, $$I=\int _0^{\sqrt{\pi}} G(x)\;\mathrm dx,\qquad G(x)=\int_x^{\sqrt{\pi}} \sin (y^2 )\; \mathrm dy.$$ Then, exchanging the order of integration yields $$I =\int _0^{\sqrt{\pi}}\sin (y^2 ) \left(\int_0^y \mathrm dx\right)\mathrm dy=\int _0^{\sqrt{\pi}}y\sin (y^2 )\; \mathrm dy=\left.-\tfrac12\cos(y^2)\right|_0^{\sqrt{\pi}}=1.$$
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