Simplifying/computing an integral

Question

I am attempting to compute some expectation value in a physics problem. I have computed that given $\psi_n(x)=\sqrt{\frac{2}{a}}\sin(n\pi x/a)$, we get $$\phi_n(k)=\frac{1}{\sqrt{2\pi}}\int_0^a\psi_n(x)e^{-ikx}dx=\\ \frac{1}{\sqrt{\pi a}}\frac{a}{a^2k^2-n^2\pi^2}e^{-ika}(-e^{ika}n\pi+n\pi\cos(n\pi)+iak\sin(n\pi))$$ My goal is to find the expectation of $k$, which is computed as $\int_{-\infty}^\infty k\phi^*_n(k)\phi(k)dk$. I have verified that my computation for $\phi_n(k)$ is correct, and hence $\phi_n^*(k)=\frac{1}{\sqrt{\pi a}}\frac{a}{a^2k^2-n^2\pi^2}e^{ika}(-e^{-ika}n\pi+n\pi\cos(n\pi)-iak\sin(n\pi))$. This leads to a rather convoluted expression for $\phi^*_n(k)\phi_n(k)$: $$\frac{1}{\pi a}\frac{a^2}{(a^2k^2-n^2\pi^2)^2}(-e^{ika}n\pi+n\pi\cos(n\pi)+iak\sin(n\pi))(-e^{-ika}n\pi+n\pi\cos(n\pi)-\\iak\sin(n\pi))$$ which results in $$\frac{1}{\pi}\frac{a}{(n^2\pi^2-a^2k^2)^2}(a^2k^2\sin^2(n\pi)+n^2\pi^2\sin^2(ak)+n^2\pi^2\cos^2(ak)-2\pi^2n^2\cos(n\pi)\cos(ak)-2\pi akn\sin(n\pi)\sin(ak)+\pi^2n^2\cos^2(n\pi))$$ which "simplifies" to the following expression $$\frac{1}{\pi}\frac{a}{(n^2\pi^2-a^2k^2)^2}(n^2\pi^2+a^2k^2\sin^2(\pi n)+\pi^2n^2\cos^2(\pi n)-2\pi^2n^2\cos(\pi n)\cos(ak)-2\pi akn\sin(\pi n)\sin(ak))$$ Relabeling to make it look a little less messy, $p=ak$ and $q=n\pi$, this reads $$\frac{1}{\pi}\frac{a}{(q^2-p^2)^2}(q^2+p^2\sin^2(q)+q^2\cos^2(q)-2q^2\cos(q)\cos(p)-2pq\sin(q)\sin(p))$$ However, this expression still seems completely unworkable to me. If I want to integrate this (multiplied by $p$, since I want the expectation) with respect to $p$, one of the integrals is going to be $$\frac{-2aq}{\pi}\sin(q)\int\frac{p^2}{(q^2-p^2)^2}\sin(p)dp$$ I do not know how to manipulate an integral like this to solve it analytically. In fact, I am not even sure if this is feasible. Did I make a mistake simplifying this expression somewhere along the way, ending up with a more convoluted integral than I really need? If not, would anyone be able to provide me with a tool to solve these integrals?

Answer 1

Using @Andrei's comment, the expression simplifies to $$ \phi^*_n\phi_n= \begin{cases} \frac{2}{\pi}a(1-\cos(p))(\frac{q}{q^2-p^2})^2,& \text{if } n\text{ is even}\\ \frac{2}{\pi}a(1+\cos(p))(\frac{q}{q^2-p^2})^2, & \text{otherwise} \end{cases} $$ from which the expectation of $k^2$ can be solved using the identity $\int_{-\infty}^\infty p^2(1\pm\cos(p))(\frac{q}{q^2-p^2})^2dp=\mp \frac{\pi q}{2}(\sin(q)+q\cos(q))$, and the expectation of $k$ using the symmetry of the integrand. Thanks to him for pointing out my oversight.