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!
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!
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 $\{I_n\}$.
$I_n = \int_0^{\pi/4} \tan^n(x) dx = \int_0^{\pi/4} \tan^{n-2}(x) \sec^2(x) dx - \int_0^{\pi/4} \tan^{n-2}(x)dx$ $I_n + I_{n-2}= \int_0^1 t^{n-2} dt = \dfrac1{n-1}$ 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 $L + L = \lim_{n \to \infty} \dfrac1{n-1} \implies L = 0$
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.
Hint: Use dominated convergence theorem