Proving $\lim\limits_{n\to\infty}\int_0^{\pi/4} \tan^n{x}\,dx=0$

Question

How would you prove that $\displaystyle\lim\limits_{n\to\infty}\int_0^{\pi/4} \tan^n{x}\,dx=0$.

It is obvious if you see the graph of $\tan^n{x}$ on $(0, \pi/4)$ as $n$ increases but i'm looking for a more algebraic way.

This result is for connecting the power reduction formula for $\displaystyle\int_0^{\pi/4} \tan^n{x}\,dx$ to Leibniz formula for $\pi$.

Answer 1

Squeezing is straightforward: $$ 0\leq \int_{0}^{\pi/4}\tan^n(x)\,dx \stackrel{x\mapsto\arctan u}{=}\int_{0}^{1}\frac{u^n}{1+u^2}\,du \leq \int_{0}^{1}u^n\,du = \frac{1}{n+1}.$$

Answer 2

Let $0<\varepsilon<\frac{\pi}{4}$. Write the integral

$$I_n=\int_0^{\pi/4}\tan^n x\,\mathrm dx=\int_0^{\pi/4-\varepsilon}\tan^n x\mathrm dx+\int_{\pi/4-\varepsilon}^{\pi/4}\tan^n x\,\mathrm dx=J_n+K_n$$

First notice the limit of $I_n$ exists, because it's decreasing and positive, since $0\le\tan^{n+1}x\le\tan^{n}x$.

Since the function $x\to\tan^n x$ is increasing, the first term $J_n$ is bounded by:

$$J_n\le(\frac{\pi}{4}-\varepsilon)\tan^n(\frac{\pi}{4}-\varepsilon)\le\frac{\pi}{4}\beta^n$$

With $0<\beta<1$.

Since $\tan^n x\le1$, The second term $K_n$ is bounded by:

$$K_n\leq\varepsilon$$

Hence, for all $\varepsilon$, there is a $\beta$ such that $0<\beta<1$ such that for all $n$,

$$0\le I_n\le \frac{\pi}{4}\beta^n+\varepsilon$$

Hence, for all $\varepsilon$,

$$0\le\lim_{n\to\infty} I_n\le\varepsilon$$

So

$$\lim_{n\to\infty} I_n=0$$

Answer 3

More than likely, this is too complex.

The antideirvative can be computed using hypergoemetric functions $$I_n=\int \tan ^{n}(x)\,dx=\frac{\tan ^{n+1}(x)}{n+1} \, _2F_1\left(1,\frac{n+1}{2};\frac{n+3}{2};-\tan ^2(x)\right)$$ making $$J_n=\int_0^{\frac \pi 4} \tan ^{n}(x)\,dx=\frac{1}{4} \left(\psi \left(\frac{n+3}{4}\right)-\psi \left(\frac{n+1}{4}\right)\right)$$ Using the asymptotics $$\psi(z)=\log \left({z}\right)-\frac{1}{2 z}-\frac{1}{12 z^2}+\frac{1}{120 z^4}+O\left(\frac{1}{z^6}\right)$$ and continuing using Taylor series $$J_n=\frac{1}{2 n}-\frac{1}{2 n^3}+O\left(\frac{1}{n^5}\right)$$

Answer 4

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With Laplace's Method:

\begin{align} \int_{0}^{\pi/4}\tan^{n}\pars{x}\,\dd x & = \int_{0}^{\pi/4}\tan^{n}\pars{{\pi \over 4} - x}\,\dd x = \int_{0}^{\pi/4}\exp\pars{n\ln\pars{\tan\pars{{\pi \over 4} - x}}}\,\dd x \\[5mm] & \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, \int_{0}^{\infty}\expo{-2nx}\bracks{1 -{4n \over 3}\,x^{3} + \mrm{O}\pars{nx^{5}}}\dd x \\[5mm] & = {1 \over 2n} - {1 \over 2n^{3}} + \mrm{O}\pars{1 \over n^{5}} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\,\bbx{\large 0} \end{align}