Fourier transform of $xf''(x)$

Question

I am very unsure of how to approach this Fourier transform and have not been able to find anything which has cleared up my confusion. I know that the transform of $xf(x)=ig'(\alpha)$, but I fail to see how I can extend this to $xf''(x)$.

Answer 1

We know that Fourier transform, $$\mathcal{F}\{f(x):u\}=F(u)=\dfrac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}f(t)e^{-iut}dt,$$ $$\mathcal{F}\{f'(x):u\}=\mathcal{F}\left\{\frac{d}{dx}f(x):u\right\}=iu~F(u)$$ $$\text{and}~~~~~~~~\mathcal{F}\left\{\frac{d^n}{dx^n}f(x):u\right\}=(iu)^nF(u)$$

So $$\mathcal{F}\left\{\frac{d^2}{dx^2}f(x):u\right\}=(iu)^2F(u)$$ Now$$\mathcal{F}\{xf''(x)\}=\dfrac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}xf''(x)e^{-iux}dx=i\dfrac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}f''(x)\dfrac{\partial}{\partial u}\left(e^{-iux}\right)dx=i\dfrac{\partial}{\partial u}\left(\dfrac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}f''(x)e^{-iux}~dx\right)=i\dfrac{d}{d u}\left[(iu)^2F(u)\right]=-i\left\{2uF(u)+u^2F'(u)\right\}$$