I'm supposed to work out the following limit:
$$\lim_{n\to\infty} \int_{0}^{\pi/2}\frac{1}{1+x \left( \tan x \right)^{n} }dx$$
I'm searching for some resonable solutions. Any hint, suggestion is very welcome. Thanks.
Answer
Note that the integrand is bounded in $[0,\pi/2]$, so if $$\lim_{n\to \infty} \frac{1}{1+x\tan^nx}$$ exists a.e. then we may apply the Dominated Convergence Theorem to show $$\lim_{n\to \infty} \int_0^{\pi \over 2}\frac{1}{1+x\tan^nx}dx = \int_0^{\pi \over 2}\lim_{n\to \infty} \frac{1}{1+x\tan^nx}dx.$$
If $x<\pi/4$ then the integrand converges to 1, and if $x>\pi/4$ then it converges to 0. Thus we have the integral equals
$$
\int_0^{\pi \over 4} 1dx + \int_{\pi \over 4}^{\pi \over 2} 0dx = \frac{\pi}{4}.
$$
No comments:
Post a Comment