Kokeboken
Table of Contents
1. First Part
1.1. Fractions
It's useful to remember how to sum fractions:
\begin{equation} \frac{1-x}{1+x} + \frac{x}{x-1} = \frac{3x-1}{x^{2}-1}. \end{equation}Going the other way, finding the partial fraction expansion for a given rational expression:
\begin{equation} \frac{2}{x^{2}-1} = \frac{1}{x-1} - \frac{1}{x + 1} \end{equation}1.1.1. Some Fun Integrals
Integration is more an art than a science. There are a variety of ways to compute the same integral. Consider
\begin{equation} I = \int^{2}_{1}\frac{4x}{x^{2}-1}\D x \end{equation}Consider the integral
\begin{equation} I = \int^{\infty}_{1}\frac{1}{x(1+x^{2})}\D x \end{equation}We can use partial fraction expansion
\begin{equation} \frac{1}{x(1 + x^{2})} = \frac{1 + x^{2} - x^{2}}{x(1 + x^{2})} = \frac{1}{x} - \frac{x}{1 + x^{2}}. \end{equation}Integration becomes
\begin{equation} \begin{split} I &= \int\frac{\D x}{x} - \int\frac{x\,\D x}{1 + x^{2}} \\ &= \log(x) - \frac{1}{2}\int\frac{2x\,\D x}{1 + x^{2}} = \log(x) - \frac{1}{2}\log(1+x^{2}) + C \end{split} \end{equation}for some constant of integration \(C\). This is a correct result, but we invoke the law of logarithms \(\log(a) + \log(b) = \log(ab)\) to handle the bounds of integration
\begin{equation} I = \frac{1}{2}\log\left(\frac{x^{2}}{1 + x^{2}}\right) + C \end{equation}Then we have
\begin{equation} \lim_{x\to\infty}\frac{1}{2}\log\left(\frac{x^{2}}{1 + x^{2}}\right) = \lim_{x\to\infty}\frac{1}{2}\log\left(\frac{1}{x^{-2} + 1}\right) = 0. \end{equation}Hence we find the integral
\begin{equation} \begin{split} I &= \lim_{x\to\infty}\frac{1}{2}\log\left(\frac{1}{x^{-2} + 1}\right) - \frac{1}{2}\log\left(\frac{1}{1 + 1}\right)\\ &= \frac{1}{2}\log(2). \end{split} \end{equation}And we're done. Is there an easier method? Yes, luckily. But it requires greater foresight to recognize. We sart instead with the somewhat unintuitive substitution \(x=1/y\) with the differential \(\D x=-\D y/y^{2}\). Then \(x\to\infty\) is the same limit as \(y\to0\) since \(y=1/x\). And since \(x\to 1\) we have \(y\to 1\). Then the integral becomes
\begin{equation} \begin{split} I &= -\int^{0}_{1}\frac{1}{(1/y)(1 + (1/y)^{2})}\frac{\D y}{y^{2}}\\ &= \frac{1}{2}\int^{1}_{0}\frac{2y}{1 + y^{2}}\D y. \end{split} \end{equation}And this can be solved easily with a substitution \(u=y^{2}\). This trick of replacing \(x=1/y\) is called an inverse substitution or bijective substitution. It's really useful for integrals with domain of integration from 1 to \(\infty\).
1.2. Trigonometric Functions
We have that
\begin{equation} \frac{\D}{\D x}\arctan(x) = \frac{1}{1+x^{2}} \end{equation}so
\begin{equation} \int^{t}_{0}\frac{1}{1+x^{2}}\D x = \arctan(t) \end{equation}for all \(t\gt 0\). Further, we find
\begin{equation} \int^{t}_{0}\frac{1}{1+x^{2}}\D x = \int^{\infty}_{1/t}\frac{1}{1+x^{2}}\D x = \arctan(t). \end{equation}The proof involves recalling the derivative of inverse functions. Let \(y=\tan(x)\). Then \(\arctan(y) = x\) and the chain rule gives us the left-hand side differentiates as
\begin{equation} \frac{\D}{\D x}\arctan(y) = \frac{\D\arctan(y)}{\D y}\frac{\D y}{\D x} \end{equation}whereas the right-hand side differentiates as
\begin{equation} \frac{\D}{\D x}x = 1. \end{equation}Hence we have
\begin{equation} \frac{\D\arctan(y)}{\D y} = \frac{1}{\D y/\D x}. \end{equation}Now recalling
\begin{equation} \frac{\D}{\D x}\tan(x) = \sec^{2}(x) = 1 - \tan^{2}(x) \end{equation}we have
\begin{equation} \frac{\D\arctan(y)}{\D y} = \frac{1}{\D y/\D x} = \frac{1}{1 - \tan^{2}(x)}=\frac{1}{1-y^{2}}. \end{equation}The inverse function theorem gives us the result
\begin{equation} \int^{t}_{0}\frac{\D x}{1 + x^{2}} = \arctan(t). \end{equation}The substitution \(u=1/x\) gives us the second result, since \(-\D u/u^{2}=\D x\) and the bounds of integration become \(x\to 0\) is \(u\to\infty\), while \(x\to t\) is \(u\to 1/t\), so
\begin{equation} \int^{t}_{0}\frac{\D x}{1 + x^{2}} = -\int^{1/t}_{\infty}\frac{1}{1 + (1/u)^{2}}\frac{\D u}{u^2}. \end{equation}Massaging the right-hand side produces the desired result.