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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.

Bita
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  • Probably, I should use the same contour ?? as https://math.stackexchange.com/questions/2261015/calculate-int-infty-infty-frac-coskx-sqrtx2a2-dx?rq=1 – Bita Dec 26 '20 at 16:51
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    How about taking $-\frac{d}{dA}$ of your integral, which lets you compute a simpler integral in exchange for solving a separable differential equation? – Diffusion Dec 26 '20 at 17:38
  • Hi Zachary, Thanks for your comment. Unfortunately, I cannot envision what you mean by taking $-\frac{d}{dA}$, I appreciate if you can elaborate more on that. – Bita Dec 26 '20 at 18:02
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    Call your integral $I(A)$. Then differentiating under the integral sign, you obtain that $-I'(A)=\int_{-\infty}^\infty dx, \exp\left(-A\sqrt{x^2+B} +Cix\right)$. – Diffusion Dec 26 '20 at 18:11
  • Ah I get what you mean. Let me see if it works. Thanks – Bita Dec 26 '20 at 18:33
  • I am now working on that but whether or not it works, I should admit, your idea was brilliant. Thanks – Bita Dec 26 '20 at 18:42

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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.

Diffusion
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