The integral is
$$\int_0^{\infty}\frac {1}{\sqrt{x}(1+x^2)}dx$$
which is to be evaluated by contour integration.
So, the integrand clearly has simple poles at $+/- i$.
But what kind of pole does the factor $\large \frac{1}{\sqrt{z}}$ have? Should I... "round up" to 1, so that $z=0$ is also a simple pole?
If what I said about the pole at $z=0$ is ok, then would a keyhole contour be advisable to use? The smaller circle would go to zero - and touch the pole -so is this an issue?
Or is there a better / correct contour to use instead?
Thanks,
We can enforce the substitution $x\to x^2$ and write the integral of interest, $I$, as
$$I=\int_{-\infty}^\infty \frac{1}{1+x^4}\,dx \tag 1$$
The integral can be evaluated in terms of its residues in the upper-half plane as
$$\begin{align} I&=2\pi i\left(\text{Res}\left(\frac{1}{1+z^4}, z=e^{i\pi/4}\right)+\text{Res}\left(\frac{1}{1+z^4}, z=e^{i3\pi/4}\right)\right)\\\\ &=2\pi i \left(\frac{1}{4e^{i3\pi/4}}+\frac{1}{4e^{i9\pi/4}}\right)\\\\ &=\frac{\pi\sqrt{2}}{2} \end{align}$$
NOTE:
If one wishes to proceed using a keyhole contour, then we have
$$\begin{align} 0&=2\int_0^\infty \frac{1}{\sqrt{x}(1+x^2)}\,dx+2\pi i\text{Res}\left(\frac{1}{\sqrt{z}(1+z^2)},z=i\right)\\\\ &+2\pi i \text{Res}\left(\frac{1}{\sqrt{z}(1+z^2)},z=-i\right)\\\\ \int_0^\infty \frac{1}{\sqrt{x}(1+x^2)}\,dx&=-\pi i\left(\frac{1}{\sqrt{e^{i\pi/2}}(2i)}+\frac{1}{\sqrt{e^{i3\pi/2}}(-2i)}\right)\\\\ &=\frac{\pi\sqrt{2}}{2} \end{align}$$