Use contour integration to evaluate the integral
$\int_{-\infty}^{\infty} \frac{dx}{x^2+3} $
Using the change of coordinates $y = \frac{1}{x}$
So am I literally just substituting this coordinate in the equation before solving it or does it have something to do with the limits instead? It's been a while since I've done contour integration if you hadn't have guessed
That substitution cannot work. It is discontinuous on the interval of integration. If you are persistent and would perform the integration with that substitution, you would have to take those limits along, which then means you'd be integrating from $0$ to $0$, which of course cannot be right. Why not straight forward integration using an arctan? The anti derivative is then $\frac{1}{\sqrt{3}}arctan\frac{x}{\sqrt{3}}$ running from $-\infty$ to $\infty$ which results in $\pi/\sqrt{3}.$ With contour integration, realize there is a pole at $x=i\sqrt{3}$. Factoring your denom gives $\frac{1}{(x+i\sqrt{3})(x-i\sqrt{3})}$ and thus the residue can be found by substituting $i\sqrt{3}$ into $\frac{1}{x+i\sqrt{3}}$ to get $RES=\frac{1}{2i\sqrt{3}}$ By Cauchy's theorem we multiply by $2i\pi$ to arrive at the same answer. These are the two methods that come to my mind. No clue what that substitution has got to do with it...