Show that these circles are linked (their linking number is $\pm 1$) There are two circles given in 3 dimensions:
$$c_0(\phi) = (\cos(\phi), \sin(\phi),0)\ \ \ \phi \in [0,2\pi[$$
$$c_1(\phi) = \frac{1}{\sqrt{2}-\sin(\phi)}(\cos(\phi), \sin(\phi), \cos(\phi))\ \ \ \phi \in [0,2\pi[$$
Show that their linking number is $\pm 1$, where the linking number is defined by:
$$L(c_0,c_1) = \frac{1}{4\pi}\int_0^{2\pi}\int_0^{2\pi}\frac{(c_0(t)-c_1(s),\dot c_0(t),\dot c_1(s))}{|c_0(t)-c_1(s)|^3}\,dt\ ds$$
Where
$$\text{if } \ a,b,c:\Bbb{R} \to \Bbb{R}^3$$
$$\text{then } \ (a(t),b(t),c(t)) =
\begin{align}\begin{vmatrix}
a_1(t) & b_1(t) & c_1(t) \\ 
a_2(t) & b_2(t) & c_2(t) \\ 
a_3(t) & b_3(t) & c_3(t) \\
\end{vmatrix}
\end{align}$$
And
$$\text{if } \ c: \Bbb{R} \to \Bbb{R}^3, \ c(t) = (c_1(t),c_2(t),c_3(t))$$
$$\text{then } \ \dot c(t) = (c_1'(t),c_2'(t),c_3'(t))$$
Usually the integral for the linking number is very difficult to calculate, so there are other options too. For example, since we know that our curves are circles, their linking number can only be $L(c_0,c_1) \in \{-1,0,+1\}$. Showing that an integral is equal to a specific value can be done with this following technique:
$$L(c_0,c_1) \stackrel{?}{=} a \in \{-1,0,+1\} \iff \\ 
\iff \forall \varepsilon \in \Bbb{R}^+: a - \varepsilon \stackrel{?}{\le} L(c_0,c_1) \stackrel{?}{\le} a + \varepsilon $$
So far I've shown that the circles don't intersect, meaning that there are no $\phi \in \ [0,2\pi[$ values for which $|c_0(\phi)-c_1(\phi)| = 0$. That was relatively easy to show. But the big question is:
$$L(c_0,c_1) \stackrel{?}{=} \pm 1$$
I would appreciate any insight on how I could possibly show this.
In other words, how can we show that the above integral is approximately equal to $1$, with perhaps numerical methods?
 A: Mathematica cannot solve your integral exactly, but it can numerically integrate it. The output of the numerical integration is 1. See the Mathematica code below.

Also, you can use Mathematica to plot the two curves, yielding the following picture.

The linking number of two components of a link can be computed from a diagram of the link. Each crossing is either a positive (or $+1$) crossing or a negative (or $-1$) crossing, as below.

The linking number of the two components is 
$$\frac{1}{2}(\#\text{positive crossings} - \#\text{negative crossings})$$
where we only count the crossings between the two components. See the wikipedia page for linking number. If you look in the picture produced by Mathematica above, you can tell that the two circles are linked (and thus have linking number $\pm 1$). By looking at the parameterizations, it shouldn't be too hard to find the orientation of each curve, and then to compute the linking number from the diagram. Both crossings in the above diagram turn out to be positive, and so the linking number is 1.
