# Solving PDE with Legendre functions

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!

• 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.
– Gary
Feb 20, 2020 at 15:02
• Thank you greatly for this reference! Feb 20, 2020 at 15:02
• @Gary I understand how the integral representation is formulated, but why is the certain $-\frac12+i\mu$ chosen for the Legendre function? Feb 21, 2020 at 14:24
• It should be clear from the derivation in the paper.
– Gary
Feb 24, 2020 at 7:27