Inverse Fourier transform to find out $\hat c_1$

Question

If we have an integration which is need to solve inversely $$a_0 e^{-r^2/R^2} = \int_0^\infty \hat{c}_1(k) \frac{\sin(k r)}{r} dk,$$ If I transform the $\sin(kr)$, then we get imaginary part.

Please solve this elaborately.

Thanks in advance.

Answer 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 $$\hat{c_1}(k)=a_0 \frac{k R^3 e^{-\frac{1}{4} k^2 R^2}}{2 \sqrt{\pi }}. $$