I was reading G. B Jeffrey's article with The Royal Society as publisher, and I came across two integrals, for which there were solutions but no procedure how they were obtained. I was trying to do it myself but failed multiple times (I didn't get very far), moreover other authors don't provide the calculation either they just refer to them as Jeffrey's integrals.
So my question is how do we compute this pair of integrals?
$$\int_{-1}^{1} \frac{P_n(x)}{(2 \cosh u - 2x)^{3/2}} dx = \frac{e^{-(n+\frac{1}{2})|u|}}{\sinh |u|}$$ $$\int_{-1}^{1} \frac{P_n(x)}{(2 \cosh u - 2x)^{1/2}} dx = \frac{e^{-(n+\frac{1}{2})|u|}}{n + \frac{1}{2}}$$
where $P_n(x)$ are Legendre polynomials.
Denote \begin{align} I_n(t)&= \int_{-1}^1 \frac{P_n(x)}{\sqrt{1-2x t+t^2}}dx\\ J_n(t)& = \int_{-1}^1 \frac{P_n(x)}{(1-2x t+t^2)^{3/2}}dx \end{align} Then, per the generating function $\frac1{\sqrt{1-2x t+t^2}}= \sum_{k=0}^\infty P_k(x)t^k $ \begin{align} \sum_{k=0}^\infty I_k(t) t^k=& \int_{-1}^1 \frac{\sum_{k=0}^\infty P_k(x)t^k}{\sqrt{1-2x t+t^2}}dx =\int_{-1}^1 \frac{1}{{1-2x t+t^2}}dx\\ =& \ \frac1t \ln\frac{1+t}{1-t} =\sum_{k=0}^\infty \frac{t^{2k}}{k+\frac12} \end{align} which leads to $I_n(t)= \frac{t^{n}}{n+\frac12}$ and \begin{align} J_n(t) &= \frac1{1-t^2} \left[I_n(t)+2t I’_n(t) \right] =\frac{2t^{n}}{1-t^2}\\ \end{align}
Utilize $I_n(t)$ and $J_n(t)$ to evaluate the two integrals \begin{align} &\int_{-1}^{1} \frac{P_n(x)}{(2 \cosh u - 2x)^{1/2}} dx \\ =& \int_{-1}^{1} \frac{P_n(x)e^{-u/2}}{(e^{-2u} - 2x e^{-u}+1)^{1/2}} dx =I_n(e^{-u})e^{-\frac u2} =\frac{e^{-(n+\frac{1}{2})u}}{n + \frac{1}{2}}\\ \\ & \int_{-1}^{1} \frac{P_n(x)}{(2 \cosh u - 2x)^{3/2}} dx= J_n(e^{-u})e^{-\frac32 u}=\frac{e^{-(n+\frac{1}{2})u}}{\sinh u} \end{align}