Evaluate the line integral $\int_{L} \frac{-y \,d x+x \,d y}{x^{2}+y^{2}}$ for a line segment $L$ 
Calculate the line integral
$$
\int_{L} \frac{-y \,d x+x \,d y}{x^{2}+y^{2}}
$$
where $L$ is the line segment from $(1,0)$ to $(0,1)$ parametrized by
$$
L(t)=(1-t)(1,0)+t(0,1), \quad 0 \leq t \leq 1
$$

I know I can calculate by normal parametrization, but the answer says it represents the argument changed; could someone please explain this?

The integral measures the change in argument along the curve $L$.
$$
\int_{L} \frac{-y \,d x+x \,d y}{x^{2}+y^{2}}=\frac{\pi}{2}
$$

And another question I have is that when I compute this integral using change of potential, I found the potential of this vector field to be $$-\arctan\frac{x}{y}+c ,$$ but that function is not defined at the endpoint $(1, 0)$ of $L$.
 A: Consider the quarter arc: $\gamma:[0,\pi/2]\to \mathbf{R}^2$ with
$$
\gamma(t) = (\cos(t),\sin(t))
$$
with the direction from $\gamma(0)$ to $\gamma(1)$. Then $\int_L Pdx+Qdy=\int_\gamma Pdx+Qdy$ by Green's theorem, where
$$
P(x,y)=\frac{-y}{x^2+y^2},\quad Q(x,y)=\frac{x}{x^2+y^2}\;,
$$
because the closed path $L-\gamma$ is contained in an open simply connected subset of $\mathbf{R}^2$ where the vector field is smooth and $P_y=Q_x$.
By working on the line integral along $\gamma$, you can easily find the expected answer $\pi/2$ mentioned in your post:
$$
\int_\gamma Pdx+Qdy=\int_{0}^{\pi/2}\bigg((-\sin(t))(-\sin(t))+\cos(t)\cos(t)\bigg)\;dt = \frac{\pi}{2}\;.
$$
The vector field $(P,Q)$ does not have a potential function on the punctured plane $\mathbf{R}^2\setminus\{(0,0)\}$. So when you work on a potential function, you need to specify the (simply connected) domain of your vector field.
A: The "change in argument" is, informally, just the total net angle that the path winds around the origin anticlockwise (a negative change in argument indicates a net clockwise winding). We can see this from a computation similar to the one that gave you a potential of a restriction of $$\omega := \frac{-y \,dx + x \,dy}{x^2 + y^2} .$$ This description of $\omega$ motivates the common notation $d\theta$ for that form, but we'll see below that $\omega$ is not exact, i.e., isn't the exterior derivative of any function, so that notation can be misleading.
Indeed, integrating $\omega$ gives the function
$$f(x, y) := \arctan \frac{y}{x} ;$$
strictly speaking $f$ is not a potential for $\omega$ but rather a potential for the restriction of $\omega$ to the domain $H := \{x \neq 0\}$ of $f$. (This phenomenon explains why you ran into trouble evaluating your potential at a point on one of the axes.) Drawing a triangle with vertices $(0, 0), (x, 0), (x, y)$ (with, say, $x > 0$) and using the usual definition of the arctangent function shows that $f(x, y)$ is just the signed angle between the positive $x$-axis and the ray from the origin through $(x, y)$. Thus, for a differentiable parametrized curve $\gamma : [a, b] \to H$ in the right half-plane, the F.T.C. gives that
$\int_\gamma \omega$ is just the total (net) angle the curve winds, starting with $\gamma(a)$ and ending at $\gamma(b)$.
To extend this notion to any path $\gamma: [a, b] \to \Bbb R^2 \setminus \{(0, 0)\}$ notice $\omega$ is rotationally symmetric, so our conclusion must hold for any path whose image is contained inside an open half-plane passing through the origin. At this point we can conclude that $\int_L \omega$ is the signed angle between $(1, 0)$ and $(0, 1)$ with respect to the origin, namely $\frac{\pi}{2}$.
But we can partition the curve $\gamma$ into pieces such that each piece is contained inside such a half-plane, hence the value of $\int_\gamma \omega$ is the sum of the net angles $\gamma$ winds over each subinterval, that is, just the total net angle $\gamma$ winds around the origin, as claimed.

Notice that integrating $\omega$ along the unit circle $S^1$ (oriented anticlockwise) thus gives
$$\int_{S^1} \omega = 2 \pi .$$ In particular, the integral of $\omega$ along some closed path is nonzero, hence the F.T.C. implies that $\omega$ is not exact, i.e., has no potential.
That said, we can still use a potential to compute the line integral over the line segment $L$. One method is to (again) use rotational symmetry: For any rotation $f \circ R$ must again be a potential for the restriction of $\omega$ to its domain, so it's enough to pick an angle of rotation such that $L$ is contained in a single half-plane, and, e.g., $-\frac{\pi}{4}$ will do. Simplifying gives the potential
$$(f \circ R)(x, y) = \arctan \frac{y - x}{x + y}$$ of the restriction of $\omega$ to the half-plane $\{x + y > 0\}$ containing $L$, so that
$$\int_L \omega = \int_L d(f \circ R) = \int_{\partial L} (f \circ R) = (f \circ R)\bigg\vert_{(1, 0)}^{(0, 1)} = \frac{\pi}{2} .$$
A: Let's write your integral as $$\int_L\vec f d\vec r$$
In Cartesian coordinates $$d\vec r=\hat x dx+\hat y dy$$  so then $$\vec f=\frac{-y}{x^2+y^2}\hat x+\frac x{x^2+y^2}\hat y$$
In polar coordinates $$d\vec r=dr\hat r+rd\theta\hat \theta$$
Here $$\hat r=\frac xr\hat x+\frac yr\hat y$$ and $$\hat\theta=-\frac yr\hat x+\frac xr\hat y$$
Then it's easy to see that in polar coordinates $$\vec f=\frac1r\hat\theta$$
Therefore $$\vec f d\vec r=d\theta$$
Therefore $$\int_L\vec f d\vec r=\int_Ld\theta=\theta_{end}-\theta_{start}$$
