Is $n=2$ the only solution of the below identity?

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 ?

1 Answer

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 $$n=2$$.