A question about Fourier Transformation Recently I try to evaluate a integral in a paper:
$$
\Gamma(x,y)=\int_{-\infty}^{\infty} \frac{dk}{2\pi} \sqrt{k^2+m^2} e^{ikx}
$$
This is the Fourier Transform of:
$$
f(k)=\sqrt{k^2+m^2}
$$
The integral seems diverges since $f(k)\rightarrow \infty$ when $k$ approach infinity, however, Mathematica give me the explicit answer:
$$
 -\frac{\sqrt{\frac{2}{\pi }} m K_1(m x)}{x}
$$
where $K_1(m x)$ is modified Bessel function. Could you show me how to evaluate this Fourier Transformation. 
 A: Mathematica is rightfully telling me that your integral doesn't converge. Indeed if your answer was correct this would pose an apparent contradiction with the following bessel identities
\begin{align}
K_o (mx) &= \int_{-\infty }^{+\infty}\frac{e^{ikx}}{\sqrt{k^2 + m^2}} \\
\text{now use }K_o' = -K_1 &\\
\Rightarrow K_o'(mx) &= \frac{1}{x}\frac{\partial}{\partial m} K_o(mx) = \frac{-m}{x}  \int_{-\infty }^{+\infty}\frac{e^{ikx}}{\left(k^2 + m^2\right)^{3/2}} = -K_1(mx) \\
\end{align}
this integral representation of $K_1$ clearly conflicts with your answer.
Another way is to integrate your expression by parts
\begin{align}
\int_{-\infty }^{+\infty}e^{ikx}\sqrt{k^2 + m^2} = i x \left[e^{ikx}\sqrt{k^2 + m^2} \right]_{-\infty}^{+\infty} - ix K_o (mx)
\end{align}
now this clearly doesn't converge, a fact also demonstrated by noticing that $\operatorname{Im}\Gamma =0$ which makes it even easier to observe that $\int_{-\infty }^{+\infty}\text{cos }kx\sqrt{k^2 + m^2} $ does not converge
EDIT: you are right mathematica indeed gave me your answer, and I was able to derive it using bessel identities as follows, however my concerns about convergence stated above are still not resolved!
\begin{align}
\int_{-\infty }^{+\infty}e^{ikx}\sqrt{k^2 + m^2} &= \int_{-\infty }^{+\infty}\left(k^2 + m^2\right)e^{ikx}\frac{1}{\sqrt{k^2 + m^2}} \\
&=  \left(m^2 - \frac{\partial ^2}{\partial x^2} \right) \int_{-\infty }^{+\infty}\frac{e^{ikx}}{\sqrt{k^2 + m^2}} \\
&=m^2 \left( K_o(mx) - K_o^{''}(mx)\right) \\
&= m^2\left( K_o(mx) + K_1'(mx)\right) \\
&= \frac{-m^2}{x} K_1(mx)
\end{align}
where in the last line I used the identity $K_\nu '(x) = -\frac{\nu}{x}K_\nu(x) - K_{\nu -1}(x)$
