Complex Integral $\int_{-\infty}^{+\infty}\frac{e^{(-A\sqrt{x^2+B}+Cix)}}{\sqrt{x^2+B}}$ I am confused calculating the below integral:
$$\int_{-\infty}^{+\infty}\frac{e^{(-A\sqrt{x^2+B}+Cix)}}{\sqrt{x^2+B}}dx$$
Where A, B, C are real and A and B are positive. I don't think I can use roots of the denominator and calculate residue at those points. Please let me know if you need further information. Thank you !
Edit: Bases on what suggested (Thanks to Zachary), I changed my focus on calculating below integral:
$$\int_{-\infty}^{+\infty}e^{(-A\sqrt{x^2+B}+Cix)}dx$$
where I considered $Cix=cos(Cx)+isin(Cx)$ considering even/odd function properties, I ended up with:
$$2\int_{0}^{+\infty}e^{(-A\sqrt{x^2+B})}cos(Cx)dx$$
At this point, I applied a change of variables as $x'=\sqrt{x^2+B}$, doing so, I come up with below:
$$2\int_{0}^{+\infty}e^{(-A\sqrt{x^2+B})}cos(Cx)dx=2\int_{0}^{+\infty}e^{(-Ax')}cos(C\sqrt{(x'^2-B)})\frac{x'}{\sqrt{x'^2-B}}dx'$$
Then, I was thinking of changing another variable $x''=sin(C\sqrt{(x'^2-B)})$ to get:
$$2\int_{0}^{+\infty}e^{(\frac{-A}{\sqrt{C}}(\sqrt{arcsin(x'')^2+BC}))}dx''$$
Not sure, if it makes sense, please advise, Many thanks.
 A: This is not a complete answer, but a long comment on an upper  bound that I found for your integral. Let $B=u^2$. Indeed, substituting $ x\mapsto u\sinh z$ (as done similarly here), $-I'(A)$ reduces to
$$2u\int_0^\infty\cos\left(uC\sinh z\right)e^{-Au\cosh(z)}\cosh(z)\,dz$$
This is roughly the Fourier Transform of the second modified Bessel function's  "density". Indeed, this function has the integral representation
$$K_1(x)=\int_{0}^\infty e^{-x\cosh(z)}\cosh(z)\,dz.$$
The triangle inequality tells us that
$$-I'(A)\le 2u\,K_1(Au).$$
We know that $-I'(A)>0$ from the integral, as well as $\lim_{A\to\infty} I(A)=0$. Hence, integrating, you find that
$$I(A)\le 2K_0(Au),$$
where $K_0(x)$ is the second modified Bessel function of null order. This shows you (although we could have seen this above) that $I$ is best viewed as a function of the dimensionless parameter $s=Au$ (as well as of $uC$ but we will ignore this dependence here).
In particular, you now know that
$$I(s)=\mathcal{O}\left(\frac{1}{\sqrt{s}}e^{-s}\right)\qquad \text{as } s\to\infty$$
thanks to the asymptotics of the Bessel functions.
I am not sure about the existence of a closed-form, but if I find one I will edit my answer.
