Verify $\int_{-\infty}^{\infty}\frac{1}{\pi\hbar}\cos\left(\frac{p|x|}{\hbar}+\delta\right)\cos\left(\frac{p'|x|}{\hbar}+\delta\right)dx=\delta(p-p')$

Question

I'm trying to verify that $$\int_{-\infty}^{+\infty}\frac{1}{\pi\hbar}\cos\left(\frac{p|x|}{\hbar}+\delta\right)\cos\left(\frac{p'|x|}{\hbar}+\delta\right)dx=\delta(p-p'),\quad(p,p'\ge0)$$ but I can't manage to show this. I'm given the following identity $$\int_{-\infty}^{\infty}\frac{e^{-ikx}}{\sqrt{2\pi}}\frac{e^{ik'x}}{\sqrt{2\pi}}dx=\delta(k-k'),\quad(-\infty<k,k'<+\infty)$$ but I get tripped up with the absolute values when trying to manipulate the expression into that form. Perhaps there's a simple trick that I'm missing? I've been struggling with this for a while so I might be stuck in a rut.

Answer 1

First: the phase number $\delta$ let's change to $\alpha,$ otherwise it will mess with the notation of the Dirac-delta.

Second: notice that, as suggested by @Thomas Russell, $$\;\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2}$$ and $$\cos z_{1}\cos z_{2}=\dfrac{(e^{iz_{1}}+e^{-iz_{1}})(e^{iz_{2}}+e^{-iz_{2}})}{4}=\dfrac{(e^{i(z_{1}+z_{2})}+e^{i(z_{1}-z_{2})}+e^{-i(z_{1}-z_{2})}+e^{-i(z_{1}+z_{2})})}{4}$$

so your integrand becomes $$\;\cos\left(\dfrac{p|x|}{\hbar}+\alpha\right)\cos\left(\dfrac{p'|x|}{\hbar}+\alpha\right)=\dfrac{1}{4}\left(e^{i(p+p')\frac{|x|}{\hbar}+2i\alpha}+e^{i(p-p')\frac{|x|}{\hbar}}+e^{-i(p-p')\frac{|x|}{\hbar}}+e^{-i(p+p')\frac{|x|}{\hbar}-2i\alpha}\right).$$

When integrating and using the Dirac-delta identities

$$\int_{-\infty}^{+\infty}e^{i\frac{p}{\hbar}|x|}dx=2\pi\hbar\delta(p),\;\;\delta(p)=\delta(-p)$$ one gets $$\int_{-\infty}^{+\infty}\dfrac{1}{\hbar\pi}\cos\left(\dfrac{p|x|}{\hbar}+\alpha\right)\cos\left(\dfrac{p'|x|}{\hbar}+\alpha\right)dx=\dfrac{1}{4\pi\hbar}\left(e^{2i\alpha}\int_{-\infty}^{+\infty}e^{i(p+p')\frac{|x|}{\hbar}}dx+\int_{-\infty}^{+\infty}e^{i(p-p')\frac{|x|}{\hbar})}dx+\int_{-\infty}^{+\infty}e^{-i(p-p')\frac{|x|}{\hbar}}dx+e^{-2i\alpha}\int_{-\infty}^{+\infty}e^{-i(p+p')\frac{|x|}{\hbar}}dx\right);$$

\begin{align}\int_{-\infty}^{+\infty}\dfrac{1}{\hbar\pi}\cos\left(\dfrac{p|x|}{\hbar}+\alpha\right)\cos\left(\dfrac{p'|x|}{\hbar}+\alpha\right)dx=\delta(p+p')\cos(2\alpha)+\delta(p-p'). \end{align}

The above formula is general, for every $p,p'$ and $\alpha$

Third: for $p,p'>0,$ then the first term in the above formula vanishes and one ends with the desire result: \begin{align}\int_{-\infty}^{+\infty}\dfrac{1}{\hbar\pi}\cos\left(\dfrac{p|x|}{\hbar}+\alpha\right)\cos\left(\dfrac{p'|x|}{\hbar}+\alpha\right)dx=\delta(p-p'). \end{align}