I am trying to solve $\int^{+\infty}_{-\infty}\frac{1}{x}dx$ I read that it is a contour integral along the semi-circle of large radius in the lower complex plane, firstly is there any justification for this and secodnly does the integral then become $\int^{\pi}_{0} i d\theta$
I expect you have to use the parameterisation $\displaystyle x = r\,e^{i\theta} \implies dx = i\,r\,e^{i\theta}\,d\theta$.