# Missing condition

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1. Jun 9, 2015

### Asteroid

I have a question about an integral taken from integral tables of Gradshteyn and Ryzhik precisely , 3,914 -1 ( pag.490 ):
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.

Last edited by a moderator: May 7, 2017
2. Jun 9, 2015

### 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.

3. Jun 10, 2015

### 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?)

4. Jun 10, 2015

### 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 $\text{Re}~\beta=0$.

5. Jun 10, 2015

### Asteroid

Thanks for all.
It is very clear. :)