I am trying to find the Fourier transform of $f_n(x)=\frac{\sin x\sin nx}{x^2}$ and have not been successful yet. I am wondering if there is an easy way of solving the following integral. $$\hat f_n(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{\sin x\sin nx}{x^2}e^{-ixt}dx$$
I tried writing $e^{-ixt}=\cos(xt)-i\sin(xt)$ and got the following ;$$\hat f_n(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{\sin x\sin (nx)\cos(xt)}{x^2}dx-\frac{i}{\sqrt{2\pi}}\underbrace{\int_{-\infty}^{\infty}\frac{\sin x\sin (nx)\sin(xt)}{x^2}dx}_{=0\text{ by symmetry}}$$
How can I proceed from here?
We have that $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\frac{\sin(1x)\sin(nx)\cos(tx)}{x^2}\,dx\\=\frac{1}{2\sqrt{2\pi}}\int_{-\infty}^{+\infty}\frac{\sin(x)\left[\sin((n+t)x)+\sin((n-t)x)\right]}{x^2}\,dx\\=\Large\scriptstyle{\frac{1}{4\sqrt{2\pi}}\int_{-\infty}^{+\infty}\frac{\cos((1-n-t)x)-\cos((1+n+t)x)+\cos((1-n+t)x)-\cos((1+n-t)x)}{x^2}\,dx} $$ and by integration by parts $$ \int_{-\infty}^{+\infty}\frac{1-\cos(\alpha x)}{x^2}\,dx = \pi|\alpha|$$ hence the graph of $\widehat{f_n}(t)$ is an isosceles trapezoid: this also follows from the fact that $\widehat{f_n}(t)$ (up to a multiplicative constant) is given by the convolution of the characteristic functions of two intervals.