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Complex integration

dwsmith

Well-known member
Feb 1, 2012
1,673
Consider
\[
\int_{-\infty}^{\infty}\frac{e^{iax}}{x^2 - b^2}dx
\]
where \(a,b>0\). The poles are \(x=\pm b\) which are on the x axis. Usually, if the poles are on the x axis, I use that the integral is
\[
2\pi i\sum_{\text{UHP}}\text{Res} + \pi i\sum_{\text{x axis}}\text{Res}\quad (*)
\]
which works in this problem http://mathhelpboards.com/analysis-50/integral-=-2pi-sum-res-uhp-pi-i-sum-res-real-axis-7576.html
However, if I use this formula on the integral above, I get the answer to be
\[
-\frac{\pi}{b}\sin(ab)
\]
when the answer is
\[
-\frac{2\pi}{b}\sin(ab)
\]
which would indicate \(2\pi i\) times the sum of the residual on the x axis. What is going wrong and when can and cannot I use the formula \((*)\)?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle PV \int^{\infty}_{-\infty} \frac{e^{iaz}}{z^2-b^2}\,dz\)

The function has only poles on the real axis at \(\displaystyle z=\pm b\)

so that becomes

\(\displaystyle PV \int^{\infty}_{-\infty}\frac{e^{iaz}}{z^2-b^2}\,dz= \pi i \lim_{z\to b}\, (z-b)\frac{e^{iaz}}{z^2-b^2}+\pi i \lim_{z\to -b}\, (z+b)\frac{e^{iaz}}{z^2-b^2}=\frac{\pi i e^{iab}}{2b}-\frac{\pi i e^{-iab} }{2b}=\frac{\pi i}{2b }(e^{iab}-e^{-iab})\)

which is equal to \(\displaystyle -\frac{\pi \sin(ab)}{b}\) . As yours .

Note this the Principle value of the integral >
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
The answer is \(-\frac{2b}{\pi}\sin(ab)\) which isn't what we both have.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
The answer is \(-\frac{2b}{\pi}\sin(ab)\) which isn't what we both have.
I cannot see how that would be the correct answer. why so sure ?
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
The answer in your textbook is not correct.

But just so you're aware, that formula is generally only applicable when the the poles on the real axis are simple poles.

But it is also applicable if none of the Laurent expansions about the poles on the real axis have terms of negative even power.

That's why $ \displaystyle \text{PV} \int_{-\infty}^{\infty} \frac{1}{x^{3}}\ dx = 0$.
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
@Random my professor says it depends on how we construct the contour around the poles. How true is that? Shouldn't the integral be the same?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
If we avoid the poles then the integral along any closed smooth path is zero.