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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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    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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    Yes, you are right. This approach is something new for me! Actually, your solution is an excellent! Thank you!2012-07-23