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?

Be careful with your mathematical phrasing: don't mix English ("the integral") and symbolic ("<") expressions to obtain things like "The integral $\int_0^2 \cdots dx <1$". — Carl Mummert, Feb 25 '18 at 18:38

score 2 · Accepted Answer · answered Feb 25 '18 at 18:31

2

HINT:

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

or

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

answered Feb 25 '18 at 18:31

Mark Viola

179,405

score 1 · Answer 2 · answered Feb 25 '18 at 18:34

1

\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*}

answered Feb 25 '18 at 18:34

user284331

55,591

score 1 · Answer 3 · answered Feb 25 '18 at 18:38

1

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

answered Feb 25 '18 at 18:38

Mohammad Riazi-Kermani

68,728

How is this different from my second HINT?? – Mark Viola Feb 25 '18 at 18:40
It is not different at all. They both say the same thing and solve the same problem. – Mohammad Riazi-Kermani Feb 25 '18 at 18:43
I posted mine 8 minutes before yours. It's bad form to post an answer that is redundant. – Mark Viola Feb 25 '18 at 18:53
I do not look at other solutions and you can solve problems faster than others. What do you suggest now? I can delete my answer or I can upvote your hint, please advise. – Mohammad Riazi-Kermani Feb 25 '18 at 19:03
I'll let you decide what you think is the correct thing to do. – Mark Viola Feb 25 '18 at 19:16

score 1 · Answer 4 · answered Feb 25 '18 at 18:44

1

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

answered Feb 25 '18 at 18:44

robjohn

345,667

Best answer! (+1) – Mark Viola Feb 25 '18 at 19:18

Jack D'Aurizio · Answer 5 · 2018-04-09T14:53:07.427

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 $$ \int_{1}^{2}\frac{\pi}{1+2\sqrt{1+x^2}}\,dx \leq \frac{7}{10}.$$

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

5 Answers5