Solving Integral(s*exp(isr)/(s^2-k^2)) ds

• October 15th 2009, 05:50 PM
Solving Integral(s*exp(isr)/(s^2-k^2)) ds
Greetings,

I'm trying to teach myself contour integration/complex analysis and finding it a little more challenging than I expected. To make a long story short, it's not obvious to me how to make the semi-circle go to 0 in the following: Integral(s*exp(isr)/(s^2-k^2)) ds from -inf to inf.

The contribution from the semi-circle is given by: integral(i*R^2*exp
(2*i*theta)*exp(i*r*R*exp(i*theta))/(R^2*exp(2*i*theta) - k^2)) from 0 to pi. As R -> infinity, this is supposed to go to 0. I don't see this.

To arrive at the equation I mentioned above, I substitute R*exp(i*theta) in for s above:
integral(R*exp(i*theta)*exp(i*r*R*exp(i*theta))*i* R*exp(i*theta)/R^2*exp(2*i*theta) - k^2)) dtheta from 0 to pi
= integral(i*R^2*exp(2*i*theta)*exp(i*r*R*exp(i*thet a))/(R^2*exp(2*i*theta) - k^2)) dtheta from 0 to pi

At this point, I want to try to get it to go to 0:
Dividing top and bottom by R^2*exp(2*i*theta):
integral(i*exp(i*r*R*exp(i*theta))/(1 - (k^2 / R^2*exp(2*i*theta))))from 0 to pi

Taking R to infinity:
integral(i*exp(i*r*R*exp(i*theta))/(1 - 0)) from 0 to pi
which finally gives:
integral(i*exp(i*r*R*exp(i*theta))) from 0 to pi, R -> inf

...which doesn't obviously go to 0 as far as I can tell.
Any help appreciated!

P.S. I can follow many arguments for seeing the integral go to 0, but the extra R in the numerator is what's giving me fits. Thanks!
• October 15th 2009, 06:01 PM
Bruno J.
Sorry my friend. I would love to help but I've had a long day and it would probably take me more energy to decypher your question than to think about its content once I've understood it. Most people here will feel the same. I suggest you learn the basics of LaTeX and post your question again; it will take you an hour to learn, you will help others help you, and you'll have the amazing skill of LaTeX for the rest of your life.

On a side note, welcome to the site, and congrats for teaching yourself complex analysis.
• October 17th 2009, 11:38 AM
Greetings,

I'm trying to teach myself contour integration/complex analysis and finding it a little more challenging than I expected. To make a long story short, it's not obvious to me how to make the semi-circle go to 0 in the following definite integral that I equate to a contour integral (not shown here):
$
\int_{-\inf}^{\inf} \frac{s e^{isr}}{s^2 - k^2}\, ds
$

The contribution from the semi-circle is given by:
$
\int_0^{\pi} \frac{iR^2 e^{2it}e^{irRe^{it}}}{R^2e^{2it} - k^2}\, ds
$

As R -> infinity, this is supposed to go to 0. I don't see this.

To arrive at the semi-circle equation I substitute $Re^{it}$ in for s and $iRe^{it}\, dt$ for ds in $\int_{-\inf}^{\inf} \frac{s e^{isr}}{s^2 - k^2}\, ds$, yielding $\int_0^{\pi} \frac{Re^{it} e^{iRe^{it}r}}{s^2 - k^2}\, iRe^{it}\, dt$ --> $\int_0^{\pi} \frac{iR^2 e^{2it}e^{irRe^{it}}}{R^2e^{2it} - k^2}\, ds$.

At this point, I want to try to get it to go to 0:
Dividing top and bottom by $R^2 e^{2it}$:
$\int_0^{\pi} \frac{ie^{irRe^{it}}}{1 - \frac{k^2}{R^2 e^{2it}}}\, ds$

The denominator goes to 1 as R goes to 0:
$\int_0^{\pi} ie^{irRe^{it}}\, ds$

...which doesn't obviously go to 0 as far as I can tell.

Any help appreciated!

P.S. I can follow many arguments for seeing the integral go to 0, but the extra R in the numerator is what's giving me fits. Thanks!
• October 17th 2009, 02:16 PM
shawsend
You need to know:

If $|f(z)|\leq \frac{M}{R^k}$ for $z=Re^{it}$ where $k>0$, then over the upper half circle:

$\lim_{R\to \infty} \int_C f(z) e^{imz}=0$

So can you show $\left|\frac{s}{s^2-k^2}\right|\leq \frac{M}{R^k}$ for sufficiently large R?

Try and find this theorem in your Complex Analysis text book, or get Schaum's Complex Variables.
• October 20th 2009, 06:55 PM