2
$\begingroup$

We should find the Cauchy principal value integral of the form $$ I=\oint \frac{dz}{(z-z_1)(z-z_2)}~, $$ where both roots $z_1$ and $z_2$ lie on the contour path. My answer is: $$ I=a \left(-\oint \frac{dz}{z-z_1}+\oint \frac{dz}{z-z_2}\right)=a(-i\pi+i\pi)=0~, $$ where $a=1/(z_2-z_1)$. However, in a book, they do not find it to be zero, though they do not explain why. Is there any case for which I should find this integral to be "not zero"? In the specific example of the book, the poles are $z_1=e^{-ik}$, $z_2=e^{ik}$, and the contour is the unit circle.

$\endgroup$
  • $\begingroup$ What does the book find it to be, and what book is that?? $\endgroup$ – Timbuc Jan 27 '15 at 9:09
1
$\begingroup$

First: putting $\;f(z):=\frac1{(z-z_1)(z-z_2)}\;$ , we obtain (since both poles are simple assuming $\;z_1\neq z_2\;$ )

$$\text{Res}_{z=z_i}(f)=\lim_{z\to z_i}\frac{z-z_i}{(z-z_1)(z-z_2)}=\begin{cases}&\;\;\;\;\;\;\;\;\;\frac1{z_1-z_2}&,\;\;i=1\\{}\\&-\frac1{z_1-z_2}=\frac1{z_2-z_1}&,\;\;i=2\end{cases}$$

Now using the lemma, and its corollary, in the most upvoted answer here, we get that the integral indeed equals zero.

$\endgroup$
  • $\begingroup$ I understood now how the book does and it doesn't look good to me. He shifts $z\to z+i\epsilon$. After this, one of the roots lies inside the contour, while the other outside. The he performs the integral just with the residue of the root inside the contour and he finds the result. Then he sends $\epsilon\to0$. If we instead shift it by $z\to z-i\epsilon$, we get the opposite result. In general, this is probably not allowed. In fact, if I have a single pole $z_1$ in the contour, by shifting it either inside or outside the contour, I have two different results: $2\pi Res\,f_{z=z_1}$ or zero. $\endgroup$ – Wizzerad Jan 27 '15 at 11:05
  • $\begingroup$ IS that a known book? $\endgroup$ – Timbuc Jan 27 '15 at 11:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.