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I did google the function and I can clearly see that it is always negative or zero, but I have no idea how I would have found this on my own. Both the logarithm and the $x\cdot \arctan(x)$ are positive.

What I did do is derive the function but that did not seem to be more obvious.

$\rho' = \frac{x}{1+x^{2}}-\arctan(x)$

This also does not look obvious to me.

$\rho'' = -\frac{2\cdot x^{2}}{(1+x^{2})^{2}}$

This is most definitely negative, thus concave (or concave downwards). Is that enough to deduce anything ?Am I missing something?

Thomas Andrews
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Kalec
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3 Answers3

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For any $x\neq 0$ we have: $$ \frac{d^2}{dx^2}\left(\log(1+x^2)-x\arctan x\right) = -\frac{2x^2}{(1+x^2)^2}<0\tag{1}$$ that proves concavity, hence the graphics of $\log(1+x^2)-x\arctan x$ always lie below the tangent in $x=0$, that is just the $x$-axis, since $\log(1+x^2)-x\arctan x$ is an even $C^\infty$ function whose value in zero is zero.

Jack D'Aurizio
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  • I should already know this but I am asking for clarity: The double integration of the original function gives me the concavity of said function ? – Kalec Mar 28 '15 at 13:22
  • @ThomasAndrews: You looked aggressive. By the way, $(1)$ proves concavity, hence the graphics of $\log(1+x^2)-x\arctan x$ always lie below the tangent in $x=0$, that is just the $x$-axis. Added to the answer. – Jack D'Aurizio Mar 28 '15 at 14:09
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Let $f(x)=\ln(x^2+1)-x\arctan x$. Since $f(0)=0$ and $f(-x)=f(x)$, it suffices to show that $f'(x)\lt0$ for $x\gt0$. So noting

$$f'(x)={2x\over x^2+1}-\arctan x-{x\over x^2+1}={x\over x^2+1}-\arctan x$$

and

$$\arctan x=\int_0^x{dt\over t^2+1}\gt\int_0^x{dt\over x^2+1}={x\over x^2+1}$$

does the trick.

Remark: This answer doesn't address the OP's question about the second derivative. That portion was edited in while I was thinking and typing. The version I was working from ended at "obvious."

Barry Cipra
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  • nice trick at the last step. need to remember all these integral reps of functions. – abel Mar 28 '15 at 13:53
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let $$y = \ln(1 + x^2 ) - x \tan^{-1} x .$$ the function is even and $y = 0$ at $x = 0 \text{ and } \lim_{x \to \infty} y = -\infty. $ taking the derivative, you find $$\frac{dy}{dx} = \frac{2x}{1+x^2} - \frac{x}{1+x^2} - \tan^{-1} x =\frac{x}{1+x^2} - \tan^{-1} x$$ we will have established that $y < 0 \text{ for all } x \neq 0$ if we can show that $$f(x) = \frac{x}{1+x^2} - \tan^{-1} x <0, x \neq 0.$$ observe $x = 0$ is the only zero of $f.$ suppose $$x > 0, f'(x) = \frac{1}{1+x^2} - \frac{2x^2}{1+x^2} -\frac 1{1+x^2} = -\frac{2x^2}{1+x^2} < 0$$ together with $f(0) = 0$ proves the claim and we are done.

abel
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