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I'm trying to find some nice proofs for the following limit

$$\lim_{n\to\infty} \int_{0}^{\pi/4}\tan^n x \ dx$$

One way is to use the integration by parts. What else can we do here? Are there some fast ways?
All answers will have my upvote. Thanks!

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    $\begingroup$ Rule of thumb solves this. $\tan x$ is between $0$ and $1$. Taking this to infinite power will push everything to zero except for the rightmost point, which stays one. But that's measure zero and doesn't contribute to the integral. There is no indeterminate expression here, just a function sequence that converges to a known discontinuous function. $\endgroup$ – orion Mar 28 '14 at 23:46
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Use the substitution $t:=\tan x$ (then $\arctan t=x$ and $dx=\frac{dt}{1+t^2}$). Then $$0\leq I_n:=\int_0^{\frac{\pi}4}\tan^n xdx=\int_0^1\frac{t^n}{1+t^2}dt\leq \int_0^1t^ndt=\left[\dfrac{t^{n+1}}{n+1}\right]_{0}^{1}=\frac 1{n+1},$$ which gives $0$ as limit.

The formula also gives the recursion relation $I_{n+2}=\frac 1{n+1}-I_n$, which can help to study the asymptotic behavior of $\{I_n\}$.

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  • $\begingroup$ that's pretty nice. $\endgroup$ – user 1591719 Sep 25 '12 at 20:28
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$$I_n = \int_0^{\pi/4} \tan^n(x) dx = \int_0^{\pi/4} \tan^{n-2}(x) \sec^2(x) dx - \int_0^{\pi/4} \tan^{n-2}(x)dx$$ $$I_n + I_{n-2}= \int_0^1 t^{n-2} dt = \dfrac1{n-1}$$ Note that $I_n$ is monotone decreasing with $n$ and is bounded below by $0$ and hence it converges.

Hence, if $\lim_{n \to \infty} I_n = L$, we have that $L + L = \lim_{n \to \infty} \dfrac1{n-1} \implies L = 0$

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    $\begingroup$ Another nice way! $\endgroup$ – user 1591719 Sep 25 '12 at 20:36
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Use the Lebesgue Dominated Convergence Theorem, or the Monotone Convergence Theorem.

EDIT: Or explicitly, given $\pi/4 > \epsilon > 0$,

$$\int_0^{\pi/4} \tan^n x\ dx < (\pi/4 - \epsilon) \tan(\pi/4 - \epsilon)^n + \epsilon $$ which is less than $2 \epsilon$ if $n$ is large enough.

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  • $\begingroup$ I didn't think of the approach after EDIT. $\endgroup$ – user 1591719 Sep 25 '12 at 20:36
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Hint: Use dominated convergence theorem

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