The Bessel function of the first kind and order n has the integral representation

$J_n(z)=i^{-n}/\pi \int_0^\pi e^{iz\cos\theta}\cos(n\theta)d\theta$

By using the Laplace integral for the Legendre polynomial $P_n(x)$,

$P_n(x)=1/\pi\int_0^\pi (x+\sqrt{x^2-1}\cos\theta)^nd\theta$

Find the generating function

$\sum_{n=0}^{\infty} \frac{P_n(x)r^n}{n!}$

in terms of $J_0$.

Ok, so I've found $J_0=1/\pi \int_0^{\pi} e^{izcos\theta}d\theta$. That's not really a earth shattering amount of work but I can't find a way to relate the two together. From just plugging the legendre polynomials into the sum I get

1 + xr + 1/4(3x^2-1) + so on

but I still don't see any relation. Any help is appreciated.

  • $\begingroup$ Start by substituting the integral expression for $P_n(x)$ into that sum, you can sum $n$ inside the integral (switch summation and integration) and compare that series with the series for an exponential, can you pick out $z$ from there and rewrite in terms of $J_0(z)$? You may have to factor out another exponential term.. $\endgroup$ Aug 14, 2013 at 10:26
  • $\begingroup$ I make it something like $e^{rx}J_0(ir\sqrt{x^2-1})$ $\endgroup$ Aug 14, 2013 at 10:37
  • $\begingroup$ Ok so I don't see it. $1/\pi \int_0^{\pi} \sum_{n=0}^{\infty} (x+\sqrt{x^2-1}\cos\theta)^nr^n/n!$ $\endgroup$
    – user90312
    Aug 14, 2013 at 11:24
  • $\begingroup$ I see a e^r that I can pull out so then $e^r/\pi \int_0^{\pi} \sum_{n=0}^{\infty} (x+\sqrt{x^2-1}\cos\theta)^n d\theta$. $\endgroup$
    – user90312
    Aug 14, 2013 at 11:32
  • $\begingroup$ The first line you wrote is correct but the second is not, you cannot sum the $n$ to create that exponential without summing everything else simultaneously. You should use $a^nb^n=(ab)^n$ on the term in parentheses and the $r$ term and then sum that to create the exponential. $\endgroup$ Aug 14, 2013 at 12:39

1 Answer 1


Define the generating function: $$G \left( x,r \right) =\sum _{n=0}^{\infty }{\frac {P_{{n}} \left( x \right) {r}^{n}}{n!}},\tag{1}$$ and substitute the integral representation of the Legendre polynomials: $$P_{{n}} \left( x \right) ={\frac {\int _{0}^{\pi }\! \left( x+\sqrt {{ x}^{2}-1}\cos \left( \theta \right) \right) ^{n}{d\theta}}{\pi }},\tag{2}$$ into $(1)$ to obtain: \begin{aligned} G \left( x,r \right) &=\sum _{n=0}^{\infty } \frac {\left(\int _{0}^{\pi }\! \left( x+\sqrt {{x}^{2}-1}\cos \left( \theta \right) \right) ^{n}{d\theta}\right){r}^{n}}{\pi n! },\\ &=\frac{1}{\pi }\int _{0}^{\pi }\! \sum _{n=0}^{\infty } \frac {\left( x+\sqrt {{x}^{2}-1}\cos \left( \theta \right) \right) ^nr^n}{n! }{d\theta},\\ &=\frac{1}{\pi }\int _{0}^{\pi }\! \text{exp}\left[ rx+r\sqrt {{x}^{2}-1}\cos \left( \theta \right)\right] {d\theta},\\ &=\frac{e^{rx}}{\pi }\int _{0}^{\pi }\! \text{exp}\left[r\sqrt {{x}^{2}-1}\cos \left( \theta \right)\right] {d\theta},\\ &=e^{rx}J_0\left(i r \sqrt{x^2-1}\right)=e^{rx}I_0\left(r \sqrt{x^2-1}\right). \end{aligned} To move from line 1 to line 2 in $(3)$ I have switched the order of summation and integration, this usually requires justification (show convergence of integral and sum e.t.c..) but I will leave that to you. To move from the penultimate line to the last I used the integral representation of the Bessel function and solved for $iz$; I dropped a minus sign in the argument of the Bessel function as the function is even.

As a check you can Taylor expand the final expression in $r$ and you will find that the coefficients are indeed given by $P_n(x)/{n!}$.

  • 2
    $\begingroup$ Ha! Or I could scroll down to find the answer. Thanks again, I was looking at that problem for days. $\endgroup$
    – user90312
    Aug 14, 2013 at 19:39
  • $\begingroup$ That's super close to the generating function $(e^{yx})*J_0(y*\sqrt{1-x^2})$ $\endgroup$
    – crow
    Oct 2 at 19:31

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