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In Srednicki, $$\begin{align*} \left[\varphi^{+}(x),\varphi^{-}(x^\prime)\right]_\mp=&\int\widetilde{\mathrm{d}k}\int\widetilde{\mathrm{d}k^\prime}\mathrm{e}^{\mathrm{i}(kx-k^\prime x^\prime)}\left[a(\boldsymbol{k}),a^\dagger(\boldsymbol{k^\prime})\right]\\ =&\int\frac{\mathrm{d}^3k}{(2\pi)^22\omega}\mathrm{e}^{\mathrm{i}k(x-x^\prime)}\tag{4.12,halfway} \end{align*}$$ where $\omega=\sqrt{\boldsymbol{k}^2+m^2}$.

Srednicki says this formula is equal to $$ \frac{m}{4\pi^2 r}K_1(mr)\tag{4.12} $$ where $K_1$ is modified Bessel function, and $r^2=(x-x^\prime)^2>0$.

The answer says this formula can be rearranged: $$\begin{align*} \int\frac{\mathrm{d}^3k}{(2\pi)^22\omega}\mathrm{e}^{\mathrm{i}k(x-x^\prime)}=&\frac{2\pi}{2(2\pi)^3} \int_0^\infty\frac{\mathrm{d}k}{\omega}k^2\int_{-1}^1\mathrm{d}\cos{\theta}\mathrm{e}^{\mathrm{i}kr\cos{\theta}}\\ =&\frac{1}{4\pi^2 r}\int_0^\infty\mathrm{d}k\frac{k\sin{kr}}{\sqrt{k^2+m^2}}. \end{align*}$$ This is just a polar conversion.

Why is the last formula equal to eq. (4.12)?

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    $\begingroup$ Hi Anzu Ariake. Welcome to Phys.SE. Did you check the known integral representations for the Bessel $K_1$ function? $\endgroup$
    – Qmechanic
    Commented Jun 27, 2021 at 7:26
  • $\begingroup$ @Qmechanic The equation he is seeking is exactly Eq. (5) in your webpage after integration by part. $\endgroup$
    – Youran
    Commented Jun 27, 2021 at 15:39
  • $\begingroup$ Oh, I understood its series expansion but didn't know the integral form. We must lead the form: $K_1(mr)=mr\int_0^\infty\frac{\cos{k}}{(k^2+m^2r^2)^{\frac{3}{2}}}\mathrm{d}k$, but I have no idia... $\endgroup$ Commented Jun 27, 2021 at 15:46

2 Answers 2

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Let's just start from the integral you got in the last part of your question. \begin{align} \frac{1}{4\pi^2 r}\int_0^\infty\mathrm{d}k\frac{k\sin{kr}}{\sqrt{k^2+m^2}}&=\frac{1}{4\pi^2r^2}\int^{\infty}_{0}\frac{-k\ \mathrm{d}(\cos{k}r)}{\sqrt{k^2+m^2}}\\&=\frac{1}{4\pi^2r^2}\int^{\infty}_{0}\cos{kr}\ \mathrm{d}\left(\frac{k}{\sqrt{k^2+m^2}}\right)\mathrm{(integration\ by\ parts)}\\&=\frac{m^2}{4\pi^2r^2}\int^{\infty}_{0}\frac{\cos kr}{\left(k^2+m^2\right)^\frac{3}{2}}\mathrm{d}k\\&=\frac{m^2}{4\pi^2}\int^{\infty}_{0}\frac{\cos kr}{\left(k^2 r^2+m^2r^2\right)^{\frac{3}{2}}}\mathrm{d}(rk)\\&=\frac{m^2}{4\pi^2}\int^{\infty}_{0}\frac{\cos t}{\left(t^2+m^2r^2\right)^{\frac{3}{2}}}\mathrm{d}t \end{align} Notice that $$K_{1}(z)=z\int^{\infty}_{0}\frac{\cos t \ \mathrm{d}t}{\left(t^2+z^2\right)^{\frac{3}{2}}}.$$ So we finally have \begin{align} \frac{m^2}{4\pi^2}\int^{\infty}_{0}\frac{\cos t}{\left(t^2+m^2r^2\right)^{\frac{3}{2}}}\mathrm{d}t=\frac{m}{4\pi^2 r}K_{1}(mr) \end{align}

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Sorry for reviving a really old question. My answer is very similar to Shangjie Zhou's, with just a small twist. I will simply apply Feynman's trick.

I will also start from the same point, but use two substitutions, $k=mq=m\sinh t$. In the following, Equation (3) and Equation (5) appear in DLMF to directly lead to Equation (6), see their Equation (10.32.6) and then Equation (7) follows from another page in DLMF, Equation (10.29.3) $$ \begin{align} \tag1 \frac1{4\pi^2r}\int_0^\infty\mathrm dk\,\frac{k\sin kr}{\sqrt{k^2+m^2}} &=-\frac1{4\pi^2r}\frac{\mathrm d\ }{\mathrm dr}\int_0^\infty\mathrm dk\,\frac{\cos kr}{\sqrt{k^2+m^2}}\\ \tag2&=-\frac1{4\pi^2r}\frac{\mathrm d\ }{\mathrm dr}\int_0^\infty m\,\mathrm dq\,\frac{\cos mqr}{m\sqrt{q^2+1}}\\ \tag3&=-\frac1{4\pi^2r}\frac{\mathrm d\ }{\mathrm dr}\int_0^\infty\mathrm dq\,\frac{\cos mqr}{\sqrt{q^2+1}}\\ \tag4&=-\frac1{4\pi^2r}\frac{\mathrm d\ }{\mathrm dr}\int_0^\infty\cosh t\,\mathrm dt\,\frac{\cos(mr\sinh t)}{\cosh t}\\ \tag5&=-\frac1{4\pi^2r}\frac{\mathrm d\ }{\mathrm dr}\int_0^\infty\mathrm dt\,\cos(mr\sinh t)\\ \tag6&=-\frac1{4\pi^2r}\frac{\mathrm d\ }{\mathrm dr}K_0(mr)\\ \tag7&=\frac m{4\pi^2r}K_1(mr) \end {align} $$ I hope that this is easier to follow.

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