Putnam Calculus problem

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  • #2
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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 [itex]\sin(x)\approx 1[/itex] and [itex]\cos(x)\approx \frac{\pi}{2}-x[/itex].

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

As fraction:
[tex]\frac{r^c}{\frac{\pi}{2} (1 - \frac{r+1}{r+2})} = \frac{2}{\pi} r^c(r+2)[/tex]
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
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Thanks mfb
 

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