Question

Q3 from this years Putnam Comp. Choose and real c and find L for the following,

$$L=lim_{r \to \infty}r^c\frac{\int_0^{\pi/2} \! x^r \sin(x) \, \mathrm{d} x}{\int_0^{\pi/2} \! x^r \cos(x) \, \mathrm{d} x}$$

Answer 1

Should say "choose a real c that evaluates L and solve for L"

Answer 2

we use gamma function?

Answer 3

I've seen these before in physics and what I did was take the taylor expansion summed to infinity for both sin and cos. Like so,

$$L=\lim_{r \to \infty} r^c \frac{\int_0^{\pi/2} x^r \sum_{i=0}^\infty \frac{(-1)^i x^{2i+1}}{(2i+1)!}}{\int_0^{\pi/2} x^r \sum_{i=0}^\infty \frac{(-1)^i x^{2i}}{(2i)!}}$$

$$=\lim_{r \to \infty} r^c \frac{\int_0^{\pi/2} \sum_{i=0}^\infty \frac{(-1)^i x^{2i+1+r}}{(2i+1)!}}{\int_0^{\pi/2} \sum_{i=0}^\infty \frac{(-1)^i x^{2i+r}}{(2i)!}}$$

$$=\sum_{i=0}^\infty \lim_{r \to \infty} r^c \frac{(2i)!}{(2i+1)!}\frac{ \frac{(\pi/2)^{2i+1+r+1}}{2i+r+2}}{ \frac{(\pi/2)^{2i+r+1}}{2i+r+1}}$$

$$=\sum_{i=0}^\infty \lim_{r \to \infty} r^c \frac{1}{(2i+1)}\frac{ (\pi/2)^{2i+r+2}}{(\pi/2)^{2i+r+1}} \frac{2i+r+1}{2i+r+2}$$

$$=\sum_{i=0}^\infty \lim_{r \to \infty} r^c \frac{1}{(2i+1)}(\pi/2)\frac{r}{r} \frac{2i/r+1+1/r}{2i/r+1+2/r}$$

$$=\sum_{i=0}^\infty \lim_{r \to \infty} r^c \frac{1}{(2i+1)}(\pi/2) \frac{2i/r+1+1/r}{2i/r+1+2/r}$$

Someone want to check and finish for me?