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Evaluate $\int_0^\infty\frac{\sin^4x}{x^2}dx$ with $\int_0^\infty\frac{\sin x}{x}dx=\frac{\pi}{2}$$$$$ I use $\int_0^\infty\int_0^\infty e^{-x^2y}\sin x dy dx=\int_0^\infty\int_0^\infty e^{-x^2y}\sin x dx dy $ to evaluate the integration, I wonder how to use $\int_0^\infty\frac{\sin x}{x}dx=\frac{\pi}{2}$ to evaluate $\int_0^\infty\frac{\sin^4x}{x^2}dx$ more efficiently.

Drinzjeng Triang
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1 Answers1

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Hint. Integrate by parts. First set $$I=\int_0^{\infty}\frac{\sin^2x}{x^2}\mathrm dx.$$ Then we have that $$\int_0^{\infty}\frac{\sin^4x}{x^2}\mathrm dx=\sin^2x\int_0^{\infty}\frac{\sin^2x}{x^2}\mathrm dx-\int_0^{\infty}2\sin x\cos x\left(\int_0^{\infty}\frac{\sin^2x}{x^2}\mathrm dx\right)\mathrm dx=I\sin^2x-\int_0^{\infty}I\sin 2x\mathrm dx.$$

Now we have that $$I=\int_0^{\infty}\frac{\sin^2x}{x^2}\mathrm dx=\int_0^{\infty}\frac{\sin x}{x}\mathrm dx=π/2.$$ To see the first equality, integrate by parts again to get $$I=\int_0^{\infty}\frac{\sin^2x}{x^2}\mathrm dx=\sin^2x\int_0^{\infty}\frac{1}{x^2}+\int_0^{\infty}\frac{\sin 2x}{x}\mathrm dx= \int_0^{\infty}\frac{\sin 2x}{2x}\mathrm d(2x)=\int_0^{\infty}\frac{\sin y}{y}\mathrm dy.$$

Allawonder
  • 13,327