I would like to receive some help on evaluating the integral $\int_{0}^{\infty} \frac{z^n\,dz}{z^{2n}+1}$ because I am really new to integration over curves. So my thoughts were: The singularities of the fraction are at $z^{2n} = -1$ hence on the unit circle there are a certain number of evenly distributed singularity points (like roots of unity). I want to evaluate my integral and avoid these singularities. Assuming integer $n\geq2$
Consider the pie slice where we have a straight line from the origin to point R on the real line; then we travel up the arc of a circle of radius R centered at $0$ to the point $Re^{i\phi}$ I am not quite sure what to put for the angle, something that depends on $n$ say $\phi<\frac{\pi}{2n}$ will avoid the singularities from earlier; then we travel from this point to $0$ via a straight line. The orientation for this closed curve is the positive orientation.
I am thinking that, by avoiding the singularities, I can say that the integral over this closed curve must be zero. After that I want to split the integral into the three parts, corresponding to the pieces of the closed curve. I have figured out so far that the integral along the arc of the circle of radius R centered at $0$ approaches $0$ as R approaches $\infty$ but I am stuck on what to do next in evaluating the remaining two integrals so I would appreciate as much guidance as possible, I really want to learn this type of integration well for a complex analysis class.
Best Answer
The reason to use the suggested contour is not to avoid the singularities (these help, actually) but to have the constituent integrals related to the given integral (or evaluated some other way).
In our case, a good choice is $\color{blue}{\phi=\pi/n}$: $$\int_0^\infty\frac{z^n\,dz}{z^{2n}+1}=I\implies\int\limits_{\substack{z=re^{i\pi/n}\\0<r<\infty}}\frac{z^n\,dz}{z^{2n}+1}=-e^{i\pi/n}I,$$ so that, applying the residue theorem (and taking $R\to\infty$), you get $$(1+e^{i\pi/n})I=2\pi i\operatorname*{Res}_{z=e^{i\pi/2n}}\frac{z^n}{z^{2n}+1}=\frac\pi{n}e^{i\pi/2n}\implies I=\frac\pi{2n}\sec\frac\pi{2n}.$$