Integral $\int_0^\infty \frac{\sqrt{\sqrt{\alpha^2+x^2}-\alpha}\,\exp\big({-\beta\sqrt{\alpha^2+x^2}\big)}}{\sqrt{\alpha^2+x^2}}\sin (\gamma x)\,dx$ I am having trouble showing this equality is true$$
\int_0^\infty \frac{\sqrt{\sqrt{\alpha^2+x^2}-\alpha}\,\exp\big({-\beta\sqrt{\alpha^2+x^2}\big)}}{\sqrt{\alpha^2+x^2}}\sin (\gamma x)\,dx=\sqrt\frac{\pi}{2}\frac{\gamma \exp\big(-\alpha\sqrt{\gamma^2+\beta^2}\big)}{\sqrt{\beta^2+\gamma^2}\sqrt{\beta+\sqrt{\beta^2+\gamma^2}}},
$$
$$
\mathcal{Re}(\alpha,\beta,\gamma> 0).
$$
I do not know how to approach it because of all the square root functions. 
It seems if $x=\pm i\alpha \ $   we may have some convergence problems because of the denominator.  Perhaps there are ways to solve this using complex methods involving the branch cut from the square root singularity.   I just do not know what to choose $f(z)$ for a suitable complex function to represent the integrand.
I also tried differentiating under the integral signs w.r.t $\alpha,\beta,\gamma$ but it did not simplify anything.  Thanks.   How can we calculate this integral? 
 A: Replace $\alpha$, $\beta$ and $\gamma$ with $a$, $b$ and $c$ respectively.
With the substitution $x=a\sinh t$, the integral can be written as:
$$\begin{aligned}
I & = \sqrt{2a}\int_0^{\infty} e^{-ab\cosh t}\sin(ac\sinh t)\sinh \left(\frac{t}{2}\right)\,dt \\
&=-\sqrt{2a}\Im\left(\int_0^{\infty} e^{-ab\cosh t}\cos\left(ac\sinh t+\frac{it}{2}\right)\,dt \right)
\end{aligned}$$ 
Thanks to sir O.L. for evaluating the final integral here: Integral: $\int_0^{\infty} e^{-ab\cosh x}\cos\left(ac\sinh(x)+\frac{ix}{2}\right)\,dx$
The result is hence,
$$\begin{aligned}
I & = -\sqrt{2a}\Im\left(e^{-\frac{i}{2}\arctan\frac{c}{b}}\sqrt{\frac{\pi}{2a\sqrt{b^2+c^2}}}e^{-a\sqrt{b^2+c^2}}\right) \\
&=\sqrt{2a}\sqrt{\frac{\pi}{2a\sqrt{b^2+c^2}}}e^{-a\sqrt{b^2+c^2}} \sin\left(\frac{1}{2}\arctan\frac{c}{b}\right) \\
&=\sqrt{\frac{\pi}{2}}\sqrt{\frac{1}{\sqrt{b^2+c^2}}}e^{-a\sqrt{b^2+c^2}}\frac{c}{\sqrt{\left(\sqrt{ b^2+c^2}+b \right)\sqrt{b^2+c^2}}} \\
&=\boxed{\sqrt{\dfrac{\pi}{2}}\dfrac{c\exp\left(-a\sqrt{b^2+c^2}\right)}{\sqrt{b^2+c^2}\sqrt{\sqrt{b^2+c^2}+b}}}
\end{aligned}$$ 
I used the following to simplify the above expression
$$\begin{aligned}
\sin\left(\frac{1}{2}\arctan\frac{c}{b}\right) &=\sqrt{\frac{1-\cos\left(\arctan\frac{c}{b}\right)}{2}}\\
&= \frac{1}{\sqrt{2}}\sqrt{1-\frac{b}{\sqrt{b^2+c^2}}}\\
&= \frac{1}{\sqrt{2}}\sqrt{\frac{\sqrt{b^2+c^2}-b}{\sqrt{b^2+c^2}}}=\frac{1}{\sqrt{2}}\frac{c}{\sqrt{\left(\sqrt{ b^2+c^2}+b \right)\sqrt{b^2+c^2}}}
\end{aligned}$$
A: As Lucian said. Take $\alpha,\beta,\gamma>0$ reals (once you are done you can extend it analytically).
$$
F(\beta):=\int_0^\infty \frac{\sqrt{\sqrt{\alpha^2+x^2}-\alpha}\,\exp\big({-\beta\sqrt{\alpha^2+x^2}\big)}}{\sqrt{\alpha^2+x^2}}\sin (\gamma x)\,dx
$$
\begin{eqnarray*}
F^\prime(\beta)&=&\int_0^\infty {\sqrt{\sqrt{\alpha^2+x^2}-\alpha}\,\exp\big({-\beta\sqrt{\alpha^2+x^2}\big)}}\sin (\gamma x)\,dx \\
&=& \int_0^\infty \alpha^{3/2}\sqrt{2}\sinh\left(\frac{t}{2}\right)\,\exp\big({-\beta\alpha\cosh(t)\big)}\sin (\gamma\alpha \sinh(t))\cosh(t)\,dt 
\end{eqnarray*}
Now you have a nice analytic integrand, you can residue formula it away.
