Why is $\int_{\mathbb{S}^1} \frac{-v\,du+u\,dv}{u^2+v^2}=2\pi$? Why not $0$?

Question

Why is

$$\int_{\mathbb{S}^1} \frac{-v\,du+u\,dv}{u^2+v^2}=2\pi$$

$$\mathbb{S}^1 = \{(u,v)\in\mathbb{R}^2 \quad st \quad u^2+v^2 =1 \}$$

I am a bit confused about this

$$ \int_{\mathbb{S}^1} \frac{-v\,du+u\,dv}{u^2+v^2}= \int_{\mathbb{S}^1} \frac{-v\,du+u\,dv}{1}= \int_{\mathbb{S}^1} -v\,du + \int_{\mathbb{S}^1} u\,dv= -vu+uv\bigg|_{\mathbb{S}^1}= 0 $$

What am I missing?

Edit:

Going my the suggestion you get $$ \int_{\mathbb{S}^1} \sin^2 \theta + \cos^2 \theta \, d\theta = \int_{\mathbb{S}^1} 1 \, d\theta = \theta_{\mathbb{S}^1} = 2\pi $$

Answer 1

Turning my comments into an answer:

Your first step of noticing $u^2+v^2 = 1$ to rewrite the denominator is a good one. However your second to last step is woefully mistaken because $u$ and $v$ are not independent variables since they solve the equation $u^2+v^2=1$. Try writing both in terms of the variable $\theta$ as $u = \cos\theta$ and $v = \sin\theta$ and see what you get.

Answer 2

Let $u=\cos t,v=\sin t$ and then $$\int_{\mathbb{S}^1} \frac{-v\,du+u\,dv}{u^2+v^2}=\int_0^{2\pi}(\sin^2t+\cos^2t)\,dt=2\pi$$

Answer 3

You can also view the polar substitution as $t=\frac vu=\tan(\theta)$

with $dt=d\left(\frac vu\right)=\dfrac{u\,dv-v\,du}{u^2}=(1+t^2)(u\,dv-v\,du)$

So $\displaystyle \int_{\mathbb S^1} \frac{u\,dv-v\,du}{u^2+v^2}=2\int_{-\infty}^{+\infty} \dfrac{dt}{1+t^2}=\bigg[\arctan(t)\bigg]_{-\infty}^{+\infty}=2\pi$

Answer 4

If $z=u+iv$ where $u,v$ are real, and $z$ moves around the circle, then $dz$ is an infinitely small change in $z$ that is at a right angle to $z$ itself. Multiplying by $i$ is a $90^\circ$ counterclockwise rotation. Therefore the quotient $dz/z$ is an infinitely small pure imaginary number, whose absolute value is $\left| dz/z\right| = \left| dz\right| = {}$an infinitely small increment of arc length. Therefore the integral evaluates to $i\times(\text{total arc length}) = 2\pi i.$

And \begin{align} & \frac{dz} z = \frac{du + i\, dv}{u+iv} = \frac{(du+i\,dv)(u-iv)}{(u+iv)(u-iv)} \\[10pt] = {} & \frac{u\,du +v\,dv + i(u\,dv-v\,du)}{u^2+v^2} = u\,du + v\,dv + i(u\,dv - v\,du). \end{align} The integral of the real part of this expression is $0$ because each variable goes from $-1$ to $1$ and back again.