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Seems to be a hard nut:

$$I=\int_{0}^{\infty}\frac{\arctan \sin^2x}{x}dx$$ Any hint?

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I'd try partial integration ($u = \arctan\sin^2x$, $v'=1/x$) to get rid of the arctan, and then hope that trigonometric identities allow to simplify the expression. –  celtschk Jul 18 '12 at 14:12
    
@celtschk This road leads to a dead end, i think. –  Martin Gales Jul 18 '12 at 14:28
    
Do you have reason to believe that this has a closed-form solution? –  joriki Jul 18 '12 at 14:31
    
@joriki I do not know. However, it should be possible to demonstrate, i think. –  Martin Gales Jul 18 '12 at 14:36
    
Have you tried to see 'Wolfram Mathematica Online Integrator'?I tried, and the answer was that probably a formula for such an integral dodoes not exist –  user36360 Jul 23 '12 at 15:54
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1 Answer

up vote 11 down vote accepted

It is not hard to show that

\begin{equation}\int_{0}^{\infty} \frac{\sin^2 x}{x} \; dx = \infty, \quad \cdots \quad (1)\end{equation}

and it is also easy to show that

$$\arctan x \geq Cx \quad \text{for} \quad 0 \leq x \leq 1 \quad \cdots \quad (2)$$

for some positive constant $C > 0$. Now it is clear that these together imply

\begin{equation}\int_{0}^{\infty} \frac{\arctan \sin^2 x}{x} \; dx \geq C \int_{0}^{\infty} \frac{\sin^2 x}{x} \; dx = \infty\end{equation}


Indeed, we first show that $(1)$ diverges. It suffices to show that

$$ \int_{2012}^{\infty} \frac{\sin^2 x}{x} \; dx = \infty. $$

By integration by parts, we have

$$ \begin{align*} \int_{2012}^{R} \frac{\sin^2 x}{x} \; dx &= \left[ \frac{1}{2} - \frac{\sin 2x}{4x}\right]_{2012}^{R} + \int_{2012}^{R} \left( \frac{1}{2x} - \frac{\sin 2x}{4x^2}\right) \; dx\\ &= \frac{1}{2}\log R + O(1), \end{align*}$$

which proves $(1)$ by letting $R\to\infty$.

Now we prove $(2)$. by examining second derivative of arc-tangent function, we find that it is concave on $[0, 1]$. Thus on this interval we have

$$\arctan x \geq (\arctan 1) x,$$

which proves $(2)$ with $C=\arctan 1 =\frac{\pi}{4}$.

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I can not get it. $\arctan x -x =-\frac{x^3}{3}+\frac{x^5}{5}...$ That means $\arctan x -x < 0$ for $0<x<1$. In this case $\arctan \sin^2x <\sin^2x$. You have not shown that the integral diverges. –  Martin Gales Jul 19 '12 at 14:12
    
@MartinGales: I made some detailed explanation to my original claim. –  sos440 Jul 20 '12 at 9:37
    
$\arctan x \geq Cx \quad \text{for} \quad 0 \leq x \leq 1 $ and $C>0$. Ok, let $C=1$ and $x=0.1$. Then $\arctan 0.1-0.1=-0.000331...$ Your second claim does not hold. –  Martin Gales Jul 20 '12 at 14:31
    
@MartinGales, Forgive me if the word 'some' is a synonym for 'every'. Up to now I have believed otherwise. –  sos440 Jul 20 '12 at 16:39
    
But i do not see your point. Why do we need introduce some C? In given case $C=1$. And the fact is that $\arctan x< x ; 0<x<1$ and $\frac{\arctan \sin^2x}{x}\leqslant \frac{ \sin^2x}{x};0 \leqslant x$. Do not get me wrong. I am inclined to believe that the integral in the question diverges but it still needs to prove. Your answer does not prove it. –  Martin Gales Jul 21 '12 at 6:51
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