## Monday, 13 October 2014

### Analysis for Realsies: 6

One of the challenge fun exercises that I gave you last week's tute B was to exactly compute the integral of $$\frac{1}{1+x^4}$$ rather than to rely on comparison techniques to show that this is integrable. How did you guys go with it? I know that I certainly didn't see the first step of this when I tried to do this integral in first year. =P

My hint was the $$1+x^4$$ was factorisable into two real polynomials, and one way to see this is to factorise it into four linear terms involving complex numbers and to pair up linear terms until you got pairings which produced real quadratic polynomials. The other way is to complete the square: \begin{align}1+x^4=&1+2x^4+x^4-2x^2\\=&(1+x^2)^2-(\sqrt{2}x)^2\\=&(1-\sqrt{2}x+x^2)(1+\sqrt{2}x+x^2).\end{align} Now, if you're comfortable with partial fractions, you should immediately notice that: \begin{align}\frac{1}{1-\sqrt{2}x+x^2}-\frac{1}{1+\sqrt{2}x+x^2}&=\frac{2\sqrt{2}x}{1+x^4}\text{, and}\\ \frac{1}{1-\sqrt{2}x+x^2}+\frac{1}{1+\sqrt{2}x+x^2}&=\frac{2+2x^2}{1+x^4}.\end{align} Since $$1=\frac{1}{2}(2+2x^2)-\frac{x}{2\sqrt{2}}2\sqrt{2}x$$, we see that: \begin{align}\frac{1}{1+x^4}=&\frac{1}{2}\left(\frac{1}{1-\sqrt{2}x+x^2}+\frac{1}{1+\sqrt{2}x+x^2}\right)\\&-\frac{x}{2\sqrt{2}}\left(\frac{1}{1-\sqrt{2}x+x^2}-\frac{1}{1+\sqrt{2}x+x^2}\right)\\=&\frac{\frac{1}{2}-\frac{x}{2\sqrt{2}}}{1-\sqrt{2}x+x^2}+\frac{\frac{1}{2}+\frac{x}{2\sqrt{2}}}{1+\sqrt{2}x+x^2}\\=& \frac{1+\sqrt{2}x}{4(1+\sqrt{2}x+x^2)}+\frac{1-\sqrt{2}x}{4(1+\sqrt{2}x+x^2)}\\&+\frac{1}{4(1+\sqrt{2}x+x^2)}+\frac{1}{4(1-\sqrt{2}x+x^2)}.\end{align} Finally, integrating the two upper-line terms into a logarithm and integrating the two lower-line terms into an inverse tan function, we get: \begin{align} \int \frac{\mathrm{d}x}{1+x^4}=&\frac{1}{4\sqrt{2}}\log|1+\sqrt{2}x+x^2|-\frac{1}{4\sqrt{2}}\log|1-\sqrt{2}x+x^2|\\&+\frac{1}{2\sqrt{2}}\mathrm{arctan}(\sqrt{2}x+1)+\frac{1}{2\sqrt{2}}\mathrm{arctan}(\sqrt{2}x-1).\end{align}Aaaaaaand here's another fun exercise: differentiate $$\sin(x)$$ from first principles! And here's a hint (it's written in white, so highlight it):

You'll probably need to show at some point that $$\lim_{x\to 0}\frac{\sin(x)}{x}=1$$...and this can totes be done by proving the following inequality $$\sin(x)<x<\tan(x)$$ for $$x\in(0,\frac{\pi}{2})$$. And...this can be done...geometrically.

=P