# How to calculate the integral of a Legendre polynomial

I would like to show that $$\int_{0}^{1}P_{l}(1-2u^{2})e^{2i\alpha u}du=i\alpha j_{l}(\alpha)h_{l}(\alpha)$$ where $$P_{l}(x)$$ are the Legendre polynomials, $$\alpha$$ is a positive constant and $$j_{l}$$ and $$h_{l}$$ are the spherical Bessel and Hankel functions of the first kind, respectively.

I can use an expansion like $$P_{l}(1-2u^2)=\sum_{k=0}^{l}{l\choose k}{-l-1\choose k}u^{2k}$$ but this leads to a solution as a sum of incomplete Gamma functions, so I'm a bit stuck.

Thanks in advance for any help.

• If you use your series and perform integration by parts repeatedly you will get $$\int\limits_{0}^{1}{{{P}_{n}}\left( 1-2{{u}^{2}} \right){{e}^{2i\alpha u}}du}=\frac{{{e}^{2i\alpha }}}{2i\alpha }\sum\limits_{k=0}^{n}{\frac{{{\left( -1 \right)}^{k}}}{k{{!}^{2}}}\frac{\left( n+k \right)!}{\left( n-k \right)!}}\left\{ -\frac{\left( 2k \right)!}{{{\left( 2i\alpha \right)}^{2k+1}}}+\sum\limits_{m=0}^{2k}{\frac{{{\left( i \right)}^{m}}\left( 2k \right)!}{{{\left( 2\alpha \right)}^{m}}\left( 2k-m \right)!}} \right\}$$ May 27, 2021 at 7:35
• In there we can start to see why they're spherical Bessel functions - just have to untangle it (easier said than done). Must be a nicer way than this i think. May 27, 2021 at 7:37

Assuming $$0 and $$w=\sqrt{u^2+v^2-2uv\cos\theta}$$, combining the Gegenbauer expansions for half-integer orders here and here gives \begin{align} \frac{\exp iw}{w}&=\sum_{n=0}^{\infty}(2n+1)\mathsf{j}_{n}\left(v\right)(-\mathsf{y}_{n}\left(u\right)+i\mathsf{j}_{n}\left(u\right))P_{n}\left(\cos\theta\right)\\ &=i\sum_{n=0}^{\infty}(2n+1)\mathsf{j}_{n}\left(v\right)\mathsf{h}_{n}^{(1)}\left(u\right)P_{n}\left(\cos\theta\right) \end{align}
The orthogonality of the Legendre polynomials reads $$$$\int_{-1}^1P_n(x)P_{\ell}(x)\,dx=\frac{2}{2\ell+1}\delta_{\ell n}$$$$ or $$$$\int_0^\pi P_n(\cos\theta)P_{\ell}(\cos\theta)\sin\theta\,d\theta=\frac{2}{2\ell+1}\delta_{\ell n}$$$$ The obtained identity is now projected on $$P_\ell(\cos\theta)$$ after multiplication by $$\sin\theta$$: \begin{align} \int_{0}^\pi \frac{\exp iw}{w}P_l(\cos\theta)\sin\theta\,d\theta=i\mathsf{j}_{\ell}\left(v\right)\mathsf{h}_{\ell}^{(1)}\left(u\right) \end{align} (only the term $$n=\ell$$ survives in the summation). Both sides of the above identity are continuous functions of $$u$$ and $$v$$, the obtained expression remains valid for $$u=v$$. Choosing $$u=v=\alpha$$, it writes $$$$\int_{-1}^1 \frac{\exp i\alpha\sqrt{2}\sqrt{1-\cos\theta}}{\alpha\sqrt{2}\sqrt{1-\cos\theta}}P_l(\cos\theta)\sin\theta\,d\theta=2i\mathsf{j}_{\ell}\left(\alpha\right)\mathsf{h}_{\ell}^{(1)}\left(\alpha\right)$$$$ Now, changing $$\cos\theta=1-2t^2$$ results in the following expression $$$$\int_0^1 e^{2i\alpha t}P_\ell(1-2t^2)\,dt=i\alpha\mathsf{j}_{\ell}\left(\alpha\right)\mathsf{h}_{\ell}^{(1)}\left(\alpha\right)$$$$ which is identical to the proposed identity.