Hi I had posted the same post 2 days ago but I am posting it again because of my bad handwriting. I apologize to the man who wanted to read my post.

I am not familiar with the tool which is used in this site. So I use another tool to write my solution.

I don't know whether my solution is right so I wanted verify it.

Moreover, I am not able to calculate the integral in the last line which is about $dx'dy'dz'$, please let me know how I can proceed to the next step.

here is my solution


To evaluate the integral of interest, we have

$$\begin{align} \frac{1}{(2\pi)^3}\int_{R^3}\frac{e^{i\vec k\cdot (\vec r-\vec r')}}{k^2}d^3\vec k&=\frac{1}{(2\pi)^3}\int_0^{2\pi}\int_0^\infty \int_0^\pi \frac{e^{i\vec k\cdot (\vec r-\vec r')}}{k^2} \,k^2\,\sin (\theta)\,d\theta\,dk\,d\phi \tag 1\\\\ &=\frac{1}{(2\pi)^2}\int_0^\infty \int_0^\pi e^{i k|\vec r-\vec r'|\cos(\theta)}\,\sin (\theta)\,d\theta\,dk\tag 2\\\\ &=\frac{1}{(2\pi)^2}\int_0^\infty 2\frac{\sin(k|\vec r-\vec r'|)}{k|\vec r-\vec r'|}\,dk \tag 3\\\\ &=\frac{1}{4\pi |\vec r-\vec r'|}\tag 4 \end{align}$$


In arriving at $(1)$ we use a spherical coordinate system $(k,\theta,\phi)$ in $\vec k$-space. Note that $d^3\vec k=k^2\,\sin(\theta)\,d\theta\,dk\,d\phi$

In going from $(1)$ to $(2)$, we rotate our $\vec k$ coordinate system so that the polar axis aligns with $\vec r-\vec r'$. Then, $\vec k\cdot (\vec r-\vec r')=k|\vec r-\vec r'|\cos(\theta)$. Noting that the integrand is independent of $\phi$, we carry out the integration over $\phi$ to produce a factor of $2\pi$.

In going from $(2)$ to $(3)$, we carry out the integration over $\theta$.

In going from $(3)$ to $(4)$, we made use of the result for the Sine Integral$\int_0^\infty \frac{sin(kR)}{kR}\,dk=\frac{\pi}{2R}$.

Had we chosen to carry out the integral in $(2)$ in the reverse order we would write

$$\begin{align} \frac{1}{(2\pi)^2}\int_0^\infty \int_0^\pi e^{i k|\vec r-\vec r'|\cos(\theta)}\,\sin (\theta)\,d\theta\,dk&=\lim_{L\to \infty}\frac{1}{(2\pi)^2}\int_0^L \int_0^\pi e^{i k|\vec r-\vec r'|\cos(\theta)}\,\sin (\theta)\,d\theta\,dk\\\\ &=\lim_{L\to \infty}\frac{1}{(2\pi)^2}\int_0^\pi \int_0^L e^{i k|\vec r-\vec r'|\cos(\theta)}\,\sin (\theta)\,dk\,d\theta\\\\ &=\lim_{L\to \infty}\frac{1}{(2\pi)^2}\int_0^\pi \frac{e^{i|\vec r-\vec r'|L\cos(\theta)}-1}{i|\vec r-\vec r'|\cos(\theta)}\,\sin(\theta)\,d\theta\\\\ &=\lim_{L\to \infty}\frac{1}{(2\pi)^2}\int_{-1}^1 \frac{e^{i|\vec r-\vec r'|L\,x}-1}{i|\vec r-\vec r'|\,x}\,dx\\\\ &=\lim_{L\to \infty}\frac{1}{(2\pi)^2}\int_{-1}^1 \frac{\sin(|\vec r-\vec r'|L\,x)}{|\vec r-\vec r'|\,x}\,dx\\\\ &=\frac{1}{(2\pi)^2}\frac{1}{|\vec r-\vec r'|}\int_{-\infty}^\infty \frac{\sin(x)}{x}\,dx\\\\ &=\frac{1}{4\pi |\vec r-\vec r'|} \end{align}$$

as expected!

  • $\begingroup$ @DKL Thank you for the best vote. Just FYI ... you can also give a vote up to questions you post. ;-)) - Mark $\endgroup$ – Mark Viola Mar 16 '16 at 23:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.