Why Does Mathematica Include a Negative Sign in Exp[]?

[sukharef]

The integral is :

[PLAIN]http://i038.radikal.ru/1108/ed/a6cd0c5e8f36.jpg

, 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

[URL]http://s006.radikal.ru/i215/1108/c9/835aa4a65f60.jpg[/URL]

2) y<0

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

[URL]http://s58.radikal.ru/i162/1108/b7/37bb3bc6170d.jpg[/URL]

So the total answer is

[URL]http://s004.radikal.ru/i208/1108/4f/c35aafd1c20b.jpg[/URL]

Let's calculate it in Math now.

[URL]http://s40.radikal.ru/i090/1108/f8/cf9f67043005.jpg[/URL]

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

 

[Dale]

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

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

 

[sukharef]

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

ok, but where is an error?

 

[jackmell]

ok, but where is an error?

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.

 

[sukharef]

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.

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

 

[jackmell]

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

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.

 

[sukharef]

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.

you are right, I'm sorry. thanks!