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

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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Use the substitution $t:=\tan x$ (then $\arctan t=x$ and $dx=\frac{dt}{1+t^2}$). Then $$0\leq I_n:=\int_0^{\frac{\pi}4}\tan^n xdx=\int_0^1\frac{t^n}{1+t^2}dt\leq \int_0^1t^ndt\leq\frac 1{n+1},$$ 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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 that's pretty nice. – Chris's wise sister Sep 25 '12 at 20:28

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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 I didn't think of the approach after EDIT. – Chris's wise sister Sep 25 '12 at 20:36

Hint: Use dominated convergence theorem

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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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 Another nice way! – Chris's wise sister Sep 25 '12 at 20:36