Difference between revisions of "The integral of sin(x) the hard way! MA181Fall2011Bell" - Rhea

Latest revision as of 08:33, 2 September 2011

We want to calculate

$ \int_0^\pi \sin x\ dx $

three days before we learn the Fundamental Theorem of Calculus, so our only tool is the limit of a Riemann sum.

So

$ \int_0^\pi \sin x\ dx\approx \sum_{n=1}^N \sin(n\pi/N)(\pi/N) $

when $ N $ is large.

Recall Euler's identity,

$ e^{i\theta}=\cos\theta + i\sin\theta. $

Hence, that Riemann sum $ R $ is the imaginary part of

$ (\pi/N)\sum_{n=1}^N e^{in\pi/N}. $

But

$ e^{in\pi/N}=\left(e^{i\pi/N}\right)^n, $

so $ R $ is just the imaginary part of a geometric sum.

The formula

$ 1+r+r^2+\dots+r^N = \frac{1-r^{N+1}}{1-r} $

lets us calculate that $ R $ is the imaginary part of

$ \frac{\pi}{N}\left(\frac{1-\left(e^{i\pi/N}\right)^{N+1}}{1-e^{i\pi/N}} -1\right). $

Now

$ \left(e^{i\pi/N}\right)^{N+1}= e^{i\pi + i\pi/N}= e^{i\pi}e^{i\pi/N}=-e^{i\pi/N} $

since

$ e^{i\pi}=\cos\pi+i\sin\pi=-1+0i=-1. $

Hence, we obtain that the Riemann sum is equal to the imaginary part of

$ \frac{\pi}{N}\left(\frac{1+e^{i\pi/N}}{1-e^{i\pi/N}}-1\right)=\frac{\pi}{N}\left(\frac{2e^{i\pi/N}}{1-e^{i\pi/N}}\right). $

Finally, use Euler's Identity and compute the imaginary part to get

$ R=\frac{\frac{\pi}{N}\sin(\pi/N)}{1-\cos(\pi/N)}, $

and then L'Hopital's rule can be used to find that the limit as N tends to infinity is equal to 2.