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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
    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
    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... $\endgroup$ Jun 27, 2021 at 15:46

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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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