I need to prove this:
$$ \mathcal F{\frac{1}{r^2}}\frac{d}{dr}r^2 \frac{dC}{dr}$$
$$= (\frac{2}{\pi})^{1/2} \int_0^\infty\frac{1}{r^2}\frac{d}{dr}r^2\frac{dC}{dr}j_0(kr)r^2dr$$
$$ =-k^2 (\frac{2}{\pi})^{1/2} \int_0^\infty C(r)j_0(kr)r^2dr$$
$$= -k^2 \hat C (k) $$
Note that the $\mathcal F$ is the Fourier Transform.
In step 3 to step 4 I know I need to do integration by parts twice, where I am guessing that :
$$u=r^2 $$ $$ du=2rdr$$
And then we do it again later where:
$$u_2 = 2r $$
$$ du_2 = 2 dr$$
I am not sure what the dv's should be, and the corresponding v.
Normally I would assume that the other term is the dv term, which here would mean:
$$dv = C(r)j_0(kr)$$
But then I have two functions of r.
Please help me understand this problem if you have time.