# Integrating sine where argument goes to infinity.

1. Sep 25, 2014

### sreerajt

After some integration, i am getting a form $e^{i \alpha\phi+i\beta\phi\sin(\phi-\phi')-i\gamma\sin\phi}$, where $\alpha, \beta, \gamma$ are constants. Now i want to apply the limit where $\phi$ ranges from 0 to $\infty$ (ya, in the argument of sine we will encounter $\infty$ which is exactly my problem). How can i do that?

2. Sep 25, 2014

### dextercioby

What's the original problem? We need to see a context. Please, post it.

3. Sep 26, 2014

### sreerajt

Okay. i am trying to derive a dispersion relation in magnetized plasma with kinetic approach. By the by i am taking only two terms, just forget about the last term in the above equation. We will encounter a term$I= \int_0^\infty cos(\phi-\phi')e^{g_1(\phi')}\,d\phi'\\$ where $g_1(\phi')= i \alpha \phi'+i\beta sin(\phi-\phi')$. Differentiating $g_1(\phi')$ with respect to $\phi'$ and then substituting back into $I_1$ will give
$I_1=\alpha\int_0^\infty e^{g_1(\phi') d\phi'}+ \gamma\int_0^\infty \frac{dg_1(\phi')}{d\phi'}e^{g_1(\phi')}d\phi'$. It is the second term that causes trouble. From second term, $\frac{dg_1(\phi')}{d\phi'}e^{g_1(\phi')}d\phi= d(e^{g_1(\phi')})$. So $I_1= firstterm + \gamma\int_0^\infty d(e^{g_1(\phi')})$. Now, i think, you can make sense out of it.

4. Sep 26, 2014

### HallsofIvy

Staff Emeritus
The trig function do NOT have a limit as the argument goes to infinity.

5. Sep 26, 2014

### sreerajt

i understand that sir. But, when you do the maths it comes into picture. And this is why i posted the question here.
Thank you...