$$\int_0^\infty\int_0^\pi\frac{k^2\left(e^{-it\sqrt{k^2+m^2}}-e^{it\sqrt{k^2+m^2}}\right)\sin(\theta)}{e^{-ikx\cos{\theta}}\sqrt{k^2+m^2}}d\theta dk$$
I saw this Integral at Quora, and I have not idea how to evaluate it. Therefore, I thought of posting it here.
How do we Evaluate this?
Best Answer
You can write:
$$\int_0^\infty\int_0^\pi\frac{k^2\left(e^{-it\sqrt{k^2+m^2}}-e^{it\sqrt{k^2+m^2}}\right)\sin(\theta)}{e^{-ikx\cos{\theta}}\sqrt{k^2+m^2}}d\theta dk$$
As:
$$\int_0^\infty\int_0^\pi\frac{k^2\left(-2 i \sin{(t \sqrt{k^2+m^2})}\right)\sin(\theta)}{e^{-ikx\cos{\theta}}\sqrt{k^2+m^2}}d\theta dk = \int_0^\infty \frac{k^2\left(-2 i \sin{(t\sqrt{k^2+m^2})}\right)}{\sqrt{k^2+m^2}} \int_0^\pi e^{ikx\cos{\theta}} \sin(\theta) d\theta dk = \int_0^\infty \frac{k^2\left(-4 i \sin{(t\sqrt{k^2+m^2})}\right)}{\sqrt{k^2+m^2}} \frac{\sin{k x}}{k x} dk=\int_0^\infty -4 ik^2 t sinc{(t\sqrt{k^2+m^2})} sinc(k x) dk= -4 t i \int_0^\infty k^2 sinc{(t\sqrt{k^2+m^2})} sinc(k x) dk $$ Now by using the relationship: $$sinc{(t\sqrt{k^2+m^2})}= \prod_{n=1}^{+\infty}(1-\frac{(t\sqrt{k^2+m^2})^2}{n^2 \pi^2})$$ And: $$sinc{(k x)}= \prod_{j=1}^{+\infty}(1-\frac{(k x)^2}{j^2 \pi^2})$$ We can write $$-4 t i \int_0^\infty k^2 sinc{(t\sqrt{k^2+m^2})} sinc(k x) dk = -4 t i \lim_{\phi \rightarrow +\infty}\prod_{n=1}^{+\infty} \prod_{j=1}^{+\infty} \int_0^\phi k^2 (1-\frac{(t\sqrt{k^2+m^2})^2}{n^2 \pi^2}) (1-\frac{(k x)^2}{j^2 \pi^2}) = -4 t i \lim_{\phi \rightarrow +\infty}\prod_{n=1}^{+\infty} \prod_{j=1}^{+\infty} \left[ \frac{\phi^3}{3}(1-(\frac{m t}{n \pi})^2)+\frac{\phi^5}{5}((\frac{m t x}{j n \pi^2})^2-(\frac{t}{n \pi})^2-(\frac{x}{j \pi})^2)+\frac{\phi^7}{7}(\frac{t x }{j n \pi^2})^2\right]$$ Which seems to diverge