What is a reference for a formula in Erdelyi et al., Tables of Integral Transforms, Vol. I, 116 (6)?

Question

In [p. $116$ $(6)$, Tables of Integral Transforms, Vol. I, Erdelyi et al.] the following formula is given:

$$\int_{0}^{\infty} {}_2 F_1(a,b;3/2;-x^2 c^2)\,x \sin(y x) dx=\dfrac{2^{-a-b+1}\pi c^{-a-b}y^{a+b-2}K_{a-b}(y/c)}{\Gamma(a)\Gamma(b)}$$

for $\Re{a}>1/2$ and $\Re{b}>1/2$, where ${}_2 F_1$ is the Gauss hypergeoemtric function and $K_\gamma$ is the Bessel function.

Then, I would like to know a reference or how to prove this formula.

Answer 1

The proposed relation can be proved by verifying the identity of the Mellin transforms of both sides.

For the lhs, we use the Mellin convolution formula: \begin{equation} \mathcal M\left[\int_{0}^{\infty}g(xy)f(x)\,dx\right]= G(s)F(1-s) \end{equation} Here, we take \begin{align} f(x)&={}_2 F_1(a,b;3/2;-x^2 c^2)\,x\\ g(x)&=\sin x \end{align} One has \begin{align} F(s)&=\int_0^\infty x^s{}_2 F_1(a,b;3/2;-x^2 c^2)\,dx\\ &=\frac{c^{-s-1}}{2}\int_0^\infty{}_2 F_1(a,b;3/2;-u)\,u^{\frac{s-1}{2}}\,du \end{align} assuming $c>0$. From the tabulated transform (DLMF) \begin{equation} \int_{0}^{\infty}x^{\sigma-1}{}_2 F_1\left({a,b\atop d};-u\right)\,du=\Gamma(d) \frac{\Gamma\left(\sigma\right)\Gamma\left(a-\sigma\right)\Gamma\left(b-\sigma\right)} {\Gamma\left(a\right)\Gamma\left(b\right)\Gamma\left(d-\sigma\right)} \end{equation} for $0<\Re(\sigma)<\min\left( \Re(a),\Re(b) \right)$. We take $\sigma=(s+1)/2,d=3/2$ to write \begin{align} F(s)&=\frac{c^{-s-1}}{2}\Gamma(3/2) \frac{\Gamma\left((s+1)/2\right)\Gamma\left(a-(s+1)/2\right)\Gamma\left(b-(s+1)/2\right)} {\Gamma\left(a\right)\Gamma\left(b\right)\Gamma\left(3/2-(s+1)/2\right)}\\ &=\frac{\sqrt{\pi}c^{-s-1}}{4} \frac{\Gamma\left((s+1)/2\right)\Gamma\left(a-(s+1)/2\right)\Gamma\left(b-(s+1)/2\right)} {\Gamma\left(a\right)\Gamma\left(b\right)\Gamma\left(1-s/2\right)} \end{align} We assume $a>b$, the condition is $-1<\Re s<2\Re b-1$.

Using the reflection formula, \begin{align} G(s)&=\mathcal M[\sin x]\\ &=\Gamma(s)\sin\left( \frac{\pi s}{2} \right)\\ &=\frac{\pi \Gamma(s)}{\Gamma(s/2)\Gamma(1-s/2)} \end{align} for $-1<\Re(s)<1$ Then, \begin{equation} G(s)F(1-s)=\frac{\pi^{3/2}c^{s-2}}{4}\frac{ \Gamma(s)}{\Gamma(s/2)} \frac{\Gamma\left(a+s/2-1\right)\Gamma\left(b+s/2-1\right)} {\Gamma\left(a\right)\Gamma\left(b\right)\Gamma\left((1+s)/2\right)} \end{equation} with the dupplication formula \begin{equation} \Gamma\left(2z\right)=\pi^{-1/2}2^{2z-1}\Gamma\left(z\right)\Gamma\left(z+\tfrac{1}{2}\right) \end{equation} we obtain \begin{equation} G(s)F(1-s)=2^{s-3}\pi c^{2-s} \frac{\Gamma\left(a+s/2-1\right)\Gamma\left(b+s/2-1\right)}{\Gamma\left(a\right)\Gamma\left(b\right)} \end{equation} This expression is valid for $-1<\Re s<\min\left( 1,2b-1 \right)$. As $b>1/2$ we can take $-1<\Re s<1$.

The Mellin transform of the rhs is \begin{align} \mathcal M\left[\dfrac{2^{-a-b+1}\pi c^{-a-b}y^{a+b-2}K_{a-b}(y/c)}{\Gamma(a)\Gamma(b)}\right]&=\frac{2^{-a-b+1}\pi c^{-a-b}}{\Gamma(a)\Gamma(b)}\mathcal M\left[y^{a+b-2}K_{a-b}(y/c)\right]\\ &=\frac{2^{-a-b+1}\pi }{\Gamma(a)\Gamma(b)}c^{s-2}\left.\mathcal M\left[K_{a-b}(.)\right]\right|_{s+a+b-2}\\ &=\frac{2^{-a-b+1}\pi }{\Gamma(a)\Gamma(b)}c^{s-2}2^{s+a+b-4}\Gamma(a+s/2-1)\Gamma(b+s/2-1)\\ &=\frac{2^{s-3}\pi }{\Gamma(a)\Gamma(b)}c^{s-2}\Gamma(a+s/2-1)\Gamma(b+s/2-1) \end{align} where we used the tabulated Mellin transform of the Bessel function DLMF. It is valid here for $\Re s>\Re|a-b|-\Re(a+b)+2$ or $\Re s>2(1-\Re b)$.

The Mellin transforms of both sides are thus identical if we take $\max\left(-1,2(1-\Re b)\right)<\Re s<1$, which proves the proposed relation.