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I am trying to calculate this contour integral using the parameterization $z(t) = e^{it}$ where $0 \leq t \leq 2\pi$. We then have $z'(t) = ie^{it}$ so our contour integral becomes

$\int_{C}f(z)dz = \int_{0}^{2\pi}f(z(t))z'(t)dt = \int_{0}^{2\pi}\frac{e^{it}}{2(e^{it})^2+1}ie^{it}dt$

Now if I perform the substitution $u = e^{it}$, then our integral becomes $\int_{1}^{1}\frac{u}{2u^2 + 1}du = 0$, however Mathematica evalutes this integral to be $i\pi$. I believe there must be some assumption i'm making here that isn't valid. Any insight would be greatly appreciated.

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    I don't why your method is wrong, but I can give you a method that produces the right result: Use the residue theorem. Inside the unit circle your function has two singularities at $z = \frac{\pm i}{\sqrt2}$, where the residue is in both cases $\frac{1}{4}$, so the integral has the value $I = 2\pi i \cdot (1/4 + 1/4) = 2\pi i(1/2) = \pi i$. I also don't see any error in your integral substition or calculation. – 2017-02-12
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    I feel like there has to be some subtlety that I'm missing. We haven't gotten to the residue theorem but I'm glad to know that there is a way to get $i\pi$. – 2017-02-12
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    If it makes you feel any better, I have done a lot of this type of stuff and this is driving me nuts. All I can really think is that $C$ was always a closed curve, you just made it more explicit. The problem is that integration over a closed curve is not valid with a function that is not analytic inside the curve, which your integrand isn't. But that shouldn't affect your solution... – 2017-02-12
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    Have you learnt Cauchy's integral formula yet? The residue theorem is a related to that (in fact gives you the same answer but in a shorter way), but you don't need it. Take a look at https://en.wikipedia.org/wiki/Cauchy%27s_integral_formula#Example , I think you can deduce your solution in the same style. – 2017-02-12
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    Nope, we just started doing contour integrals. All that we've learned is Green's Theorem and the method that involves parameterizing the curve. I guess we do also have the Cauchy-Goursat Theorem, but that definitely isn't applicable here. Since our function isn't analytic in domain bounded by the curve I avoided Green's Theorem, so that left me with parameterization. – 2017-02-12

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The error comes in the $u$ substitution. Note that after some work,$$\int_0^{2\pi}\frac{e^{it}}{2(e^{it})^2+1}ie^{it}dt=\int_0^{2\pi} \frac{-\sin 2t}{5 + 4\cos 2t}dt + i\int_0^{2\pi}\frac{2+\cos 2t}{5 + 4\cos 2t}dt$$

I.e., its real and imaginary parts are just ordinary real integrations. But when you substituted $u = e^{it}$, you introduced a variable of integration that was not real again. In fact, you undid the very "substitution" that is used to define complex integration in the first place. Since your $u$ substitution is a complex variable, it has to be treated as a complex integration again, which means it is not an integration along an interval in the real line, but along a contour in $\Bbb C$ (in particular, that same unit circle $C$ you started with). It may start and end at $1$, but it takes a loop to do so.

Integration around a closed contour only reliably gives $0$ if the function is analytic on the curve and its interior (Cauchy's theorem), but this function has two poles inside.

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    This clears up a lot thank you. – 2017-02-12