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Why is $\displaystyle{{1 \over 4}\int_{-1}^{1}\int_{-1}^{1} \mathrm{e}^{\mathrm{i}t\left(x + y\right)}\ \left[1 + xy\left(x^{2} - y^{2}\right)\right]\,\mathrm{d}x\,\mathrm{d}y = {\sin^{2}\left(t\right) \over t^{2}}}$ ?.

How do I solve this integral ?. The answer must be right and WolframAlpha gives the same solution.

But calculating $\displaystyle{\int_{-1}^{1}\mathrm{e}^{\mathrm{i}t\left(x + y\right)} \left[1 + xy\left(x^{2} - y^{2}\right)\right]\mathrm{d}x}$ first will probably give my a really complicated term, so I guess there must be some kind of tricky subsitution or identity that I can't see right now.

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    haye you tried a change of coordinates $x+y=R$,$x-y=r$? – 2017-01-19

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The square $[-1,1]^2$ is symmetric with respect to its diagonals, hence the given integral equals the same integral with the variables $x$ and $y$ exchanged. By cancellation it follows that

$$ \iint_{(-1,1)^2}e^{it(x+y)}(1+xy(x^2-y^2))\,dx\,dy = \iint_{(-1,1)^2}e^{it(x+y)}\cdot 1\,dx\,dy $$ and by Fubini's theorem the last integral is the square of $\int_{-1}^{1}e^{itz}\,dz$, i.e. $4\frac{\sin^2 t}{t}$.

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    Could you give me one more hint about the cancellation part? So $\iint_{(-1,1)^2}e^{it(x+y)}(1+xy(x^2-y^2))\,dx\,dy = \iint_{(-1,1)^2}e^{it(x+y)}(1+xy(y^2-x^2))\,dx\,dy$ But how do I conclude your identity from there? – 2017-01-19
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    @PeterGarder: $$ xy(x^2-y^2)+yx(y^2-x^2) = 0$$ and twice the integral is the sum of those integrals. – 2017-01-19
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    Thank you, it's a nice and elegant solution! – 2017-01-19
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    @PeterGarder: you're welcome. – 2017-01-19