I'm preparing notes for an undergrad physics course I'm going to be teaching soon. Unfortunately, this sort of stuff was taught to me only in a very handwavy sort of way ("you take these physical principles and these equations and then skip over a bunch of maths and you eventually get here"). I do realise why it is done in such a way, as the maths are pretty dense (for an undergrad, anyway) and the result is what's important, but, well, I've always had a soft spot for mathematical rigour.

Without further ado, here's where I'm stuck: I'm trying to show that $$\int_0^\pi e^{i\rho\cos(\theta)}\left(\cos(\theta)\,\frac{\text{d}P_\ell^1\left(\cos(\theta)\right)}{\text{d}\theta}+\frac{P_\ell^1\left(\cos(\theta)\right)}{\sin(\theta)}\right)\sin(\theta)\,\text{d}\theta = \frac{2\ell(\ell+1)}{\rho}\,\frac{\text{d}}{\text{d}\rho}\left(\rho j_\ell(\rho)\right),$$ where $P_\ell^1$ is the associated Legendre function of order $_\ell^1$ and $j_\ell$ is the spherical Bessel function of the first kind of order $\ell$. (In case anyone is interested, the above is equation 4.35 of Bohren C F & Huffman D R (1983): Absorption and scattering of light by small particles, Wiley. I've been following that book for this part of the course, and this is the first time I've been this stuck.)

I've been trying to use the identities $$\int_0^\pi\left(P_\ell^m\left(\cos(\theta)\right)\right)^2\sin(\theta)\,\text{d}\theta = \frac{2(\ell+m)!}{(2\ell+1)(\ell-m)!},\\ \int_{-1}^1\frac{P_\ell^m(x)^2}{1-x^2}\,\text{d}x = \frac{(\ell+m)!}{m(\ell-m)!},\\ 2i^\ell j_\ell(\rho) = \int_0^\pi e^{i\rho\cos(\theta)}P_\ell\left(\cos(\theta)\right)\sin(\theta)\,\text{d}\theta,$$ where $P_\ell$ is the Legendre polynomial of order $\ell$ and is equal to $P_\ell^0$ by definition of the $P_\ell^m$. (Obviously the third one has to be used at some point; the others might or might not be, but I've tried using them.)

I've also tried using the relationship $$P_\ell^1\left(\cos(\theta)\right) = \frac{\text{d}P_\ell\left(\cos(\theta)\right)}{\text{d}\theta}$$ and the fact that the $P_\ell^m$ satisfy the differential equation $$\frac{\text{d}^2 P_\ell^m\left(\cos(\theta)\right)}{\text{d}\theta^2}+\frac{\cos(\theta)}{\sin(\theta)}\,\frac{\text{d}P_\ell^m\left(\cos(\theta)\right)}{\text{d}\theta}+\left(\ell(\ell+1)-\frac{m^2}{\sin^2(\theta)}\right)P_\ell^m\left(\cos(\theta)\right) = 0$$ (and the $P_\ell$ satisfy that with $m=0$).

I've tried integrating by parts, combining the above identities in different ways, using some of the various recurrence relations between the $P_\ell^m$ and between the $P_\ell$ and between the $j_\ell$... but, as I've said, after a whole week of dealing with that one integral, I'm still stuck.

Any help would be appreciated greatly.


1 Answer 1


Tthe associated Legendre polynomials are first converted into Legendre polynomials, \begin{align} I(\rho)&=\int_0^\pi e^{i\rho\cos(\theta)}\left(\cos(\theta)\,\frac{dP_\ell^1\left(\cos(\theta)\right)}{d\theta}+\frac{P_\ell^1\left(\cos(\theta)\right)}{\sin(\theta)}\right)\sin(\theta)\,d\theta \\ &=\int_0^\pi e^{i\rho\cos(\theta)}\left(\cos(\theta)\,\frac{d^2P_\ell\left(\cos(\theta)\right)}{d\theta^2}+\frac1{\sin(\theta)}\frac{dP_\ell\left(\cos(\theta)\right)}{d\theta}\right)\sin(\theta)\,d\theta \end{align} From the Legendre differential equation given in the OP (with $m=0$), we have \begin{equation} \frac{d^2 P_\ell\left(\cos(\theta)\right)}{d\theta^2}+\frac{\cos(\theta)}{\sin(\theta)}\,\frac{dP_\ell\left(\cos(\theta)\right)}{d\theta}+\ell(\ell+1)P_\ell\left(\cos(\theta)\right) = 0 \end{equation} and thus \begin{align} \cos(\theta)&\,\frac{d^2P_\ell\left(\cos(\theta)\right)}{d\theta^2}+\frac1{\sin(\theta)}\frac{dP_\ell\left(\cos(\theta)\right)}{d\theta}\\ &=\sin(\theta)\frac{dP_\ell\left(\cos(\theta)\right)}{d\theta}-\ell(\ell+1)\cos(\theta)P_\ell\left(\cos(\theta)\right) \end{align} Then, \begin{equation} I(\rho)=\int_0^\pi e^{i\rho\cos(\theta)}\left( \sin(\theta)\frac{dP_\ell\left(\cos(\theta)\right)}{d\theta}-\ell(\ell+1)\cos(\theta)P_\ell\left(\cos(\theta)\right)\right)\sin(\theta)\,d\theta \end{equation} Now, by changing $x=\cos\theta$, with $\frac{dP_\ell\left(\cos(\theta)\right)}{d\theta}=-\sin\theta\frac{dP_\ell(x)}{dx}$ \begin{equation} I(\rho)=\int_{-1}^1e^{i\rho x}\left( (x^2-1)\frac{dP_\ell(x)}{dx}- \ell(\ell+1)xP_\ell\left(x\right)\right)\,dx \end{equation} From the recurrence relation \begin{equation} \frac {x^{2}-1}{n}\frac {d}{dx}P_{n}(x)=xP_{n}(x)-P_{n-1}(x) \end{equation} we have \begin{equation} I(\rho)=-\int_{-1}^1e^{i\rho x}\left(\ell P_{\ell-1}(x)+ \ell^2xP_\ell\left(x\right)\right)\,dx \end{equation} and from the given spherical Bessel function relation, \begin{align} 2i^\ell j_\ell(\rho) &= \int_0^\pi e^{i\rho\cos(\theta)}P_\ell\left(\cos(\theta)\right)\sin(\theta)\,d\theta\\ &=\int_{-1}^1e^{i\rho x}P_{\ell}(x)\,dx\\ 2i^{\ell-1}\frac{dj_{\ell}(\rho)}{d\rho}&=\int_{-1}^1e^{i\rho x}xP_{\ell}(x)\,dx \end{align} Then, \begin{equation} I(\rho)=-2i^{\ell-1}\ell j_{\ell-1}(\rho)-\ell^22i^{\ell-1}\frac{dj_{\ell}(\rho)}{d\rho} \end{equation} One of the recurrence relations for the spherical Bessel functions reads \begin{equation} j_{n}^{\prime}(z)=j_{n-1}(z)-((n+1)/z)j_{n}(z) \end{equation} Thus \begin{align} I(\rho)&=-2i^{\ell-1}\ell(\ell+1)\left( \frac1\rho j_\ell(\rho)+\frac{dj_\ell(\rho)}{d\rho} \right)\\ &=\frac{-2i^{\ell-1}\ell(\ell+1)}{\rho}\frac{d}{d\rho}\left( \rho j_\ell(\rho) \right) \end{align} This result, which seems to be numerically correct, is different from the proposed one by a factor of $-i^{\ell-1}$. For $\ell$ even, the present result shos that the integral is purely imaginary as is expected, contrarily to the proposed identity. Could the discrepancy be related to a particular choice of the branch cut for the Legendre function?

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    $\begingroup$ This is awesome, thanks! I have no idea how you managed to do all of that. :) I just followed your steps carefully and got the same result as you. I wouldn't be surprised if the book were wrong; it's a great resource for learning and one of my favourite optics textbooks, but it wouldn't be the first mistake I come across. $\endgroup$
    – Rain
    Feb 3, 2022 at 7:12

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