# How prove this $\frac{1}{2\pi h}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{\frac{-i(p-p')x}{h}}x^n\varphi{(p')}dxdp'$

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 $$\left(\frac{\partial}{\partial p}\right)^n\phi(p)=\left(\frac{-i}{h}\right)^n \frac{1}{2\pi h}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\phi(p')x^ne^{\displaystyle \frac{i(p'-p)x}{h}}dxdp'$$

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