# Calculus of Residues

After blowing off the cobwebs after a couple of years I have been looking at some of the notes that I made some years back on some courses that I took at Warwick on Complex Analysis and Vector Analysis.

Integration has always been one of my favourite areas of mathematics. At A-Level I learned lots of different techniques for calculating some interesting integrals – but A-Level only just skims the surface when it comes to integration and it can be difficult for A-Level students (through no fault of their own) to appreciate the significance of integration. Integration by Parts, Integration by substitution and reduction formula are all great but there are still many integrals which require more advanced techniques to calculate. Contour integrals and the calculus of residues can often come to the rescue.

Contour integrals are a way of passing difficult integrals over a real-interval such as $$\int_{-\infty}^{\infty}\!{\dfrac{x^{2}}{1+x^{4}}\mathrm{d}x}$$ into the complex plane and taking advantage of the Cauchy integral theorem and the calculus of residues. I remember how it felt when I first learned the formula for integration by parts because it meant that I was able to find integrals that were previously impossible for me to calculate – even though I have done contour integrals before it has been very exciting for me to re-discover them. Looking through one of my books I came across this problem – show that for $a>1$

$$\int_{0}^{2\pi}\!\frac{\mathrm{sin}2\theta}{(a+\mathrm{cos}\theta)(a-\mathrm{sin}\theta)}\;\mathrm{d}\theta = -4\pi\left(1-\frac{2a\sqrt{a^{2}-1}}{2a^{2}-1}\right)$$

After spending a good deal of one of my afternoons wrestling with the algebra I managed to arrive at a solution which you can download here as a pdf. Here is a graph of the integrand in the case when $a=2$

As you can see from the diagram, the area bounded by the curve and the $x$-axis certainly exists but trying to calculate this integral using A-Level techniques is going to be incredibly difficult if not impossible (if anyone can do it then I would love to see the solution). Unfortunately there are and always will be integrals that cannot be calculated analytically – this is just the way it is and there is no getting around it but contour integrals certainly allows you to calculate a huge range of integrals that previously would have been seemingly impossible.