3
$\begingroup$

How do I integrate $$\int_{1/b}^{b} \frac {\arctan(x)+ \arctan\left(\frac{1}{x}\right)}{x} dx \text{ ?} $$

$\endgroup$
4
  • 1
    $\begingroup$ Hint : What is $arctan(1/x) $ in terms of $arccot$ $\endgroup$ – Mann Jun 14 '15 at 17:38
  • $\begingroup$ Oh thanks, sorry, I see this is too easy, should I delete the question? $\endgroup$ – Stuart Miller Jun 14 '15 at 17:43
  • $\begingroup$ No, don't delete it. Feel free to write out the answer below and maybe we can use this as a reference for the next time someone asks a version of this integral (which happens every couple of weeks). $\endgroup$ – Simon S Jun 14 '15 at 17:45
  • $\begingroup$ Noo, do not delete. If you have got the answer, just answer your own question. It will help the community. $\endgroup$ – Mann Jun 14 '15 at 17:46
5
$\begingroup$

Deriving $\arctan$, you'll see that $\arctan x+\arctan\frac1x$ is constant. Passing to the limit for $x\to\pm\infty$ it's easy to see that $\arctan x+\arctan\frac1x=\operatorname{sgn}(x)\frac{\pi}2$. Thus, supposing $b>0$, you'll have $$ \int_\frac1b^b\frac{\arctan x+\arctan\frac1x}x\,dx=\frac{\pi}2\int_{\frac1b}^b\frac1x\,dx $$ from this it should be easy, I guess.

$\endgroup$
3
  • $\begingroup$ You can see $\arctan(x)+\arctan(1/x)$ is $\frac{\pi}{2}$ (for positive $x$) by drawing a right triangle with legs $1$ and $x$. Then $\arctan(x/1)$ and $\arctan(1/x)$ are the two non-right angles. $\endgroup$ – alex.jordan Jun 14 '15 at 18:44
  • $\begingroup$ That sum is constant on the interval $(0,\infty)$ and is a different constant on the interval $(-\infty,0)$. There is a gap in the domain at $0$. However, one should not leave the impression that calculus is needed for that. Let $x$ and $1$ be the two legs of a right triangle. Then one of the acute angles is $\arctan x$ and the other is $\arctan(1/x)$, so of course those add up to $\pi/2$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 14 '15 at 18:49
  • $\begingroup$ I'd thought to write "the derivative is $0$ so the function is constant on the connected components of its domain, which is $]-\infty,0[\cup]0,+\infty[$", but I thought that what I wrote was indeed clear enough. $\endgroup$ – Joe Jun 14 '15 at 20:16
2
$\begingroup$

HINT:

$\arctan(x)+\arctan(1/x)= \pi/2$ for $x>0$

$\endgroup$
0
0
$\begingroup$

$$\arctan \left( \frac{1}{x} \right)=\mathrm{arccot} \left( {x} \right)$$ so the numerator is $$\frac{\pi}{2}.$$

$\endgroup$
1
  • 1
    $\begingroup$ See my comment above. $\endgroup$ – Bernard Jun 14 '15 at 17:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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