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.

  • $\begingroup$ If you use your series and perform integration by parts repeatedly you will get $$\int\limits_{0}^{1}{{{P}_{n}}\left( 1-2{{u}^{2}} \right){{e}^{2i\alpha u}}du}=\frac{{{e}^{2i\alpha }}}{2i\alpha }\sum\limits_{k=0}^{n}{\frac{{{\left( -1 \right)}^{k}}}{k{{!}^{2}}}\frac{\left( n+k \right)!}{\left( n-k \right)!}}\left\{ -\frac{\left( 2k \right)!}{{{\left( 2i\alpha \right)}^{2k+1}}}+\sum\limits_{m=0}^{2k}{\frac{{{\left( i \right)}^{m}}\left( 2k \right)!}{{{\left( 2\alpha \right)}^{m}}\left( 2k-m \right)!}} \right\}$$ $\endgroup$ May 27, 2021 at 7:35
  • $\begingroup$ 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. $\endgroup$ May 27, 2021 at 7:37

1 Answer 1


Assuming $0<u<v$ 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 \begin{equation} \int_{-1}^1P_n(x)P_{\ell}(x)\,dx=\frac{2}{2\ell+1}\delta_{\ell n} \end{equation} or \begin{equation} \int_0^\pi P_n(\cos\theta)P_{\ell}(\cos\theta)\sin\theta\,d\theta=\frac{2}{2\ell+1}\delta_{\ell n} \end{equation} 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 \begin{equation} \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) \end{equation} Now, changing $\cos\theta=1-2t^2$ results in the following expression \begin{equation} \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) \end{equation} which is identical to the proposed identity.


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