Complex Analysis/Residue Theory/Some Consequences

Simplifying Integrals
Given the following integral:

$$\int_a^b \frac{1}{x^2-1} dx$$

Now with partial fractions (or the residue theorem) we can split this up into a series of mono-pole terms, which would allow us to use substitution and receive logarithmic answers:

$$\int_a^b \frac{\frac{1}{2}}{x-1} dx + \int_a^b \frac{-\frac{1}{2}}{x+1} dx$$

$$\frac{1}{2} \ln (x-1) |_a^b - \frac{1}{2} \ln (x+1) |_a^b$$

Cauchy's Residue Theorem
Cauchy's Residue Theorem is a VERY important result which gives many other results in Complex Analysis and Theory, but more importantly to us, is that it allows us to calculate integration with only residue, that is, we can literally integrate without actually 'integrating'. Note: Its derivation is in Complex Analysis, which is listed as a prerequisite for these more advanced tricks.

This is the actual (general) theorem:

Let &Gamma; be a simple closed positively oriented contour. If f is analytic in some simply connected domain D containing &Gamma; and $$z_0$$ is any point inside &Gamma;s, then:

$$f^{n}(z_0)= \frac{n!}{2 \pi i} \int_{\Gamma} \frac{f(\gamma)}{(\gamma-z)^{n+1}} dz$$

Upon first look, this has absolutely nothing to do with residues, but mathematicians are very abstract and tricky people.

Take a general function, call it $$g(z)$$, just so we don't confuse it with the function in Cauchy's Integral Formula. $$g(z)$$ can be made into a Laurent Series:

$$g(z) = \sum_{n=-\infin}^{\infin} a_n (z-z_0)^n$$

Now, we integrate over a contour &Gamma; of $$g(z)$$, keeping in mind that $$g(z)$$ has been 'Laurentized':

$$\int_{\Gamma} \sum_{n=-\infin}^{\infin} a_n (z-z_0)^n dz$$

Also by Complex Analysis, the parts of the series that are analytic and zero and thus dropped out (which is actually ANOTHER result by Cauchy), this leaves the sum of the integrals containing powers on the bottom, now by carrying out the differentiations of $$g(z)$$ and applying the General Cauchy Integral Formula (The proof is tedious but you can do the Laurent series and check yourself), you will come upon the Cauchy Residue Theorem (Cauchy really did do a lot of this stuff, a running joke in Complex Analysis classes is, "Isn't every proof done by Cauchy?"):

$$\int_{\Gamma} g(z) = 2 \pi i \sum_{j=1}^{n} \mathrm{Res}(z_j)$$

Read over that equation a few times, make sure you really grasp what it's saying, to do an integral, you need only calculate the residues. Does it seem useless because you're concerned with only the real number line? You're not being creative enough. Take the line integral with one part over the real-number line, and the other lines over the complex such that they are easily computable, as a general example:

$$\int_{\Gamma} g(z) = \int_\text{Real Line} g(z) + \int_\text{A simple line that closes the loop} g(z) = 2 \pi i \sum_{j=1}^{n} \mathrm{Res(z_j)}$$

Unfortunately, this too borders the line with a book on Complex Analysis, since these 'simple lines' are discussed therein, but just to not leave you hanging, here's a typically simple one so you can try out integration without integrating on your own:

If $$f(z)$$ is the quotient of two polynomials such that the degree of the lower polynomial is at least two more than the numerator polynomial then $$\lim_{\rho \rightarrow \infin} \int_{C_{\rho^+}} f(z) dz = 0$$.

Where $$C_{\rho^+}$$ is a half-circle on the plane. This allows you to close some loops made by integrals over the real-number line and try out this method for yourself.

And that's where we leave off!