Closed contour integral: $\int_{\mathbb{c}}\frac{ z}{2z^{2}+1} dz$ where the contour is the unit circle

Question

first and foremost please excuse my English.

given $∫_c \frac{{z}}{2z^{2}+1}dz$ where the contour is the unit circle. so c = $e^{it}$ from 0 to $2\pi$.

since the contour is the unit circle we can say that $f(z(t)) = \frac{e^{it}}{2*(e^{it})^{2}+1}$ and $z'(t) = i*e^{it}$

We know that $\int_a^{b} f(z(t))*z'(t) = \int _c f(z) dz$

so then we just substitute what we know and we get:$$ \int _0 ^{2\pi } \frac{e^{it}}{2*(e^{it})^{2}+1} * i*e^{it} dt $$

$$ i\int _0 ^{2\pi } \frac{e^{2it}}{2e^{2it}+1} dt $$

we let $ u = 2e^{2it}+1 $ and $du = 4ie^{2it}$ and we get : $$i\int _0 ^{2\pi } \frac{e^{2it}}{u} \frac{du}{4ie^{2it}}$$ $$ \frac{1}{4}\int _0 ^{2\pi } \frac{1}{u}du $$

we solve and see that:

$$ i\int _0 ^{2\pi } \frac{e^{2it}}{2e^{2it}+1} dt = \frac{1}{4}(log (1+2e^{4\pi i})-log(1+2e^{0 i})) = 0 $$

is this correct? this was a problem on my final and when I computed this contour integral on wolfram alpha I got $\pi$i?

any explanation would by much appreciate it. I understand I could have done this problem with Cauchy's Integral Formula. Our class did not get up to residue calculus since this is an undergraduate course. Many thanks in advance.

Answer 1

No, the answer is $\pi i$. Since

$$\frac{z}{2z^2 + 1} = \frac{1/4}{z - i/\sqrt{2}} + \frac{1/4}{z + i/\sqrt{2}}$$

then

$$\int_c \frac{z}{2z^2 + 1}\, dz = \frac{1}{4}\int_c \frac{1}{z - i\sqrt{2}}\, dz + \frac{1}{4}\int_c \frac{1}{z + i/\sqrt{2}}\, dz = \frac{1}{4}(2\pi i) + \frac{1}{4}(2\pi i) = \pi i$$

using Cauchy's integral formula in the second to last step.

Answer 2

The solution presented in the OP might appear paradoxical. The substitution $u=1+e^{i2t}$ seems both natural and innocuous. Yet, inappropriate application of the substitution led to the integral

$$\int_{t=0}^{t=2\pi}\frac{1}{u}\,du=0$$

So, what went wrong here?

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In THIS ANSWER, I showed that for any rectifiable curve $\gamma$ on $\mathbb{C}\setminus\{0\}$, that begins at $1$ and ends at $re^{i\theta}$

$$\int_\gamma \frac1z\,dz=\log(r)+i(\theta +2\pi k) \tag 1$$

where $k$ is the net number of times $\gamma$ crosses the positive real axis from the fourth quadrant to the first quadrant. (For additional explanation, SEE THIS ANSWER).

Note that $(1)$ can be easily generalized by replacing the starting point at $z=x$. In that case, we simply replace $\log(r)$ with $\log(r/x)$.

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To make things clear, we apply $(1)$ to the problem of interest. Observe that for $t\in [0,2\pi]$, the substitution $u=1+e^{2it}$ results in a contour $\gamma$ defined by $|u-1|=2$, that has one net crossing (at $t=\pi$) of the positive real axis in the $u$-plane.

Then, using the aforementioned modification of $(1)$, we find that with $x=r=3$, $\theta=2\pi$, and $k=1$

$$\int_{|u-1|=2}\frac1u\,du =i4\pi$$

Finally, we have

$$\begin{align} \oint_{|z|=1} \frac{z}{2z^2+1}\,dz&=i\int_0^{2\pi}\frac{e^{i2t}}{2e^{i2t}+1}\,dt\\\\ &=\frac14 \oint_{|u-1|=2}\frac1u\,du\\\\ &=i\pi \end{align}$$

as was to be shown!

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