Complex Analysis Integral by Substitution

[Poopsilon]

Im trying to take the integral, using substitution, of \int_0^1\frac{2\pi i[cos(2\pi t) + isin(2\pi t)]dt}{cos(2\pi t)+isin(2\pi t)}

So I set u=cos(2\pi t)+isin(2\pi t)du=2\pi i[cos(2\pi t) + isin(2\pi t)]dt

Yet when I change the endpoints of the integral I get from 1 to 1, which doesn't make sense. Since the fraction is equal to 2πi I could just take the integral directly and get 2πi, yet for some reason doing it by substitution gives me 0. This integral was a pull back from the integral of 1/z around a closed curve, so it seems that for some reason by doing the u-du substitution the integral "forgets" that there is a singularity on the interior of the curve, what is going on here?

 

[jackmell]

Im trying to take the integral, using substitution, of \int_0^1\frac{2\pi i[cos(2\pi t) + isin(2\pi t)]dt}{cos(2\pi t)+isin(2\pi t)}

So I set u=cos(2\pi t)+isin(2\pi t)du=2\pi i[cos(2\pi t) + isin(2\pi t)]dt

Yet when I change the endpoints of the integral I get from 1 to 1, which doesn't make sense. Since the fraction is equal to 2πi I could just take the integral directly and get 2πi, yet for some reason doing it by substitution gives me 0. This integral was a pull back from the integral of 1/z around a closed curve, so it seems that for some reason by doing the u-du substitution the integral "forgets" that there is a singularity on the interior of the curve, what is going on here?

Setting u=\cos(2\pi t)+i\sin(2\pi t) for 0\leq t\leq 1 changes the path of integration from a straight line over the real axis to a closed contour about the origin with radius one. So your integral is equivalent to:

\mathop\oint\limits_{|u|=1} \frac{du}{u}=2\pi i

 

[Dick]

And the endpoints of the integration of du/u are 1 and 1. But that doesn't mean the integral is zero. The antiderivative of 1/u is log(u) and that's multivalued. You can't just insert the limits without considering the path.