Complex integration

  • Thread starter fauboca
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  • #1
158
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Without using Cauchy's Integral Formula or Residuals, I am trying to integrate

[tex]\int_{\gamma}\frac{dz}{z^2+1}[/tex]

Around a circle of radius 2 centered at the origin oriented counterclockwise.

[tex]\frac{i}{2}\left[\int_0^{2\pi}\frac{1}{z+i}dz-\int_0^{2\pi}\frac{1}{z-i}dz\right][/tex]

[tex]\gamma(t)=2e^{it}, \quad \gamma'(t)=2ie^{it}[/tex]

The answer is zero. I am supposed to get each integral to be [itex]2\pi i[/itex] which is 0 when subtracted.

I know it is related to the fact that [itex]\int_{\gamma}\frac{1}{z}dz = 2\pi i[/itex].

And using u-sub isn't correct since any closed path would be zero when that isnt true. The integral of 1/z shows that not all closed paths will be zero.
 

Answers and Replies

  • #2
70
1
Try using the identity:

[tex]\int \frac {du}{a^2 + u^2} = \frac{1}{a}arctan\frac{u}{a} +c[/tex]
 
  • #3
158
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Try using the identity:

[tex]\int \frac {du}{a^2 + u^2} = \frac{1}{a}arctan\frac{u}{a} +c[/tex]
That is nice and a lot easier than the route I was taking. Without the substitution of the integral, would it sill have the bounds of 0 and 2pi though? I know it is over gamma but the bounds were for 2e^{it}.
 
  • #4
65
0
The circle starts and ends at the same point. Your bounds are actually from 2 to 2. The integral identity applied to your problem requires bounds in z, not in terms of its parametrization.
 

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