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$f(x)=\sin x/x$, \begin{align*}\int_0^{\pi/2}f(x)f(\dfrac\pi2-x)dx&=\int_0^{\pi/2}\dfrac{\sin x\cos x}{x(\pi/2-x)}dx=2\int_0^{\pi/2}\dfrac{\sin 2x}{2x(\pi-2x)}dx\\ &=\dfrac2\pi\int_0^{\pi/2}\sin2x\bigg(\dfrac1{2x}+\dfrac1{\pi-2x}\bigg)dx\\ &=\dfrac1\pi\int_0^\pi\dfrac{\sin u}{u}+\dfrac{\sin u}{\pi-u}du\\ &=\dfrac2\pi\int_0^\pi\dfrac{\sin u}udu\text{ (Symmetry.)}\\ &=\dfrac2\pi\int_0^\pi f(x)dx\end{align*}
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Tue, 21 Apr 2020 14:11 GMT