Residue Theorem for Real Integral: Where did I go Wrong? So the integral is
$$\int_{0}^{\infty} \frac{\sqrt{x}}{x^2 + 4x + 5} d x.$$
I did keyhole integration avoiding the positive real axis. Let this be $C$.
I defined the integrand as a complex valued function, and found that the poles are at $-2-i$ and $-2+i$.
Computing the residues, we have
$$Res(f, -2+i) = -\frac{\sqrt{-2-i}}{2i}$$
and
$$Res(f, -2+i) = \frac{\sqrt{-2+i}}{2i}.$$
So then
$\int_C f(z) = 2\pi i (\frac{\sqrt{-2+i}}{2i}-\frac{\sqrt{-2-i}}{2i}) = \pi (\sqrt{-2+i}-\sqrt{-2-i}).$
Now, after doing the keyhole part, the integral over the large circle and the small circle goes to 0 and so the only thing left are the integral over the line segment connecting the small circle to the large circle at angle $\epsilon$ and the integral over the line segment connecting the large circle to the small circle at angle $2\pi - \epsilon$. Letting the radius of the small circle go to 0 and the radius of the large circle to infinity, we find that the two integrals are equal and
$$\int_C f(z) dz = 2\int_{0}^{\infty} f(x) dx.$$
But
$$\int_{0}^{\infty} f(x) dx = \frac{\pi}{2} (\sqrt{-2+i}-\sqrt{-2-i}),$$
which is an imaginary number.
I've been trying to find my mistake but I can't seem to. Where did I go wrong? Wolfram tells me that the answer is $\sqrt{\frac{1}{2}(\sqrt{5} - 2)}\pi$. My guess is that it's somewhere in the residue theorem step... Please help!
 A: With a keyhole contour you give $z$ a phase from $0$ to $2\pi$, not $-\pi$ to $\pi$, because of how the contour is traversed anticlockwise, so the phase of phase $\sqrt{z}$ is in $[0,\,\pi)$. (Logarithms provide a similar need for caution.) Write $w:=\sqrt{-2+i}$ as $x+iy$ with $x,\,y\in\Bbb R$ so$$x^2+y^2=\sqrt{5},\,x^2-y^2=-2\implies x^2=\frac{\sqrt{5}-2}{2}.$$Since $w^2$ has obtuse phase, $w$ has acute phase, so $x>0$. What's more, $\sqrt{-2-i}=-w^\ast$ (you took it as $w^\ast$), so since $z^2+4z+5=(z-w^2)(z-w^{\ast2})$ the integral is$$\frac{2\pi i}{1-e^{2\pi i\cdot\frac32}}\frac{w+w^\ast}{w^2-w^{\ast2}}=\pi x=\pi\sqrt{\frac{\sqrt{5}-2}{2}}$$as expected, because $w^2-w^{\ast2}=2i\Im(-2+i)=2i$.
A: The issue is with the square root. How do you define the $\sqrt z$ for a complex number? If you say $$\sqrt z=a+bi$$and you find $a$ and $b$ such that $$z=(a+bi)^2$$
then $a'=-a$ and $b'=-b$ will also be a solution. Therefore $$\sqrt z=\pm(a+bi)$$ Then using the same convention as you used, the answer from Wolfram corresponds to $$\sqrt{-2+i}-(-\sqrt{-2-i})=\sqrt{2(\sqrt 5-2)}$$
EDIT:
A different option is to use your integral from $-\infty$ to $\infty$. The negative part will be a purely complex number. So all you need to do is to continue the contour on the upper half, and you need to calculate only one residue. Then take the real part, and you got the answer.
