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.


closed as off-topic by Emilio Pisanty, John Rennie, Martin, ACuriousMind, Sebastian Riese Sep 17 '15 at 18:02

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  • $\begingroup$ Quite intristing. The identity is false but it is true in the sense of the analytic continuation in the exponent of $k^2+m^2$... $\endgroup$ – Valter Moretti Sep 17 '15 at 7:06
  • 2
    $\begingroup$ ...So the result is the Fourier transform in the distributional sense. $\endgroup$ – Valter Moretti Sep 17 '15 at 7:14
  • $\begingroup$ To complement on @ValterMoretti comment, the function $\sqrt{k^2+m^2}$ can be seen as a $\mathscr{S}'$ distribution, for the integral $\int \sqrt{k^2+m^2}f(k)dk$ is finite whenever $f\in\mathscr{S}$, i.e. it is a function of rapid decrease. The Fourier transform $\mathscr{F}$ is a bijection of $\mathscr{S}'$ into itself, so it makes sense to calculate $\mathscr{F}(\sqrt{k^2+m^2})$. This cannot be done however directly as a standard integral. $\endgroup$ – yuggib Sep 17 '15 at 9:41
  • $\begingroup$ Ah..Is it so-called generalized function? It is beyond my knowledge..Anyway,thanks a lot, both of you! $\endgroup$ – JQ Skywalker Sep 17 '15 at 10:48
  • $\begingroup$ This seems to be a pure math question. $\endgroup$ – ACuriousMind Sep 17 '15 at 12:47

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 $K_\nu '(x) = -\frac{\nu}{x}K_\nu(x) - K_{\nu -1}(x)$

  • $\begingroup$ Indeed I try to use "Integrate" to calculate it and it doesn't converge. But if you use the command "FourierTransform" $$\text{FourierTransform}\left[\sqrt{k^2+m^2},k,x,\text{Assumptions}\to \{x>0,m>0\}\right]$$to perform it and it gave me the answer above… $\endgroup$ – JQ Skywalker Sep 17 '15 at 10:34
  • $\begingroup$ check edit for derivation of result, without explanation of convergence... $\endgroup$ – Ali Moh Sep 17 '15 at 18:05

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