show that this integral:

$$\dfrac{1}{2\pi h}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{\dfrac{-i(p-p')x}{h}}x^n\varphi{(p')}dxdp'=\left(ih\dfrac{\partial }{\partial p}\right)^n\varphi{(p)}$$

where $i^2=-1$

maybe this use integration by parts? But I fell very hard,and I can't prove it.

I think first we must this $$\int_{-\infty}^{\infty}e^{\dfrac{-i(p-p')x}{h}}x^ndx=?$$ and follow it can use

integration by parts?

But I consider sometimes can't have this resulut

Thank you


$$\phi(p)=\int_{-\infty}^{\infty}\phi(p')\delta(p'-p)dp'\\=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\phi(p')e^{\displaystyle i(p'-p)x}dxdp'$$ The last step follows from the Fourier expansion of $\delta(\cdot)$. Now, put $x/h$ instead of $x$ in the inner integral to get $$\phi(p)=\frac{1}{2\pi h}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\phi(p')e^{\displaystyle \frac{i(p'-p)x}{h}}dxdp'$$ Differentiate with respect to $p$ both sides $n$ times to get $$\left(\frac{\partial}{\partial p}\right)^n\phi(p)=\left(\frac{-i}{h}\right)^n \frac{1}{2\pi h}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\phi(p')x^ne^{\displaystyle \frac{i(p'-p)x}{h}}dxdp'$$

  • $\begingroup$ Hello,why is first step is true? $\endgroup$ – china math May 14 '14 at 9:51
  • $\begingroup$ Sorry, the $1/{2\pi}$ factor should come in step 2 after the fourier expansion. Thanks for pointing out. $\endgroup$ – Samrat Mukhopadhyay May 14 '14 at 9:55
  • $\begingroup$ Oh,Thank you +1,But maybe this problem have other methods $\endgroup$ – china math May 14 '14 at 10:37
  • $\begingroup$ Maybe there are, but the presence of the differentiation makes this method very handy to use. $\endgroup$ – Samrat Mukhopadhyay May 14 '14 at 10:45
  • 1
    $\begingroup$ $\delta(\cdot)$ is called the dirac delta function and the above property follows from the definition of the delta function. See here. $\endgroup$ – Samrat Mukhopadhyay May 14 '14 at 11:02

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