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.

## Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett