I'm studying a paper which has a PDE of the form

$$\frac{\partial p}{\partial L}(L,n) = A\bigg((n^2-1)\frac{\partial p}{\partial n}(L,n)\bigg),\quad n>1,$$ with a Dirac delta initial condition $p(L=0,n) = \delta(n-1).$ Also note that $$n = \frac{2-\tau}{\tau},\quad \tau\in[0,1],\quad n\in[1,\infty).$$ To solve this PDE they then denote $P_{-\frac12+i\mu}(n),$ $n\geq1,\mu\geq 0$ the Legendre function of the first kind, which is the solution of

$$\frac{d}{dn}(n^2-1)\frac{d}{dn}P_{-\frac12+i\mu}(n) = \bigg(\mu^2+\frac14\bigg)P_{-\frac12+i\mu}(n),\quad P_{-\frac12+i\mu}(1)=1.$$ They then say that the Legendre function has the integral representation $$P_{-\frac12+i\mu}(n) = \frac{\sqrt{2}}{\pi}\cosh{(\pi\mu)}\int_{0}^{\infty}\frac{\cos{(\mu\tau)}}{\sqrt{\cosh{(\tau)}+n}}d\tau.$$

I'm completely stuck to how they get this integral representation, and why they would choose a Legendre polynomial of this form. I would really appreciate help on how to obtain this integral representation, thank you!

  • 1
    $\begingroup$ The Legendre functions are basically particular cases of the Gauss hypegeomteric function (see dlmf.nist.gov/14.3.E6). Your integral is formula (11) at the bottom of page 156 in authors.library.caltech.edu/43491/1/Volume%201.pdf There is an explanation in that book on how to arrive at this representation. $\endgroup$
    – Gary
    Feb 20, 2020 at 15:02
  • $\begingroup$ Thank you greatly for this reference! $\endgroup$ Feb 20, 2020 at 15:02
  • $\begingroup$ @Gary I understand how the integral representation is formulated, but why is the certain $-\frac12+i\mu$ chosen for the Legendre function? $\endgroup$ Feb 21, 2020 at 14:24
  • $\begingroup$ It should be clear from the derivation in the paper. $\endgroup$
    – Gary
    Feb 24, 2020 at 7:27


You must log in to answer this question.