Proposicion 4.2. of the book "An introduction to pseudo differential operators by Wong

Question

Hi! Why $$(2\pi)^{-n/2} \int e^{-ix\xi}(D^{\alpha}\varphi)(x)dx=(2\pi)^{-n/2}\int \xi^{\alpha}e^{-ix\xi}\varphi(x)dx?$$

I tried using part integration with $u=e^{-ix\xi}$ and $D^{\alpha}\varphi(x)=dv$ but it did not work

Answer 1

Integrate by parts $\alpha$ times. For example, the first one integrates $(D^\alpha \varphi)dx$ to $D^{\alpha-1} \varphi$ and differentiates $e^{-ix \xi}$ to $-\xi e^{-x\xi}$ thus getting $$ \int e^{-ix \xi} (D^\alpha \varphi)dx = e^{-x\xi} \left(D^{\alpha-1} \varphi\right) + \int \xi e^{-x\xi}\left(D^{\alpha-1} \varphi\right)dx $$ and doing this $\alpha$ times reduces the differential by one and increments the power of $\xi$ by one as well, so you end up with $$\int \xi^\alpha e^{-x\xi} \varphi dx$$ as expected.

It remains to show the non-integral term is zero but that is likely a consequence of integration over $\mathbb{R}^n$...