# Inverse Fourier transform to find out $\hat c_1$

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.

## 1 Answer

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

• Can you give me the mathematica code please? – Complex Guy Sep 1 '13 at 16:36
• Integrate[ a0*Sqrt[r]*Exp[-r^2/R^2]*BesselJ[1/2, r k]*r, {r, 0, Infinity}] – user1337 Sep 1 '13 at 17:00
• $\frac{\text{a0} \sqrt{\frac{2}{\pi }} \text{If}\left[\text{Re}\left[R^2\right]>0,\frac{e^{-\frac{1}{4} k^2 R^2} k \sqrt{\pi }}{4 \left(\frac{1}{R^2}\right)^{3/2}},\text{Integrate}\left[e^{-\frac{r^2}{R^2}} r \text{Sin}[k r],\{r,0,\infty \},\text{Assumptions}\to \text{Re}\left[R^2\right]\leq 0\right]\right]}{\sqrt{k}}$ it gives the above, do we need to integrate further? – Complex Guy Sep 1 '13 at 17:19
• If you assume that $R>0$ the result simplifies considerably. – user1337 Sep 1 '13 at 17:29
• so the right side of the part $Re[R^2] \leq 0$ will be omitted? I'm little bit confused. Can you please do the full step in mathematica then I can catch the code and logic. Thanks in advance for doing this. – Complex Guy Sep 1 '13 at 17:33