I'm trying to solve this problem using complex analysis:
$\int_0^\infty\frac{\sqrt{x}}{x^2-2x+3}\, dx$
I'm using a contour that avoids the branch cut along the +x axis of the complex plane, looking like a block capital C.
Ive shown that the two circular portions of the contour drop to zero as you shrink the inner circular contour and grow the outer one.
The remaining two straight portions of the contour sum to be twice that of the intergral I'm trying to solve. The problem is the the poles of the above formula give residues of:
$\frac{1}{2} i \sqrt{\frac{1}{2}-\frac{i}{2}} $
and
$\frac{-1}{2} i \sqrt{\frac{1}{2}+\frac{i}{2}} $
Here's my problem: $2 \pi i \sum \text{Residues}$ is an completely imaginary number, giving my integral no real value.
Am I using the wrong contour, or am I just missing something really simple and the value of my integral is just zero?
Thanks for any help
I will suppose that you cut along $\Bbb R^+$ and that you picked $\lim_{\varepsilon \to 0^+}\sqrt{1+ \varepsilon i} = +1$, so that $\sqrt x$ always has a positive imaginary part.
Then your contour includes two poles, at $x = 1 \pm \sqrt{2}i$ and where $\sqrt x$ have positive imaginary parts in both cases. You most likely picked one with positive imaginary part and another one with negative imaginary part instead.
The residue at $x$ is $\sqrt{x}/(x-\bar x)$, and summing them gives $(\sqrt x - \sqrt {\bar x}) /(x- \bar x) $.
If $\sqrt x = a+ib$ with $b>0$ then $\sqrt{\bar x} = -a+ib$, so $\sqrt x - \sqrt {\bar x} = 2a$, so the sum of the residues is $\pm ai/\sqrt{2}$, which is purely imaginary (the sign depends on which $x$ you choose when computing the square root).
After multiplying by $2i\pi$ you get $\pm a\pi\sqrt 2$, a real number.
And really everything would be more simple and apparent if you made a change of variable $y = \sqrt x$