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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.

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According to numerical simulation, the integrand looks like a monotone decreasing sequence (in $n$, of course). – Siminore Jun 22 '12 at 13:32
It seems to go to $\int_0^1 \frac 1 {1+u^2}=\frac \pi 4$ – Pedro Tamaroff Jun 22 '12 at 13:33
@Peter Tamaroff: yes, but the graph looks like a rectangle when n is large. – user 1618033 Jun 22 '12 at 13:36
I suggest splitting the integral in two parts: from $0$ to $\pi/4$, and from $\pi/4$ to $\pi/2$. The second integral should converge to zero. – Siminore Jun 22 '12 at 13:37
@Chris There's no fuzz about that. – Pedro Tamaroff Jun 22 '12 at 13:37
up vote 8 down vote accepted

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 $$ \int_0^{\pi \over 4} 1dx + \int_{\pi \over 4}^{\pi \over 2} 0dx = \frac{\pi}{4}. $$

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