In my research the following equation appeared:

$$\frac{1}{4\pi}\int_{0}^{1}\frac{t^{s-1}(1-t)^{s-1}}{(\rho-t)^s}dt=\int_0^{\infty} f(a) Q^{i\sqrt{a}}_{s-1}(2\rho-1) da,$$

where $\rho,s>1$, $Q^{\mu}_{\nu}$ is the associeted Legendre function of second kind and $f(a)$ is to be found.

All my attempts to solve this equation failed. Maybe it is much to difficult to solve. However, I'm wondering if anyone knows similar equations appearing somewhere else in mathematics and if there are fruitful results regarding integrals similar to the right hand side with various $f(a)$? It looks somehow similar to an integral transform, e.g. like the Mehler Fock transform.

Best wishes

  • $\begingroup$ lower index of $Q$ looks like $s-1$, is not it? $\endgroup$ Mar 1 '15 at 13:31
  • $\begingroup$ You are right. I have changed it in the formula. $\endgroup$
    – Jop Kop
    Mar 1 '15 at 18:24

[Disclaimer: the below text is not mine, but by Vladimir Petrov, who does not have MO account]

Consider generalized Mehler-Fock transform: $$ \begin{cases} F(\xi,\,\mu)&=\intop_1^\infty f(y)P^{-\mu}_{-1/2+i\xi}(y)\,dy,\ \ \ 0\le\Re\mu<1;\\ f(x)&={1\over\pi}\intop_0^\infty \xi\sinh\pi\xi\,\Gamma(\mu+1/2+i\xi)\Gamma(\mu+1/2-i\xi) P^{-\mu}_{-1/2+i\xi}(x)F(\xi,\,\mu)\,d\xi.\,\,(*) \end{cases} $$

We have $$\intop_0^\infty f(\alpha)\,Q_{s-1}^{i\sqrt{\alpha}}(z)\,d\alpha=A(s,z).\,\,(1)$$ Change bariables to $\xi=\sqrt{\alpha}$ and use Wipple transform (Bateman, Erd\'elyi vol. 1, p. 3.3.1, formula (14)). We get equation $$\sqrt{2\pi}\intop_0^\infty g(\xi)\xi e^{-\pi\xi}\,\Gamma(s+i\xi)\,(z^2-1)^{-1/4}\,P_{-1/2-i\xi}^{1/2-s}\left({z\over\sqrt{z^2-1}}\right)\,d\xi =A(s,z).\,\,(2)$$

Now set $z/\sqrt{z^2-1}=x$. Then $z^2-1=(x-1)^{-1}.$ And the answer is (calculations have to be checked):

$$\sqrt{2\pi^3}\,g(\xi){e^{-\pi\xi}\over\sinh\pi\xi\,\Gamma(s-i\xi)}=\intop_1^\infty A\left(s,{x\over x-1}\right)\,(x-1)^{-1/4}P_{-1/2-i\xi}^{1/2-s}(x)\,dx \,\,(3)$$

Something should be said about analytical continuation on $s$ to the right half-plane. Transform $(*)$ is defined in the strip $0<\Re\mu<1,$ but the same formula holds for integer $\mu.$ Since Legendre function is analytical in upper index in the right halfplane, Carlson theorem implies that the function holds in the whole right halfplane.


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