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Here is the challenge integral appearing on a physics book:

To prove the following equality $$\iiint_{\mathbb{R}^3}\frac{1}{(2\pi)^3}\frac{e^{-i\vec{q}\cdot\vec{r}}}{|\vec{q}|^2+m^2}dq_xdq_ydq_z=\frac{e^{-m|\vec{r}|}}{4\pi|\vec{r}|} $$ Here, $\vec{r}=x\vec{i}+y\vec{j}+z\vec{k}$, $m$ is non-negative.

I think this should be a Fourier transformation, specifically, I find that change the sign of $\vec{q}$ can keep the left-hand side unchanged. Hence, one can have the integrand being $\frac{1}{(2\pi)^3}\frac{e^{i\vec{q}\cdot\vec{r}}}{|\vec{q}|^2+m^2}$, which is now a typical inverse Fourier transform with the transforming function being $\frac{1}{|\vec{q}|^2+m^2}$. However, when I do the transformation part by part, i.e. do $q_x$ then $q_y$ then $q_z$, after transforming $q_x$, the integrand will have a square root term which will be hard to continue. Then I get stuck in this way. On the other hand, I try to express the whole integral in the spherical coordinate, notice, one can show that this triple integral is unchanged under any rotation (specifically, given a rotation matrix $R$, then $\vec{q}\cdot\vec{r}$ and $|\vec{q}|^2$ is unchanged under action of $R$, and notice $d^3q:=dq_xdq_ydq_z$ under the transformed frame $d^3q’=det\,R d^3q$ where $R\in SO_3\Rightarrow det\, R=1$, hence it is unchanged). One can choose direction of $\vec{r}$ to be alone $z$-axis. Then, if my calculation is correct (which seems not true), one need to calculate $\int_0^\pi d\varphi e^{-m|\vec{r}|\cos(\varphi)}$ which does not have a simple original function. Can someone gives me some hint? It can be other solution method or the spherical one. For the spherical one, could you point out which part is easy to make mistakes. Thanks in advance.

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  • $\begingroup$ I didn't read past the first line. What does sign of $\vec q$ mean? If you mean changing $\vec q$ to $-\vec q$, how does that not change the LHS? $\endgroup$ – Ted Shifrin Feb 12 '18 at 2:24
  • $\begingroup$ @TedShifrin I mean the sign before $\vec{q}$. Since for a standard definition of Fourier transform, the kernel is $e^{-i\omega t}$, while the inverse Fourier transform has the kernel $e^{i\omega t}$. Hence, the sign matters. $\endgroup$ – Hamio Jiang Feb 12 '18 at 2:26
  • $\begingroup$ But isn't there a negative sign when you transform the triple integral? When you send $(x,y,z)$ to $(-x,-y,-z)$, there is a negative in the Jacobian. This is not a single rotation. $\endgroup$ – Ted Shifrin Feb 12 '18 at 2:27
  • $\begingroup$ @TedShifrin Yes, typically, there will be a minus sign, but you also change the integral domain say $\int_{\infty}^{-\infty}$, then you use the minus sign to swap the upper and lower bound which leads you $\int_{-\infty}^{\infty}$; hence, the minus sign is absorbed into the integral domain. $\endgroup$ – Hamio Jiang Feb 12 '18 at 2:31
  • $\begingroup$ OK. Typically, one has the absolute value of the Jacobian when one writes down the change of variables theorem. You're right. My apologies in misconstruing. $\endgroup$ – Ted Shifrin Feb 12 '18 at 2:33
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Caution. Within the purview of mathematics, the integral is not absolutely convergent and thus an appropriate interpretation must be introduced. In this case, the integral can be considered as the limit

$$ \lim_{R\to\infty} \frac{1}{(2\pi)^3} \int_{B_R(0)} \frac{e^{-i\langle q,r\rangle}}{\|q\|^2+m^2} \, \mathrm{d}q. $$

Now let me convince you that spherical coordinates works here. By the rotational symmetry we may assume that $r = (0, 0, \|r\|)$. Then

\begin{align*} \frac{1}{(2\pi)^3} \int_{B_R(0)} \frac{e^{-i\langle q,r\rangle}}{\|q\|^2+m^2} \, \mathrm{d}q &= \frac{1}{(2\pi)^3} \int_{0}^{R} \int_{0}^{\pi} \frac{e^{-i\rho \|r\| \cos \phi}}{\rho^2 + m^2} \, 2\pi\rho^2 \sin\phi \, \mathrm{d}\phi\mathrm{d}\rho \\ &= \frac{1}{(2\pi)^2} \int_{0}^{R} \frac{\rho^2}{\rho^2 + m^2} \left[ \frac{e^{-i\rho \|r\| \cos \phi}}{i\rho\|r\|} \right]_{0}^{\pi} \,\mathrm{d}\rho \\ &= \frac{1}{2\pi^2\|r\|} \int_{0}^{R} \frac{\rho \sin(\rho \|r\|)}{\rho^2 + m^2} \,\mathrm{d}\rho \\ &\xrightarrow[R\to\infty]{} \frac{1}{2\pi^2\|r\|} \int_{0}^{\infty} \frac{\rho \sin(\rho \|r\|)}{\rho^2 + m^2} \,\mathrm{d}\rho. \end{align*}

Now the last integral can be evaluated in various ways, including contour integral, and we are led to

$$ \lim_{R\to\infty} \frac{1}{(2\pi)^3} \int_{B_R(0)} \frac{e^{-i\langle q,r\rangle}}{\|q\|^2+m^2} \, \mathrm{d}q = \frac{1}{2\pi^2\|r\|} \cdot \frac{\pi}{2} e^{-m\|r\|} = \frac{e^{-m\|r\|}}{4\pi \|r\|}. $$


Alternatively, one may plug

$$ \frac{1}{\|q\|^2 +m^2} = \int_{0}^{\infty} e^{-(\|q\|^2+m^2)s} \, \mathrm{d}s$$

and interchange the order of integration to obtain

\begin{align*} \frac{1}{(2\pi)^3} \int_{\mathbb{R}^3} \frac{e^{-i\langle q,r\rangle}}{\|q\|^2+m^2} \, \mathrm{d}q &= \frac{1}{(2\pi)^3} \int_{0}^{\infty} e^{-m^2s} \prod_{k=1}^{3} \left( \int_{\mathbb{R}} e^{-s q_k^2 - iq_k r_k} \, \mathrm{d}q_k \right) \, \mathrm{d}s \\ &= \frac{1}{(2\pi)^3} \int_{0}^{\infty} e^{-m^2s} \prod_{k=1}^{3} \left( \sqrt{\frac{\pi}{s}} e^{-r_k^2/4s} \right) \, \mathrm{d}s \\ (s=\|r\|^2/4t^2) \qquad &= \frac{1}{2\pi^{3/2}\|r\|} \int_{0}^{\infty} e^{ - t^2-\frac{m^2\|r\|^2}{4t^2}} \, \mathrm{d}t. \end{align*}

Again, one may want to introduce some modification of this argument to justify applying the Fubini's theorem to a non-absolutely convergent integral. In this case, Gaussian regularization is enough. Finally, the last integral can be computed by the Glasser's master theorem.

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  • $\begingroup$ Thanks! I think I made an mistake since I forget there is a $sin\varphi$ in the spherical coordinate. $\endgroup$ – Hamio Jiang Feb 12 '18 at 3:24
  • $\begingroup$ By the way, I think I still could do the Fourier transform to get some thing directly from manuals (though its calculation is based on the contour integral as you said.) $\endgroup$ – Hamio Jiang Feb 12 '18 at 3:26

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