# Putnam Calculus problem

1. Sep 6, 2012

2. Sep 6, 2012

### Staff: Mentor

Probably by clever testing. For large r, the integrals get dominated by large x close to pi/2 (where x^r >> 1). Both integrals will diverge to positive infinity. Approximate $\sin(x)\approx 1$ and $\cos(x)\approx \frac{\pi}{2}-x$.

Numerator: $r^c \int_0^{\pi/2} x^r dx = r^c \frac{1}{r+1} \left(\frac{\pi}{2}\right)^{r+1}$
Denominator: $\int_0^{\pi/2} x^r (\frac{\pi}{2}-x) dx = \frac{1}{r+1} \left(\frac{\pi}{2}\right)^{r+2} - \frac{1}{r+2} \left(\frac{\pi}{2}\right)^{r+2}$

As fraction:
$$\frac{r^c}{\frac{\pi}{2} (1 - \frac{r+1}{r+2})} = \frac{2}{\pi} r^c(r+2)$$
There is only one way to make it finite and positive in the limit r->inf, this is c=-1, and it directly leads to L=2/pi.

You can even do this a bit more formal with upper and lower limits for sin and cos and prove the limit in the process.

3. Sep 6, 2012

Thanks mfb