# I How did this integral come into being?

1. May 31, 2016

### Josh1079

Hi, this is actually from a text of Krane's Introductory Nuclear Physics.
I don't really understand how did he arrive (3.5) from (3.2) and (3.4), and the reason for setting q ⋅ r = qrsinθ is also confusing.

This is the first time I post and I'm not really sure whether this is the correct section for this kind of question, so if it's the wrong place to post please just tell me. Thanks!

2. May 31, 2016

### blue_leaf77

Substitute (3.4) into (3.2) to obtain
$$F(\mathbf q) \propto \int e^{i\mathbf q \cdot \mathbf r} \int \frac{\rho(\mathbf r')}{|\mathbf r - \mathbf r'|} dv' dv = \int \rho(\mathbf r') \left( \int \frac{e^{i\mathbf q \cdot \mathbf r}}{|\mathbf r - \mathbf r'|} dv \right)dv'$$
Compute the integral in the parentheses first by a change of variable $\mathbf u = \mathbf r - \mathbf r'$,
$$\int \frac{e^{i\mathbf q \cdot \mathbf r}}{|\mathbf r - \mathbf r'|} dv = e^{i\mathbf q \cdot \mathbf r'} \int \frac{e^{i\mathbf q \cdot \mathbf u}}{u} dv$$
where now $dv = u^2 \sin \theta du\,d\theta\,d\phi$.Here $\theta$ is the angle between $\mathbf q$ and $\mathbf u$, thus $\mathbf q \cdot \mathbf u = qu \cos \theta$ (I think it should be cosine instead of sine as your source states). The integral above becomes
$$e^{i\mathbf q \cdot \mathbf r'} \int e^{iqu \cos \theta} u \sin \theta du\,d\theta\,d\phi = -e^{i\mathbf q \cdot \mathbf r'} \int_0^\infty \int_0^\pi \int_0^{2\pi} e^{iqu \cos \theta} u du\,d\cos\theta\,d\phi$$
Now solve this last integral.
Hint: Solve the integral over $\theta$ first. The integral over $\phi$ is trivial. As for the remaining integral over $u$, employ a limiting condition by multiplying the integrand with a function of the form $e^{-bu}$, then apply the limit $b\to 0$ after calculating the integral.

3. May 31, 2016

Question I have-(I tried working this earlier and got something similar to @blue_leaf77 calculation above), but I got that there should be a $q^2$ in the denominator of equation (3.5) which will be a $q^3$ in the denominator of equation (3.6). Did I compute it incorrectly? I'm double-checking my algebra. I'm hoping the author has (3.5) and (3.6) correct, but I'm not so sure. (If you compute blue_leaf77's last integral with the substitution $u'=qu$, you get a $q^2$ in the denominator.)

Last edited: May 31, 2016
4. May 31, 2016

### blue_leaf77

Yes you are right I think there should be $1/q^2$ in equation (3.5).

5. May 31, 2016

### Josh1079

Thank you so much @blue_leaf77!!!