How to prove $$\frac{1}{2n+2}<\int_0^{\frac{\pi}{4}}\tan^nx\,dx< \frac{1}{2n}$$ Set $A_n=\int_0^{\frac{\pi}{4}}\tan^nx\,dx$, then we have $A_n+A_{n+2}=\frac{1}{n+1}$ and we have $A_{n+2} < A_n$ ,so we can get $$\frac{1}{2n+2}< \int_0^{\frac{\pi}{4}}\tan^nx\,dx < \frac{1}{2n-2}$$ But how to show that$$\int_0^{\frac{\pi}{4}}\tan^nx\,dx < \frac{1}{2n}$$


Change variable to $t = \tan x$, we have

$$I_n \stackrel{def}{=}\int_0^{\pi/4} \tan^n x dx = \int_0^1 \frac{t^n}{1+t^2} dt$$

Notice for $t \in (0,1)$, we have $\frac{1 + t^2}{2} < 1$. This implies

$$I_n > \int_0^1 \frac{t^n}{1+t^2}\cdot\frac{1+t^2}{2} dt = \frac12\int_0^1 t^n dt = \frac{1}{2(n+1)}$$

On the other direction, AM $\ge GM$ tell us $t = \sqrt{1 \cdot t^2} \le \frac{1+t^2}{2}$ and the inequality is strict when $t \ne 1$. This leads to

$$I_n = \int_0^1 \frac{t^{n-1}}{1+t^2} t dt < \int_0^1 \frac{t^{n-1}}{1+t^2}\cdot \frac{1+t^2}{2} dt = \frac12 \int_0^1 t^{n-1} dt = \frac{1}{2n}$$


For $n=0$ there is nothing to prove. For $n=1$, this is $\log\sqrt{2}<\frac12$, which is obviously true.

We have $$ \tan^{n+1}x<\frac12(\tan^n x+\tan^{n+2}x)\text{ for }x\in(0,\pi/4) \tag{1} $$ because $y\in\mathbb{R}^+\mapsto c^y\in\mathbb{R}^+$ is strictly convex for $c>0$. So you get $A_{n+1}<\frac1{2(n+1)}$, or equivalently $A_n<\frac1{2n}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.