1
$\begingroup$

I want to know whether the following integral converge in the region $k\to \infty$ or not.
$$ \int\frac{d^4k}{(2\pi)^4}\,\frac{1}{k^4}e^{-i\mathbf{k}\cdot\mathbf{\epsilon}} =-\frac{1}{4\pi^3}\int_0^{\pi}d\theta \sin^2\theta\int_0^{\infty} dk\, \frac{1}{k}\, e^{-ik\,\epsilon\cos\theta} $$ When $\cos \theta \ne 0$,
$$ \int_0^{\infty} dk\, \frac{1}{k}\, e^{-ik\epsilon\cos\theta}= \begin{cases} \displaystyle \int_0^{\infty} dk \frac{1}{k}e^{-ik} & (\cos \theta>0) \\ \displaystyle \int_0^{\infty} dk \frac{1}{k}e^{ik} & (\cos \theta<0) \end{cases} $$ It converges in the region $k\to \infty$ because $\int_0^{\infty} dk \frac{\cos k}{k}$ and $\int_0^{\infty} dk \frac{\sin k}{k}$ converges in the region $k\to \infty$.
When $\cos \theta =0$,
$$\int_0^{\infty} dk\, \frac{1}{k}\, e^{-ik\epsilon\cos\theta}=\int_0^{\infty} dk \frac{1}{k}$$ It diverges in the region $k\to \infty$. But its contribution to the original integral is only at $\theta=\pi/2$ in $\int_0^{\pi}d\theta$. That's a sort of $\infty \cdot 0$ problem.
How should we treat this pole in the multiple integral?
Thanks.

$\endgroup$
2
$\begingroup$

To know if the integral converges, you can explicitly do the integration in $\theta$. Using the change of variables $\cos\theta = x$ you get $$ \int_0^\pi d\theta\sin^2\theta e^{\alpha\cos\theta} = -\int_{-1}^1dx\sqrt{1-x^2}e^{\alpha x} = -\frac{\pi}{\alpha}I_1(\alpha) $$ where $I_1$ is the modified Bessel function of order $1$. Substituting for your case $\alpha = -i\epsilon k$ and introducing back into your total integral we have $$ -\frac{1}{4\pi^3}\int_0^\pi d\theta\sin^2\theta\int_0^\infty dk\frac{1}{k}e^{-ik\epsilon\cos\theta} = \frac{1}{4\pi^2}\int_0^\infty dk\frac{J_1(k\epsilon)}{\epsilon k^2} $$ which does never converge as it has a second order pole at $k=0$ that the Bessel function is unable to compensate (equivalently to the function $\sin x/x^2$). The divergence is logarithmic as, according to Mathematica, $$ \lim_{\delta \to 0}\int_\delta^\infty \frac{J_1(k\epsilon)}{\epsilon k^2} \to -\frac{1}{2}\log(\delta\epsilon) =\infty $$ Hope it helps!

$\endgroup$
  • $\begingroup$ According to this Mathematica Online $$ \int_a^{\infty} \frac{J_1(x)}{x^2}dx = \frac{1}{32} \left( a^2 {}_2F_3 \left(1,1;2,2,3;-\frac{a^2}{4} \right) + 8 (-2\log a-2\gamma+1+\log4) \right) ,$$ where $a=\delta\epsilon$. ( Integrate[BesselJ[1, x]/x^2, {x, a, Infinity}] ) How should we treat $$ \lim_{a\to 0} a^2\, {}_2F_3 \left(1,1;2,2,3;-\frac{a^2}{4} \right) \quad ? $$ $\endgroup$ – GotchaP Jul 15 '17 at 3:51
  • 1
    $\begingroup$ In the Wolfram documentation (functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F3/…) you can find lots of info about this function and, specifically, its value at zero argument, $$ _2F_3({a_1,a_2},{b_1,b_2,b_3},0) = 1 $$ The limit you seek is therefore zero and the divergence of the integral is logarithmic. $\endgroup$ – photonQ Jul 16 '17 at 21:29
  • $\begingroup$ Thanks so much! Owing to you, I've realized the real usefullness of the Wolfram website. $\endgroup$ – GotchaP Jul 17 '17 at 2:44

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.