Interesting exam question

[Nobby]

asymptote at x=1

[Lasercookie]

asymptote at x=1

Oops didn't notice that. Outside of the VCE course, but you can evaluate it as an improper integral.









Since

[lzxnl]

The problem is, if we're going to integrate with complex numbers, we now have to specify a path to integrate from 0 to 2. This can get quite tricky in the complex plane; our "antiderivative" sqrt(x^2-1) isn't differentiable along the negative real axis, so we can't really integrate. That's the problem with square roots; the complex valued function f(z)=sqrt(z) is discontinuous along the entire negative real axis as if we approach -1 from above (with regards to the imaginary component) and from below, the limit as z=>-1 is different.

[Mao]

I don't think you can do the first integral, since is undefined for (unless our codomain is complex).

And if we are indeed looking at the complex integral, then we need to start worrying about paths and poles, something definitely outside of specialist (or even a mathematics minor).

[lzxnl]

I don't think we have a pole at x=1 here. Although can we integrate around the pole by taking the integration path to consist of a tiny arc that bypasses x=1? I'm not sure if taking the arc to go above x=1 and below x=1 will make a difference.