Solving a Laplace Transform Integral

[rbwang1225]

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\begin{document}

I met some problem in the following integration; does anyone know how to solve it?

\begin{equation}
\frac{2m}{h^2}\frac{1}{4\pi}\int{d^3xe^{i\mathbf{k\cdot r}}\frac{V_0e^{-\mu r}}\mu r}e^{ikx}.\
\end{equation}
The key part that can not be solved is:

\begin{equation}
\int_0^{\infty}\int_0^{\pi}e^{(ik-\mu) r}e^{ikr\cos\theta}\sin\theta d\theta dr.\
\end{equation}

I found it can be treated as a Laplace transform of

\begin{equation}
\frac{e^{ikr}\sin(kr)}r
\end{equation}

And the solution is

\begin{equation}
i\tanh^{-1}[\frac{k}{k+i\mu}].
\end{equation}

Any help would be appreciated!
\end{document}

 

[Stephen Tashi]

The LaTex on the forum is somewhat buggy. You must use the "preview" function and then also tell your browser to reload the page. (Before reloading the page, the "preview" may show nonsensical or obsolete LaTex.)

So let me convert your post to LaTex and we'll see if a Laplace transform expert will answer.

-----


I met some problem in the following integration; does anyone know how to solve it?

<br /> \frac{2m}{h^2}\frac{1}{4\pi}\int{d^3xe^{i\mathbf{k\cdot r}}\frac{V_0e^{-\mu r}}\mu r}e^{ikx}.<br />

The key part that can not be solved is:

<br /> \int_0^{\infty}\int_0^{\pi}e^{(ik-\mu) r}e^{ikr\cos\theta}\sin\theta d\theta dr.\<br />

I found it can be treated as a Laplace transform of

<br /> \frac{e^{ikr}\sin(kr)}r<br />

And the solution is

<br /> i\tanh^{-1}[\frac{k}{k+i\mu}].<br />

Any help would be appreciated!
-------------------

 

[rbwang1225]

How do I use "preview" function?

Thanks for your help!

 

[Char. Limit]

At the bottom of the posting window, you have two buttons, "Submit Reply" and "Preview Post". The preview function is the right button. And if you need it, reloading your browser is F5 on most browsers.

 

[rbwang1225]

Ha,I got it! Thank you very much.

 

[henry_m]

That looks something like the Born approximation for the Yukawa potential to me! I think you might have got yourself an extraneous factor in there somehow... Did you mean
<br /> \frac{2m}{\hbar^2}\frac{1}{4\pi}\int{d^3x\frac{V_0e^{-\mu r}}{\mu r}e^{i\vec{k}.\vec{x}}}<br />
or something similar?

In your change of variables I think you lost another factor of r, which is what made things difficult. You needn't worry about such things as Laplace transforms here. It looks like you managed the theta integral perfectly well, but your lost factor of r made the radial integral much harder than it should have been. Check your algebra carefully.

 

[rbwang1225]

No, my consideration is not the Born approximation for the Yukawa potential. If you have Sakurai's book, please check Eq. (7.2.12). I want to confirm the statement (7.2.13) by doing direct calculation, since I don't understand what did he mean by "replace e^{ikr&#039;} by 1".

 

[henry_m]

Ah OK, yes I've got Sakurai. The integral in (7.2.12) is:
<br /> \frac{2m}{\hbar^2}\frac{1}{4\pi}\int{d^3x\frac{V_0 e^{-\mu r}}{\mu r}\frac{e^{ikr}}{r}}<br />
so I still think you have an extra term. The approximation made is e^ikr=1, which is valid for k much less than 1/r, and the exponential means that the integral will only contribute for r less than roughly 1/mu, so it's a good approximation for k much less than mu.

 

[rbwang1225]

You have lost this term e^{-i \mathbf k \cdot \mathbf r}. After doing the calculation, I take the limit k->0 and get the wanted form (7.2.13).

Regards