The integral I need to solve is:
$$\int^{\pi}_{0} P_{l}(\cos(\theta))\delta(\theta-\frac{\pi}{2})\sin(\theta)d\theta.$$
where $\theta$ is the angle between $z$-axis and $x-y$ plane. Since I don't know the values of $l$, I thought this integral could be written as
$$\int^{1}_{-1} P_{l}(\cos(\theta))\delta(\cos(\theta))d(\cos(\theta))$$
Am I correct ? If yes, how can I evaluate the latter ?
Thanks in advance.
Using \begin{align} \int_{0}^{\pi} f(\theta) \, \delta\left(\theta - \frac{\pi}{2}\right) \, d\theta &= f\left(\frac{\pi}{2}\right) \\ P_{2n}(0) &= \frac{(-1)^{n} \, \Gamma\left(n + \frac{1}{2}\right)}{\sqrt{\pi} \, n!} \\ P_{2n+1}(0) &= 0 \end{align} then $$\int^{\pi}_{0} P_{n}(\cos(\theta))\delta(\theta-\frac{\pi}{2})\sin(\theta)d\theta = P_{n}\left(\cos\left(\frac{\pi}{2}\right)\right) \, \sin\left(\frac{\pi}{2}\right) = P_{n}(0)$$ and leads to \begin{align} \int^{\pi}_{0} P_{2n}(\cos(\theta))\delta(\theta-\frac{\pi}{2})\sin(\theta)d\theta &= \frac{(-1)^{n} \, \Gamma\left(n + \frac{1}{2}\right)}{\sqrt{\pi} \, n!} \\ \int^{\pi}_{0} P_{2n+1}(\cos(\theta))\delta(\theta-\frac{\pi}{2})\sin(\theta)d\theta &= 0. \end{align}