Integrate Complex Function w/o Cauchy's or Residuals

[fauboca]

Without using Cauchy's Integral Formula or Residuals, I am trying to integrate

\int_{\gamma}\frac{dz}{z^2+1}

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

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

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

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

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

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

 

[Totalderiv]

Try using the identity:

\int \frac {du}{a^2 + u^2} = \frac{1}{a}arctan\frac{u}{a} +c

 

[fauboca]

Try using the identity:

\int \frac {du}{a^2 + u^2} = \frac{1}{a}arctan\frac{u}{a} +c

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}.

 

[alanlu]

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.