# Simple integral in Mathematica

1. Aug 23, 2011

### sukharef

The integral is :

, where "y" - coordinate, "kx" - component of wave vector

if we calculate this integral on a paper with a pen and using compex variables theory we obtain:

1) y>0

here, in the capacity of contour we take semicircumference in the upper half plane

2) y<0

here, in the capacity of contour we take semicircumference in the lower half plane

Let's calculate it in Math now.

I wonder where Math gets "-" in Exp[] ?

Last edited by a moderator: May 5, 2017
2. Aug 23, 2011

### Staff: Mentor

The - sign is correct. You made an error doing it by hand. To check do a numeric integration and compare to the analytical answers.

3. Aug 23, 2011

### sukharef

ok, but where is an error?

4. Aug 24, 2011

### jackmell

Looks to me you got the y's mixed up. When the semicircle is over the upper half-plane, I obtain an integral asymptotic to:

$$\int_{0}^{\pi} e^{R\sin(t) y} dt$$

and that converges when y<0. Similar dif for the lower half-plane.

5. Aug 24, 2011

### sukharef

i'm sorry, but where is "-" in Exp[] ?

6. Aug 24, 2011

### jackmell

Let $q=Re^{it}$ over the semi-circle in the upper half-plane, and I want to know the bounds on the integral. So, I can write that it's less than:

$$\int_0^{\pi}\left|\frac{e^{-iRe^{it}y}}{R^2e^{2it}+k^2}Rie^{it}\right|dt$$

and really, it's going to be dominated by the real part of the exponent in the numerator right? So asymptotically, I (think), it's going to approach:

$$\int_0^{\pi}\frac{e^{-iRe^{it}y}}{R}dt$$

$$\int_0^{\pi}\frac{e^{-iR(\cos(t)+i\sin(t))}}{R}dt$$

So now, I'm only interested in the absolute value of that and that's dependent on it's real part:

$$\int_0^{\pi}e^{R\sin(t)}dt$$

But I did that really quick and sloppy. Would need to double-check it and do a better job with inequalities and all if I were turning it in for a grade.

7. Aug 24, 2011

### sukharef

you are right, i'm sorry. thanks!