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It is known that $\displaystyle\int_{-\infty}^{\infty} \frac{\sin \left( x\right )}{x} \mathrm{d}x = \int_{-\infty}^{\infty} \frac{\sin ^ 2\left( x\right )}{x^2} \mathrm{d}x$, I'm interested to know whether $n=2$ the only solution of this equation:$\displaystyle\int_{-\infty}^{\infty} \frac{\sin \left( x\right )}{x} \mathrm{d}x = \int_{-\infty}^{\infty} \frac{\sin ^ n\left( x\right )}{x^n} \mathrm{d}x$ ? And if it is how I can prove that ? I have tried trigonometric transformation for $ \sin^n(x)$ but I didn't come up to $n=2$, What I guess is for $n$ is even we have :$\sin^n(x)=\cos^n(x)-1$ but this can't do anything with the titled identity ?Any help ?

zeraoulia rafik
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    For arbitrary $n$ the integral is computed here: https://math.stackexchange.com/questions/307510/a-sine-integral-int-0-infty-left-frac-sin-x-x-rightn-mathrmdx. – Martin R Jul 05 '20 at 08:46

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Note that for all $x\ne0$ we have $$0\le\left|\frac{\sin{(x)}}x\right|\lt1$$ and hence we have for $n\in\mathbb{N}$ with $n\ge3$ \begin{align} \int_{-\infty}^\infty\left(\frac{\sin{(x)}}x\right)^n\mathrm{d}x &\le\int_{-\infty}^\infty\left|\frac{\sin{(x)}}x\right|^n\mathrm{d}x\\ &\lt\int_{-\infty}^\infty\left|\frac{\sin{(x)}}x\right|^2\mathrm{d}x\\ &=\int_{-\infty}^\infty\left(\frac{\sin{(x)}}x\right)^2\mathrm{d}x\\ \end{align} So the equality you mention only holds for $n=2$.

Peter Foreman
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