View source for The integral of sin(x) the hard way! MA181Fall2011Bell - Rhea

We want to calculate

<math>\int_0^\pi \sin x\ dx</math>

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

So

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

when <math>N</math> is large.

Recall Euler's identity,

<math>e^{i\theta}=\cos\theta + i\sin\theta.</math>

Hence, that Riemann sum <math>R</math> is the imaginary part of

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

But

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

so <math>R</math> is just the imaginary part of a geometric sum.

The formula

<math>1+r+r^2+\dots+r^N = \frac{1-r^{N+1}}{1-r}</math>

lets us calculate that <math>R</math> is the imaginary part of

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

Now

<math>\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}</math>

since

<math>e^{i\pi}=\cos\pi+i\sin\pi=-1+0i=-1.</math>

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

<math>\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).</math>

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

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

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