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 $n$.
 A: Not an answer
Residue theorem can be used to calculate the integral
\begin{equation}
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
\end{equation} 
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
\begin{equation}
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)
\end{equation} 
where $k_p$ are the 4 poles of
\begin{equation}
\phi(k)=\frac{\mathrm{J}_{n+\frac{1}{2}}(k)\, \mathrm{e}^{-\mathrm{j}\,k\,x}}{\sqrt{k}\,(k^4-\lambda^4)}
\end{equation} 
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
\begin{equation}
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]
\end{equation} 
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
\begin{equation}
J_{n+1/2}\left(ze^{\mathrm{j}\pi }\right)=e^{\mathrm{j}(n+1/2)\pi }J_{n+1/2}\left(z\right)
\end{equation} 
we obtain the proposed result.
A: It seems that the solution for $-1<x<1$ 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?
