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?


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!

  • $\begingroup$ 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
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    $\begingroup$ 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

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