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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.2012-07-18
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    @celtschk This road leads to a dead end, i think.2012-07-18
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    Do you have reason to believe that this has a closed-form solution?2012-07-18
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    @joriki I do not know. However, it should be possible to demonstrate, i think.2012-07-18
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    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 exist2012-07-23

1 Answers 1

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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. In this case $\arctan \sin^2x <\sin^2x$. You have not shown that the integral diverges.2012-07-19
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    @MartinGales: I made some detailed explanation to my original claim.2012-07-20
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    $\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.2012-07-20
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    @MartinGales, Forgive me if the word *'some'* is a synonym for *'every'*. Up to now I have believed otherwise.2012-07-20
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    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 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.2012-07-21
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    @MartinGales, It is a standard usage in analysis to introduce some constant like $C$, whose value is not exactly pinpointed but whose existence is guaranteed in the context, whenever the exact nature of $C$ is unimportant. This is because we just want to concentrate on the very logic itself that leads to the conclusion. That's why I just introduced $C$. If you are not pleased, you may alter all the $C$ in my proof to $\frac{\pi}{4}$, whose applicability is assured by the last paragraph of my detailed explanation.2012-07-21
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    Thus here, we are not saying that $C$ can be $1$. We are just saying that there exists a positive real number $C$ satisfying $\arctan x \geq C x$ on $[0, 1]$, and in my detailed explanation, I have shown that actually $\frac{\pi}{4}$ is a viable choice of $C$. I hope that this explanation may resolve your adhesion to $C = 1$.2012-07-21
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    Hmm... I need to reflect. There seems a kind of deep content.2012-07-21
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    Yes, you are right. This approach is something new for me! Actually, your solution is an excellent! Thank you!2012-07-23