Close formula of Yukawa Potential in 3 dimension by integrating $\int_{0}^{\infty}{\frac{1}{(4\pi t)^{\frac{d}{2}}}e^{-\frac{x^2}{4t} - \mu^2t}}dt$

Question

In PDE, Yukawa potential can be calculate as $$G^\mu(x) = \int_{0}^{\infty} \frac{1}{(4\pi t)^{\frac{d}{2}}}e^{-\frac{x^2}{4t} - \mu^2t}dt$$

When $d = 3$ we can get the close formula $G^\mu(x) = \frac{1}{4\pi|x|}e^{-\mu|x|}$.

How should we do the integral?

Yukawa potentials are sometimes/often calculated using complex analysis and residue theorem. Maybe that can be applied here also — Mathphys meister, Dec 23 '19 at 17:47
I can not find any reference showing the process of integrating. The result is included in the analysis textbook written by Lieb. — EggTart, Dec 23 '19 at 17:51

Jacky Chong · Answer 1 · 2019-12-27T09:54:16.447

Let \begin{align} I(\mu, x) = \int^\infty_0 dt\ \frac{1}{(4\pi t)^{3/2}}\exp\left( -\frac{x^2}{4t}-\mu^2 t\right). \end{align} Then we see that \begin{align} I(\mu, x) = \int^\infty_0 dt\ \frac{1}{(4\pi t)^{3/2}}\exp\left( -\frac{x^2}{2}\left(\frac{1}{\sqrt{2t}}-\frac{\mu}{|x|} \sqrt{2t}\right)^2\right)\exp\left(-\mu |x|\right). \end{align} Set $u = \frac{1}{\sqrt{2t}}$ and $\alpha = \frac{\mu}{|x|}$ then it follows \begin{align} I(\mu, x) = \int^\infty_0 du\ \exp\left(-\frac{1}{2}x^2\left(u-\frac{\alpha}{u}\right)^2\right)(2\pi)^{-3/2}\exp(-\mu|x|) = \frac{e^{-\mu|x|}}{(2\pi)^{3/2}}J(\mu, x). \end{align} Hence it suffices to evaluate $J(\mu, x)$. Set $z=\alpha/u$ then we have \begin{align} J(\mu, x) = \int^\infty_0 \frac{dz}{z^2}\ \alpha\exp\left(-\frac{1}{2}x^2\left(z-\frac{\alpha}{z}\right)^2\right) \end{align} which means \begin{align} 2J(\mu, x) =&\ \int^\infty_0 du\ \left(1+\frac{\alpha}{u^2} \right)\exp\left(-\frac{1}{2}x^2\left(u-\frac{\alpha}{u}\right)^2\right)\\ =&\ \int^\infty_0 d\left(u-\frac{\alpha}{u} \right)\exp\left(-\frac{1}{2}x^2\left(u-\frac{\alpha}{u}\right)^2\right)\\ =& \int^\infty_{-\infty} dw\ \exp\left(-\frac{1}{2}x^2w^2\right)= \sqrt{2\pi}\frac{1}{|x|}. \end{align} Hence it follows \begin{align} I(\mu, x) = \frac{e^{-\mu|x|}}{(2\pi)^{3/2}}J(\mu, x) = \sqrt{\frac{\pi}{2}}\frac{1}{|x|}\frac{e^{-\mu|x|}}{(2\pi)^{3/2}} = \frac{e^{-\mu |x|}}{4\pi |x|}. \end{align}

Additional: Observe \begin{align} \int^\infty_0 dt\ \frac{1}{(4\pi t)^{5/2}}\exp\left( -\frac{x^2}{4t}-\mu^2 t\right) =&\ \int^\infty_0 dt\ \frac{1}{(4\pi t)^{3/2}}\left(-\frac{1}{2\pi x}\right)\frac{d}{dx}\exp\left( -\frac{x^2}{4t}-\mu^2 t\right)\\ =&\ \left(-\frac{1}{2\pi x}\right)\frac{d}{dx}I(\mu, x) = \frac{e^{-\mu|x|}(\mu|x|+1)}{8\pi^2 |x|^3}. \end{align} Using this observation, we can recover all expression of $G^\mu(x)$ for odd dimension $d$.

Here you set $u=1/\sqrt{2t}$, then $dt=-u^3du$ right? Why there is no "-" after changing variable? — Hermi, Mar 24 '20 at 05:51

score 1 · Answer 2 · answered Dec 24 '19 at 06:11

Using a CAS $$G^\mu(x) = \int_{0}^{\infty} \frac{1}{(4\pi t)^{\frac{d}{2}}}e^{-\frac{x^2}{4t} - \mu^2t} dt$$ is given, before any simplication or assumptions by $$G^\mu(x)= (2 \pi )^{-\frac d2} \left(\frac{\mu ^2}{x^2}\right)^{\frac{d-2}{4}} K_{\frac{d-2}{2}}\left(\sqrt{x^2 \mu ^2}\right)$$ which gives your result for $d=3$.

Bessel functions only appear for even values of $d$.

Close formula of Yukawa Potential in 3 dimension by integrating $\int_{0}^{\infty}{\frac{1}{(4\pi t)^{\frac{d}{2}}}e^{-\frac{x^2}{4t} - \mu^2t}}dt$

2 Answers2