Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
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
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, $\arctan x\lt 2x$.

share|cite|improve this answer

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

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

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

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


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

share|cite|improve this answer

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 $\infty$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.