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Let $x=(x_{1},..., x_n)\in \mathbb R^n$, $\gamma, \beta$ is multi-index of $n-$tuple of nonnegative integers. Put $x^{\gamma }= x_1^{\gamma_1}\cdots x_{n}^{\gamma_n}.$ Let $f\in \mathcal{S}(\mathbb R^n)$ (Schwartz space).

Let $$\displaystyle I_{\gamma}(x):=\int_{\mathbb R^n} x^{\gamma}e^{it\cdot x}t^{\beta}f(t)\ dt. $$

Can we say $$\displaystyle I_{\gamma}(x)=\int_{\mathbb R^n}e^{it\cdot x}\sum_{\alpha\leqslant \gamma}\binom{\gamma}{\alpha}\partial^{\alpha}f(t)t^{\beta-\alpha}\frac{\beta !}{(\beta-\alpha)!}\ dt?$$

(I guess, I should use Integrating by parts and using Leibniz formula, but I do not how do this explicitly in this situation)

Motivation: I am trying to understand this answer

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abcd
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  • Have you tried some simple examples? –  Dec 06 '16 at 16:37
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    Haha. Definitely not what I was expecting to see, based on the question title. –  Dec 06 '16 at 16:38
  • @Winther: Thanks: I am trying to understand this answer – abcd Dec 06 '16 at 17:45
  • Good, now its more clear what is done here. There are three ingredients here: 1) if $n$ is an integer then $\frac{d^n}{dt^n} e^{it\cdot x} = i^n x^ne^{it\cdot x}$ (abusing notation). Use this with $n=\gamma$ to rewrite $x^\gamma e^{it\cdot x}$. 2) Move all the derivatives over to $[t^\beta f(t)]$ using integration by parts 3) Use Leibnitz formula $\frac{d^\gamma t^\beta f}{dx^\gamma} = \sum_{\alpha = 0}^\gamma {\gamma \choose \alpha} f^{(\alpha)}(t) \frac{d^{\gamma - \alpha}}{dt^{\gamma - \alpha}} t^\beta$. Try to find a general expression for $\frac{d^k}{dx^k} t^n$ to deal with the last term. – Winther Dec 06 '16 at 18:00

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