# Inverse Fourier Transform of a half-integer Bessel function

Is there an analytical solution for the following inverse Fourier transform?

$$f(x)=\frac{1}{\sqrt{2\pi}}\,\mathrm{j}^n \int_{-\infty}^\infty\frac{1}{\sqrt{k}\,(k^4-\lambda^4)}\mathrm{J}_{n+\frac{1}{2}}(k)\, \mathrm{e}^{-\mathrm{j}\,k\,x}\,\mathrm{d}k$$

with $$n \in \mathbb{N_0}$$, $$\mathrm{Re}(\lambda)>0$$, $$\mathrm{Im}(\lambda)<0$$, $$x \in \mathbb{R}$$ and $$\mathrm{J}_\bullet$$ is the Bessel function of the first kind.

Edit: I found out (through a completely different approach) that

$$f(x)=-\frac{1}{4\,\lambda^3}\left(\mathrm{e}^{-\lambda\,x}\,\mathrm{j}^n\,\sqrt{\frac{2\pi\mathrm{j}}{\lambda}}\mathrm{J}_{n+\frac{1}{2}}\!\!\left(\frac{\lambda}{\mathrm{j}}\right)+\mathrm{e}^{-\mathrm{j}\,\lambda\,x}\,\mathrm{j}^{n+1}\,\sqrt{\frac{2\pi}{\lambda}}\mathrm{J}_{n+\frac{1}{2}}\!\!\left(\lambda\right)\right)\quad\text{for}\quad x\geq1$$ and $$f(x)=-\frac{1}{4\,\lambda^3}\left(\mathrm{e}^{\lambda\,x}\,\mathrm{j}^n\,\sqrt{-\frac{2\pi\mathrm{j}}{\lambda}}\mathrm{J}_{n+\frac{1}{2}}\!\!\left(-\frac{\lambda}{\mathrm{j}}\right)+\mathrm{e}^{\,\mathrm{j}\,\lambda\,x}\,\mathrm{j}^{n+1}\,\sqrt{-\frac{2\pi}{\lambda}}\mathrm{J}_{n+\frac{1}{2}}\!\!\left(-\lambda\right)\right)\quad\text{for}\quad x\leq-1$$

Now the question remains, how to solve it for $$-1\leq x \leq1$$. I am accually surprised that the integral even changes. Any idea, how to get the solution for $$x\geq1$$ and $$x\leq-1$$ directly from the integral above?

Edit2: Thanks to Mathematica I got the solution for $$x=0$$

$$f(0)=\frac{\sqrt{\pi}\,(-1)^{n+1}}{8\,\lambda^{7/2}}\left((2+2\,\mathrm{j})\, \mathrm{I}_{n+\frac{1}{2}}\!\!\left(\mathrm{j}\,\lambda\right)+2\,\sqrt{2}\, \mathrm{I}_{n+\frac{1}{2}}\!\!\left(\lambda\right)+\lambda^{3/2}\left(_1\tilde{F}_2\left(1;\frac{3}{2}-\frac{n}{2},2+\frac{n}{2};-\frac{\lambda^2}{4}\right)-\,_1\tilde{F}_2\left(1;\frac{3}{2}-\frac{n}{2},2+\frac{n}{2};\frac{\lambda^2}{4}\right)\right)\right)$$

with $$\mathrm{I}_\bullet$$ is the modified Bessel function of the first kind and $$_1\tilde{F}_2\left(\cdots;\cdots;\bullet\right)$$ the regularized hypergeometric function. The solution for $$x=0$$ vanishes for odd $$n$$.

Residue theorem can be used to calculate the integral $$$$f(x)=\frac{1}{\sqrt{2\pi}}\,\mathrm{j}^n \int_{-\infty}^\infty\frac{1}{\sqrt{k}\,(k^4-\lambda^4)}\mathrm{J}_{n+\frac{1}{2}}(k)\, \mathrm{e}^{-\mathrm{j}\,k\,x}\,\mathrm{d}k$$$$ When $$\Re x\le -1$$, we close the contour by the large upper half-circle. This additionnal contribution vanishes owing to Jordan lemma ($$\left|J(it)\right|\sim e^{\left|t\right|}/\sqrt t$$), we have then $$$$f(x)=\sqrt{2\pi}\mathrm{j}^{n+1}\sum_{k_p/\Im k_p>0}\operatorname{Res}\left( \phi(k),k=k_p \right)$$$$ where $$k_p$$ are the 4 poles of $$$$\phi(k)=\frac{\mathrm{J}_{n+\frac{1}{2}}(k)\, \mathrm{e}^{-\mathrm{j}\,k\,x}}{\sqrt{k}\,(k^4-\lambda^4)}$$$$ defined on the complex plane cut along the negative real axis. The poles are $$\lambda,\mathrm{j}\lambda,-\lambda,-\mathrm{j}\lambda$$.

As $$-\pi/2<\operatorname{Arg}(\lambda)<0$$, only $$\mathrm{j}\lambda$$ and $$-\lambda$$ have to be taken into account in the summation. Their residues are \begin{align} \operatorname{Res}\left( \phi(k),k=\mathrm{j}\lambda\right)&=\frac{\mathrm{J}_{n+\frac{1}{2}}(\mathrm{j}\lambda)\, \mathrm{e}^{\lambda\,x}}{4\sqrt{\mathrm{j}\lambda}\,\mathrm{j}^3\lambda^3}\\ &=-\frac{1}{4\mathrm{j}\lambda^3}\sqrt{\frac{-\mathrm{j}}{\lambda}}\mathrm{J}_{n+\frac{1}{2}}(\mathrm{j}\lambda)\, \mathrm{e}^{\lambda\,x}\\ \operatorname{Res}\left( \phi(k),k=-\lambda\right)&=\frac{\mathrm{J}_{n+\frac{1}{2}}(-\lambda)\, \mathrm{e}^{\mathrm{j}\lambda\,x}}{4\sqrt{-\lambda}\,(-\lambda)^3}\\ &=-\frac{1}{4\lambda^3}\sqrt{\frac{1}{-\lambda}}\mathrm{J}_{n+\frac{1}{2}}(-\lambda)\, \mathrm{e}^{\mathrm{j}\lambda\,x} \end{align} and thus $$$$f(x)=-\frac{\sqrt{2\pi}\mathrm{j}^{n}}{4\lambda^3}\left[\sqrt{\frac{-\mathrm{j}}{\lambda}}\mathrm{J}_{n+\frac{1}{2}}(\mathrm{j}\lambda)\, \mathrm{e}^{\lambda\,x} +\mathrm{j}\sqrt{\frac{1}{-\lambda}}\mathrm{J}_{n+\frac{1}{2}}(-\lambda)\, \mathrm{e}^{\mathrm{j}\lambda\,x} \right]$$$$ which is identical to the quoted result for $$x\le-1$$.

For $$\Re x\ge1$$, the same calculation can be made by closing the contour downwards. It can be easier to calculate the complex conjugate and use the previous result. Using the symmetry property $$$$J_{n+1/2}\left(ze^{\mathrm{j}\pi }\right)=e^{\mathrm{j}(n+1/2)\pi }J_{n+1/2}\left(z\right)$$$$ we obtain the proposed result.

• Very interesting approach. Now I am wondering, why isn't it working for $-1<x<1$. It seems to me that you did not restrict $x$ in this way. Numerical integration at certain points show me though that the result is not correct for this range of $x$ – Michael_K Apr 9 at 21:36
• I am surprised by your numerical results, as I can't see why $x=\pm 1$ would be a boundary of some domain relevent for the function. On the contrary, $\Re x=0$ determines its behavior for $\Im k\to \pm\infty$. – Paul Enta Apr 10 at 9:21
• @Michael_K The absolute value of $J_\nu(i t)$ grows as $e^{|t|}/\sqrt {|t|}$, the Jordan lemma is not applicable when $|x| < 1$. – Maxim Apr 10 at 15:00
• @Maxim You are right, of course. Sorry for the mistake. I correct this answer in case it could be useful, even if it does not answer the OP. – Paul Enta Apr 10 at 16:32

It seems that the solution for $$-1 is given by

$$f(x)=-\frac{1}{4\,\lambda^3}\left(-\sqrt{\frac{2}{\lambda\,\pi}}\left((-1)^n\,\mathrm{e}^{-x\lambda}+\mathrm{e}^{x\lambda}\right) \mathrm{K}_{n+1/2}(\lambda)+\\\sum_{k=0}^n\frac{(k+n)!\,(1+(-1)^k)}{2^k\,\lambda^{k+1}\,k!\,(n-k)!}\,\, {_2\mathrm{F}_1}\left(k-n;k+n+1;k+1;\frac{1-x}{2}\right)\right)-\frac{\mathrm{j}}{4\,\lambda^3}\left(-\sqrt{\frac{2}{\mathrm{j}\,\lambda\,\pi}}\left((-1)^n\,\mathrm{e}^{-x\,\mathrm{j}\,\lambda}+\mathrm{e}^{x\,\mathrm{j}\,\lambda}\right) \mathrm{K}_{n+1/2}(\mathrm{j}\,\lambda)+\\\sum_{k=0}^n\frac{(k+n)!\,(1+(-1)^k)}{2^k\,(\mathrm{j}\,\lambda)^{k+1}\,k!\,(n-k)!}\,\, {_2\mathrm{F}_1}\left(k-n;k+n+1;k+1;\frac{1-x}{2}\right)\right)$$

which I reconstructed from results given in here.

This is not very useful though, since I can not use this result for large $$n$$ (the fraction in the sum takes very large values) even though the total value of $$f(x)$$ remains nicely bounded for large $$n$$. Any ideas how to resolve this?