- #1
Poopsilon
- 294
- 1
Im trying to take the integral, using substitution, of [tex]\int_0^1\frac{2\pi i[cos(2\pi t) + isin(2\pi t)]dt}{cos(2\pi t)+isin(2\pi t)}[/tex]
So I set [tex]u=cos(2\pi t)+isin(2\pi t)[/tex][tex]du=2\pi i[cos(2\pi t) + isin(2\pi t)]dt[/tex]
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?
So I set [tex]u=cos(2\pi t)+isin(2\pi t)[/tex][tex]du=2\pi i[cos(2\pi t) + isin(2\pi t)]dt[/tex]
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?
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