- #1
HasuChObe
- 31
- 0
Does anyone know how to prove this identity? I don't quote understand why the associated Legendre function is allowed to have arguments where [itex]|x|>1[/itex].
[tex]
h_n(kr)P_n^m(\cos\theta)=\frac{(-i)^{n+1}}{\pi}\int_{-\infty}^\infty e^{ikzt}K_m(k\rho\gamma(t))P_n^m(t)\,dt
[/tex]
where
[tex]
\gamma(t)=\begin{cases}
\sqrt{t^2-1} & |t|\ge 1 \\
-i\sqrt{1-t^2} & |t|<1
\end{cases} \\
\rho=\sqrt{x^2+y^2} \\
r=\sqrt{\rho^2+z^2} \\
\cos\theta=\frac{z}{r}
[/tex]
and [itex]k[/itex] is the wavenumber, [itex]h_n(x)[/itex] is the spherical hankel function of the first kind, and [itex]K_m(x)[/itex] is the modified cylindrical bessel function of the second kind.
[tex]
h_n(kr)P_n^m(\cos\theta)=\frac{(-i)^{n+1}}{\pi}\int_{-\infty}^\infty e^{ikzt}K_m(k\rho\gamma(t))P_n^m(t)\,dt
[/tex]
where
[tex]
\gamma(t)=\begin{cases}
\sqrt{t^2-1} & |t|\ge 1 \\
-i\sqrt{1-t^2} & |t|<1
\end{cases} \\
\rho=\sqrt{x^2+y^2} \\
r=\sqrt{\rho^2+z^2} \\
\cos\theta=\frac{z}{r}
[/tex]
and [itex]k[/itex] is the wavenumber, [itex]h_n(x)[/itex] is the spherical hankel function of the first kind, and [itex]K_m(x)[/itex] is the modified cylindrical bessel function of the second kind.
Last edited: