Integrals from Gradshteyn & Ryzhik: Real Part Condition Necessary?

[Asteroid]

I have a question about an integral taken from integral tables of Gradshteyn and Ryzhik precisely , 3,914 -1 ( pag.490 ):
http://www.lepp.cornell.edu/~ib38/tmp/reading/Table_of_Integrals_Series_and_Products_Tablicy_Integralov_Summ_Rjadov_I_Proizvedennij_Engl._2.pdf
The condition to use the result of the integral regards the real part of the two parameters .
But if we do not have the real part but only imaginary part ,what we can do ?
I can not find an integral which permits to overcome this difficulty.
I can not understand the role of this condition, it is really necessary or in this case I can use the integral without it?

Thanks to all.

 

[fzero]

It's pretty clear that the requirement that $\text{Re}~\beta>0$ is required for the integral to be finite. $\text{Re}~\gamma >0$ is used to avoid the branch cut that we usually take for Bessel functions on the negative real axis. You might want to check that your integral is well-defined for the range of parameters that you have.

 

[Asteroid]

Thanks for answer, in fact I want check the integral validity for complex parameters only without real part.
using beta e gamma as complex parameters only, i can use the result ?
I.e. since I do not have real part I can neglect the condition?
(As we do with gaussian integral, in fact this type of integral is defined for real but it can be use also for complex parameter, right?)

 

[fzero]

Let's let $\beta = i a$ and $\gamma = i c$, then we have the integral

$$ I = \int_0^\infty e^{i a \sqrt{x^2-c^2} } \cos(bx) dx.$$

I am suspicious that this could converge to a finite value, since as $x\rightarrow \infty$, the integrand is oscillating between $-1$ and $1$. It is clear to see that the closely related integral

$$ \int_0^\infty e^{i a x } \cos(bx) dx$$

is not convergent for this reason, by direct integration and then examining the limit. On the other hand, the integral

$$ \int_0^\infty e^{-x} e^{i a x } \cos(bx) dx$$

does converge and this is why that formula can be trusted when $\text{Re}~\beta>0$ (and $\gamma$ is such that the argument of the Bessel function is away from the branch cut).

I would not recommend extending the book formula to $\text{Re}~\beta=0$.

 

[Asteroid]

Thanks for all.
It is very clear. :)