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.


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

  • $\begingroup$ Can you give me the mathematica code please? $\endgroup$ – Complex Guy Sep 1 '13 at 16:36
  • $\begingroup$ Integrate[ a0*Sqrt[r]*Exp[-r^2/R^2]*BesselJ[1/2, r k]*r, {r, 0, Infinity}] $\endgroup$ – user1337 Sep 1 '13 at 17:00
  • $\begingroup$ $\frac{\text{a0} \sqrt{\frac{2}{\pi }} \text{If}\left[\text{Re}\left[R^2\right]>0,\frac{e^{-\frac{1}{4} k^2 R^2} k \sqrt{\pi }}{4 \left(\frac{1}{R^2}\right)^{3/2}},\text{Integrate}\left[e^{-\frac{r^2}{R^2}} r \text{Sin}[k r],\{r,0,\infty \},\text{Assumptions}\to \text{Re}\left[R^2\right]\leq 0\right]\right]}{\sqrt{k}}$ it gives the above, do we need to integrate further? $\endgroup$ – Complex Guy Sep 1 '13 at 17:19
  • $\begingroup$ If you assume that $R>0$ the result simplifies considerably. $\endgroup$ – user1337 Sep 1 '13 at 17:29
  • $\begingroup$ 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. $\endgroup$ – Complex Guy Sep 1 '13 at 17:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.