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I can't find 2 functions to bound this integral in the intervals from $0$ to $1$ and from $1$ to $+\infty$, so I be sure that it converges.

$$\int_0^{\infty} \frac {\arctan x} {x^{3/2}}\, dx$$

any idea?


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Do you mean $\int_0^{\infty} \frac {\arctan x}{x^{1.5}}$? – Calvin Lin Jan 6 '13 at 16:36
yes, I don't know why it's look like this in my post – user1816377 Jan 6 '13 at 16:37
Learn how to use LaTeX properly. We are fixing your post, so don't undo the fix. – Calvin Lin Jan 6 '13 at 16:42
sorry, thank you – user1816377 Jan 6 '13 at 16:43

3 Answers 3

up vote 6 down vote accepted

You would like to show that $\displaystyle \int\frac{\arctan x\,dx}{x^{3/2}}$ converges. Breaking up at (say) $1$ is a good idea.

The integral from $1$ to $\infty$ is no problem, since $0\lt \arctan x\lt \frac{\pi}{2}$. So the integral from $1$ to $\infty$ converges by comparison with $\displaystyle \int_1^\infty \frac{\pi/2}{x^{3/2}}\,dx$.

Now examine $\arctan x$ near $0$. Note that $0\le \arctan x\le x$ for $x\ge 0$. This can be shown in various ways. For example, let $f(x)=x-\arctan x$. Then $f(0)=0$, and $f'(x)=1-\frac{1}{1+x^2}\ge 0$ if $x\gt 0$. Thus $f$ is an increasing function on $[0,\infty)$.

So in the interval from $0$ to $1$, we have $\frac{\arctan x}{x^{3/2}}\le \frac{x}{x^{3/2}}=\frac{1}{x^{1/2}}$. But we know that $\displaystyle\int_0^1 \frac{dx}{x^{1/2}}$ converges.

Remark: There are other ways to examine the behaviour of $\arctan x$ for $x$ small positive. We could use the power series representation $\arctan x=x-\frac{x^3}{3}+\cdots$.

Or else we can use something like L'Hospital's Rule to show that $\lim_{x\to 0}\frac{\arctan x}{x}=1$. That implies that if $x$ is positive and close enough to $0$, we have $\frac{\arctan x}{x}\lt 2$, that is, $\arctan x\lt 2x$.

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$$\int_0^{\infty} \frac {\arctan x} {x^{3/2}}\, dx=$$

$$\int_0^{\infty} \frac {1} {x^{3/2}} \int_0^{x}\frac {1} {1+t^2}\, dt\, dx=$$

$$\int_0^{\infty} \frac {1} {1+t^2} \int_t^{\infty} \frac {1} {x^{3/2}} \, dx \, dt=$$

$$\int_0^{\infty} \frac {1} {1+t^2} . \frac {2} {t^{1/2}} \, dt=$$


$$\int_0^{\infty} \frac {1} {1+u^4} . \frac {2} {u} \, 2 u du=4\int_0^{\infty} \frac {1} {1+u^4} \, du$$

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On the domain $[0,1]$, use the fact that $\arctan x\le x$ there; and on the domain $[1,\infty)$, use the fact that $\arctan x$ increases to the limit $\pi/2$ as $x$ increases to $\infty$.

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