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I've split the integral around $z_1 = 1 - i$ and $z_2 = 1+ i$ using the contours $C_1$ and $C_2$:

$ \int_{|C|=2} g(z) dz = \int_{C_1} g(z) dz + \int_{C_2} g(z) dz$

In this case, $g(z)$ for $C_1$ is $\displaystyle \frac{\frac{1}{z-z_1}}{z-z_2}$ and $g(z)$ for $C_2$ is $\displaystyle \frac{\frac{1}{z-z_2}}{z-z_1}$. Using Cauchy's Thm I get $\displaystyle \frac{1}{z_2 - z_1}$ for the first one and $\displaystyle \frac{1}{z_1 - z_2}$ for the second. But evaluating

$\displaystyle \int_{|C|=2} g(z) dz = 2\pi i \left(\frac{1}{z_2 - z_1} + \frac{1}{z_1 - z_2} \right) = 0$

I'm not interested in other methods, just this particular version where you split the contour $C$ into $C_1$ and $C_2$ My issue with this answer is that it doesn't make sense. When evaluating an integral like $\displaystyle \int_{\infty}^{-\infty} \frac{dx}{x^2 + 2x + 2}$ you must let $\displaystyle g(z) = \frac{1}{z^2 + 2z + 2}$ and then integrate over the contour $C$. I would've thought that the answer would be give me $\pi$

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  • $\begingroup$ So your question is ...? I haven't check your calculations but there is nothing wrong on having as an answer $\int_\gamma g dz=0$ even if $g$ is not analytic inside $\gamma$ $\endgroup$
    – Alonso Delfín
    May 17, 2015 at 1:36
  • $\begingroup$ The answer is supposed to be $\pi$ $\endgroup$
    – NavyColors_Blue
    May 17, 2015 at 1:38

2 Answers 2

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The roots of the denominator are actually at $z_{\pm}=-1\pm i$. Then the integral is,

$$\oint_{|z|=2} \frac{dz}{(z-z_-)(z-z_+)} = \frac1{z_+-z_-} \oint_{|z|=2} dz \left (\frac1{z-z_+} - \frac1{z-z_-} \right )$$

By Cauchy-Goursat, it should be clear that, since both poles are inside the circle $|z|=2$, the integral is zero.

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  • $\begingroup$ But wait, how can that be? I thought that to evaluate an integral like $\int_{\infty}^{-\infty} \frac{dx}{x^2 + 2x + 2}$ you must let $g(z) = \frac{1}{z^2 + 2z + 2}$ and then integrate over the contour $C$. $\endgroup$
    – NavyColors_Blue
    May 17, 2015 at 1:41
  • $\begingroup$ @NavyColors_Blue NO , yo must integrate over the upper semicircle of $|z|=R$ and the interval [-R,R] and then make $R\to \infty$, The result follows by looking that the integral vanishes on the semicircle as $R \to \infty$ $\endgroup$
    – Alonso Delfín
    May 17, 2015 at 1:46
  • $\begingroup$ @NavyColors_Blue: I have no idea what you are talking about. You seemed to state that you wanted an integral over a circle of radius $2$, so what this has to do with an integral over the real line i cannot say. Please explain what it is you want $\endgroup$
    – Ron Gordon
    May 17, 2015 at 1:51
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Take $R>2$ and let $\gamma_R$ be the upper semicircle of $|z|=R$ and the real interval $[-R, R]$, take $\gamma_R$ positively oriented:

enter image description here

Since $z=1+i$ is the only singularity inside $\gamma_R$ for all $R>2$,then $$ \int_{\gamma_R} \frac{dz}{z^2+2z+2} = 2\pi i \left(\frac{1}{z-(1+i)}\right)=2\pi i \left(\frac{1}{2i}\right) = \pi \ \ \forall \ R>2 $$ On the other side if $\Gamma_R=Re^{it}$ for $t\in[0,\pi]$, ($\Gamma_R$ is just the upper semicircle) then $\gamma_R=\Gamma_R \cup [-R,R]$, thus $$ \int_{\gamma_R} \frac{dz}{z^2+2z+2} = \int_{\Gamma_R}\frac{dz}{z^2+2z+2} + \int_{-R}^R \frac{dx}{x^2+2x+2} $$ Since the integral over $\Gamma_R$ vanishes when $R \to \infty$ then $$ \pi = \lim_{R \to \infty }\int_{\gamma_R} \frac{dz}{z^2+2z+2} = \lim_{R \to \infty }\int_{\Gamma_R}\frac{dz}{z^2+2z+2} + \lim_{R \to \infty }\int_{-R}^R \frac{dx}{x^2+2x+2} = \int_{-\infty}^\infty \frac{dx}{x^2+2x+2} $$

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    $\begingroup$ My question is why take over the upper semi-circle? And what do you mean by $\Gamma$ being part of the upper semi-circle - I thought [-R,R] was the entire upper semi-circle? $\endgroup$
    – NavyColors_Blue
    May 17, 2015 at 2:21
  • $\begingroup$ @NavyColors_Blue I have made an edit an add an image to help. You now see why is necessary to take the upper circle and the real interval? This is the only way you can relate $\int_{\gamma_R}$ with $\int_{-R}^R$ $\endgroup$
    – Alonso Delfín
    May 17, 2015 at 2:37
  • $\begingroup$ I see the mechanics behind your reasoning but it still unusual - the "why just the upper semicircle" bothers me. $\endgroup$
    – NavyColors_Blue
    May 17, 2015 at 2:43
  • $\begingroup$ Ok I want to help here. Why would you take the whole circle ? Look that since you want to calculate a real integral the whole circle dose not passes trough the real line. However, if you take only a semi circle you can add the real line using $[-R,R]$, observe that the same result is obtain by taking the lower semi circle. Is this clear enough ? Do follow all the other steps on my answer ? @NavyColors_Blue $\endgroup$
    – Alonso Delfín
    May 17, 2015 at 2:50

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