Residue calculus

Description

Residue calculus is a powerful tool where the Residue theorem is used to calculate a real integral.

The idea is to make the real integration set into a complex closed curve. We can calculate the integral of the complete closed curve and subtract the part we calculated too much of.

Integral of a rational function

We want to calculate the following integral: ${\int }_{\infty }^{\infty }\frac{p(x)}{q(x)}dx$ Where $p,q$ are polynomials, $q$ has no real zeroes and the degree of $q$ is by $2$ bigger to ensure integrability. The integral can be calculated by following the steps:

1. We calculate the location and degree of the zeros of $q$. They are the Poles of $\frac{p(x)}{q(x)}$.
2. We define the two paths ${\gamma }_1:[-R,R]\to \mathbb{C},t\mapsto t\text{ and }{\gamma }_2:[0,\pi ]\to \mathbb{C},t\mapsto e^{it}$ Notice that the paths form a closed loop and if $R$ is chosen big enough then all poles in the upper half space will sit in the interior.
3. Concatenation of the curve yields
4. ${\int }_{{\gamma }_1\ast {\gamma }_2}\frac{p(x)}{q(x)}dx={\int }_{{\gamma }_1}\frac{p(x)}{q(x)}dx+{\int }_{{\gamma }_2}\frac{p(x)}{q(x)}dx$ The left half can be calculated by adding up all Residues. The right summand goes to zero when $R$ goes to infinity. This is due to the degree difference of denominator and denumerator.

Integral of a sine/cosine function on $[0,2\pi ]$

We want to calculate the integral of a rational function dependent on $\sin$ or $\cos$. ${\int }_{\infty }^{\infty }\frac{p(\cos x)}{q(\cos x)}dx$ For example: ${\int }_0^{2\pi }\frac{1}{5+3\cos t}dt$ To do this we transform the integral into a form, where a circle appears. We do the following steps:

1. Do the following replacements: $\cos (t)=\frac{1}{2}(e^{it}+e^{-it})\text{ and }\sin (t)=\frac{1}{2i}(e^{it}-e^{-it})$
2. Do an inverse transformation from the parametrised integral back to a curve $\gamma (t)=e^{it}$. Multiply $\frac{{\gamma }^{\prime }(t)}{i\gamma (t)}=\frac{i{e}^{it}}{i{e}^{it}}$ if needed. We obtain ${\int }_{\gamma }\frac{1}{5+\frac{3}{2}(z+z^{-1})}\frac{1}{iz}dz$
3. The integral can be calculated by counting up the residues inside the circle.

If the integral is a combination of polynomials and sine/cosine periodicity is lost and we need other methods:

Integral of a sine/cosine function and polynomial

We want to calculate the integral of a function which is of the form ${\int }_{-\infty }^{\infty }\sin x\frac{p(x)}{q(x)}dx\text{or}{\int }_{-\infty }^{\infty }\cos x\frac{p(x)}{q(x)}dx$ The integral will be the real of imaginary part of a complex function, which can be integrated more easily. We apply the following steps.

1. Notice ${\int }_{-\infty }^{\infty }\cos x\frac{p(x)}{q(x)}dx=Re{\int }_{-\infty }^{\infty }{e}^{ix}\frac{p(x)}{q(x)}dx$. We will calculate the right hand side.
2. Define ${\gamma }_1:[-R,R]\to \mathbb{C},t\mapsto t$. We complete ${\gamma }_1$ into a simple closed curve $\gamma$, either by adding an arc (like before) or a half square curve (if there is no apparent degree difference). This is expected to be $0$ if $R$ goes to infinity
3. We calculate the residues inside $\gamma$.
4. The difference of the two integrals gives us the right hand result. Taking the real value gives us the desired result.

Sometimes we only want to calculate one residue. Then a piece of pizza might be best.

Integral using a Pizza piece

We want to calculate an integral of the form ${\int }_0^{\infty }f(x)dx$
Using a pizza piece we only need to calculate one residue. This can be useful if the residues are hard to calculate or if we have a variable number of residues.

1. Define three curse which together form a pizza piece with twice the angle of the first residue
2. Calculate the residue
3. The left side of the pizza piece cant be calculated explicitly but it is often related to the right side by a factor. The side at the back goes to $0$ if $R$ goes to infinity.