Proving that the integral $\int_1^2 \frac{\arctan(x)}{x} \,dx$ is less than $1$

Question

I am looking to prove that: $$\int_1^2 \frac{\arctan(x)}{x} \,dx< 1$$

Using Darboux higher sum i can do that pretty easily, however, in order to use Darboux sum (which is given by plugging the value $1$ to the function) I also need to show that the function is descreasing, and things got a little messed up there..

any help?

Answer 1

HINT:

$$\int_1^2 \frac{\arctan(x)}{x}\,dx\le \int_1^2 \arctan(x)\,dx$$

or

$$\int_1^2 \frac{\arctan(x)}{x}\,dx\le \arctan(2) \int_1^2 \frac1x\,dx$$

Answer 2

\begin{align*} \int_{1}^{2}\dfrac{\tan^{-1}x}{x}dx\leq\int_{1}^{2}\dfrac{\tan^{-1}2}{x}dx=(\tan^{-1}2)(\log 2)<1. \end{align*}

Answer 3

Note that for $x\in [1,2],$ we have $$ 0.79 \le \tan^{-1}(1)<tan^{-1}(x)< tan^{-1}(2) \le 1.11$$

Thus $$\int_1^2 \frac{arctan(x)}{x} \,dx< 1.11 \int_1^2 \frac{1}{x} \,dx=1.11 ln(2)\le 0.77 <1.$$

Answer 4

Hint: It is shown geometrically in this answer, for $-\frac\pi2\lt x\lt\frac\pi2$, $$ \frac{\tan(x)}x\gt1 $$ This means that $$ \frac{\arctan(x)}x\lt1 $$

Answer 5

Alternative approach: $f(x)=\frac{\arctan x}{x}$ is a convex function on $[1,2]$, hence by the Hermite-Hadamard inequality $$ \int_{1}^{2}\frac{\arctan x}{x}\,dx \leq \frac{1}{2}\left(\frac{\arctan 1}{1}+\frac{\arctan 2}{2}\right)=\frac{1}{2}\arctan(2+\sqrt{5})<\frac{\pi}{4}. $$ Second alternative approach: by the Shafer Fink inequality $\frac{\arctan x}{x}\leq \frac{\pi}{1+2\sqrt{1+x^2}}$, hence the integral is bounded by $$ \int_{1}^{2}\frac{\pi}{1+2\sqrt{1+x^2}}\,dx \leq \frac{7}{10}.$$