Verification of Integral identity The purpose of this thread is to 
show that 
\begin{equation}
\int_0^\infty \left(U^{-1}\Lambda_l\middle)\middle(\frac{r}{r'}\right)\cdot f(r')\frac{dr'}{r'}=\\=(n-1)\frac{2\sqrt{2}\pi}{\sqrt{(2l+1)}}\int_{0}^\infty dr'\int_{-1}^1 dx\frac{P_l^0(x)r'^2f(r')}{\left(rr'\right)^{1/2}\left(r^2+\frac{2}{m+1}rr' x+r'^2\right)}
\end{equation}
Consider the Mellin transform 
$$(Uf)(s) = \frac{1}{\sqrt{2\pi}} \int_0^{\infty} f(r) r^{-is+1/2} \ dr$$
and inverse Mellin transform $$(U^{-1}g)(s)= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} g(r) r^{is-3/2} \ dr.$$
Consider the multiplication operator (2.31b) (I chose $l=0$, $n=2$ in the formula stated in the paper)
$$(\mathcal M f)(r)=2\pi \int_0^{\infty}\int_{-1}^1 \frac{r'^2 f(r') }{(r r')^{1/2}(r^2+r'^2+rr'x)}\ dx \ dr' $$
then it is claimed that $U \mathcal M U^{-1}$ 
 A: The Author says that it must be:
$$U^{-1}\{\Lambda_l\cdot(Uf)\}(r)=\cal{(\hat M_lf)}(r)$$
So let's verify it. 
By  the convolution theorem for the Mellin transformation it follows that:
$$U^{-1}\{\Lambda_l\cdot(Uf)\}(r)=\int_0^\infty \left(U^{-1}\Lambda_l\middle)\middle(\frac{r}{r'}\right)\cdot f(r')\frac{dr'}{r'}$$
The Mellin inverse transformation of $(\text{A}.1)$ (with the misprint corrected)
$$\Lambda_l(s)=(n-1)\sqrt{\frac{1}{(2l+1)\pi}}2\pi\int_0^\infty d\xi\int_{-1}^1 dx\frac{\xi^{-is}}{\xi^2+\frac{2}{m+1}\xi x+1}P_l^0(x)$$
that can be rewritten as
$$\Lambda_l(s)=(n-1)\sqrt{\frac{1}{(2l+1)\pi}}2\pi\int_0^\infty d\xi\int_{-1}^1 dx\frac{\xi^{-is+1/2}\xi^{-1/2}}{\xi^2+\frac{2}{m+1}\xi x+1}P_l^0(x)$$
is simply given by:
$$(U^{-1}\Lambda_l)(r)=(n-1)\sqrt{\frac{1}{(2l+1)\pi}}2\pi\sqrt{2\pi}\int_{-1}^1 dx\frac{\xi^{-1/2}}{\xi^2+\frac{2}{m+1}\xi x+1}P_l^0(x)$$
from which:
$$\left(U^{-1}\Lambda_l\middle)\middle(\frac{r}{r'}\right)=(n-1)\sqrt{\frac{1}{(2l+1)\pi}}2\pi\sqrt{2\pi}\int_{-1}^1 dx\frac{r'^2}{\left(\frac{r}{r'}\right)^{1/2}\left(r^2+\frac{2}{m+1}rr' x+r'^2\right)}P_l^0(x)$$
and then, by substitution:
\begin{equation}
\int_0^\infty \left(U^{-1}\Lambda_l\middle)\middle(\frac{r}{r'}\right)\cdot f(r')\frac{dr'}{r'}=\\=(n-1)\frac{2\sqrt{2}\pi}{\sqrt{(2l+1)}}\int_{0}^\infty dr'\int_{-1}^1 dx\frac{P_l^0(x)r'^2f(r')}{\left(rr'\right)^{1/2}\left(r^2+\frac{2}{m+1}rr' x+r'^2\right)}
\end{equation}
that is just $(2.31\text{b})$.
That proves the assertion.
Now go from the bottom up in this sequence of steps and you will get the answer.
