Inverse Fourier transform to find out $\hat c_1$

Solution 1:

Note that your integral can be rewritten as $$\int_0^\infty \hat{c_1}(k) \sqrt{\frac{\pi}{2 k r}} J_\frac{1}{2}(kr)k \, \mathrm{d}k= r^{-1/2}\int_0^\infty \hat{c_1}(k) \sqrt{\frac{\pi}{2 k}} J_\frac{1}{2}(kr)k \, \mathrm{d}k $$ which is $r^{-1/2}$ times the inverse Hankel transform of order $\frac{1}{2}$ of the function $\hat{c_1}(k) \sqrt{\frac{\pi}{2k}}.$

Multiplying both sides by $r^{1/2}$ and taking the direct Hankel transform, we find that $\hat{c_1}(k) \sqrt{\frac{\pi}{2k}}$ is exactly the Hankel transform of $a_0 \sqrt{r} e^{-r^2/R^2}$, which is (according to Mathematica) $$\frac{a_0 \sqrt{k} e^{-\frac{1}{4} k^2 R^2}}{2 \sqrt{2} \left(\frac{1}{R^2}\right)^{3/2}}. $$ Dividing by the factor $\sqrt{\frac{\pi}{2k}}$ we find at last $$\hat{c_1}(k)=a_0 \frac{k R^3 e^{-\frac{1}{4} k^2 R^2}}{2 \sqrt{\pi }}. $$