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I try to evaluate this integrat

$$\int_1^{\sqrt{3}}\frac{1}{1+x^2}dx$$

It seems simple.

$$\int_1^{\sqrt{3}}\frac{1}{1+x^2}dx=\arctan(\sqrt{3})-\arctan(1)$$

My question is what exact number it is?

Should it be $\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}$?

Or should it be $(k_1\pi+\frac{\pi}{3})-(k_2\pi+\frac{\pi}{4})=(k_1-k_2)\pi+\frac{\pi}{12}$?

$k_1,k_2=0,\pm 1,\pm2,... $

I think the definite integral should be A number, but there seems can be many numbers for the results.

I think I may miss some very basic concption here.

Thank you very much for help.

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    $\begingroup$ $\arctan x$ takes values only on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. $\endgroup$
    – bames
    Commented Mar 5, 2018 at 2:20
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    $\begingroup$ $\arctan x$ for real arguments $x$ refers to values taken on the principal branch. $\endgroup$
    – bames
    Commented Mar 5, 2018 at 2:28
  • $\begingroup$ Ok, thanks, I did not see this before. $\endgroup$
    – Deep North
    Commented Mar 5, 2018 at 2:29
  • $\begingroup$ Bames may add that whatever pi length you choose to define arctan on.. result is always same. While what op did is not correct. If you define arctan to take values from 90 to 270, answer will be same. $\endgroup$
    – King Tut
    Commented Mar 5, 2018 at 2:37
  • $\begingroup$ @king then you will have a different answer $\endgroup$
    – Deep North
    Commented Mar 5, 2018 at 2:44

2 Answers 2

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$$ \arctan\sqrt3 - \arctan 1 = \frac \pi 3 - \frac \pi 4 = \frac{4\pi}{12} - \frac{3\pi}{12} = \frac{(4-3)\pi}{12} = \frac \pi {12}. $$ One of a number of ways to see that this need not involve any of the "nonprincipal" values of the arctangent is this: $$ \text{If } 1 \le x \le \sqrt 3 \text{ then } \frac 1 2 \ge \frac 1 {1+x^2} \ge \frac 1 4, $$ $$ \text{so } \frac{\sqrt 3-1}2 \ge \int_1^{\sqrt 3} \frac{dx}{1+x^2} \ge \frac {\sqrt3-1} 4. $$

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  • $\begingroup$ No I didnt and this answer is wrong. $\endgroup$
    – King Tut
    Commented Mar 5, 2018 at 3:51
  • $\begingroup$ @KingTut : What the phrase "No I didn't" refers to is unclear, since this answer says nothing about what you did, nor does it mention you, and a statement that the answer is wrong could be helpful, but only if it's specific. $\endgroup$ Commented Mar 5, 2018 at 18:21
  • $\begingroup$ Your answer was wrong and you are trying to cover it up by changing 'cannot' to 'need not'. My question is that arctan can be defined over any fixed range of length pi. Say 90 degree to 270 degrees. It's just adding a constant to antiderivative of $1/(1+x^2)$. Hope you understand what I say. $\endgroup$
    – King Tut
    Commented Mar 6, 2018 at 2:07
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We know that $\tan(x)$ is many one as its periodic with period $\pi$. As a result for defining an inverse we need to fix our range to a fixed length $\pi$.

We can define $\arctan_1(x) : \Bbb R \to (k\pi-\tfrac{\pi}{2}, k\pi + \tfrac{\pi}{2}), k \in \Bbb Z$. We get our standard $\arctan(x)$ function by putting $k=0$.

Now we see that the integral

$$\begin{align} \int_{1}^{\sqrt{3}} \dfrac{1}{1+x^2} dx &=\arctan_1(x) |_{1}^{\sqrt{3}}\\ &= k\pi + \tfrac{\pi}{3} -(k\pi + \tfrac{\pi}{4}) \\ &= \frac{\pi}{12} \end{align}$$

is independant of branch of definition. This is self evident since $\arctan_1(x) = \arctan(x) + k\pi$.

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  • $\begingroup$ Upvote, but I think it is not necessary they have the same k $\endgroup$
    – Deep North
    Commented Mar 5, 2018 at 3:50
  • $\begingroup$ I was just showing that the antiderivatives only differ by constant so no difference on definite integral (sic.) as in my comment. $\endgroup$
    – King Tut
    Commented Mar 5, 2018 at 3:52
  • $\begingroup$ @DeepNorth Answer perpetuated by Michael is plain wrong. $\endgroup$
    – King Tut
    Commented Mar 5, 2018 at 3:56
  • $\begingroup$ But it seems correct to me, such as $\frac 1 2 \ge \frac 1 {1+x^2} \ge \frac 1 4$, then we just integrate from $1$ to $\sqrt{3}$ $\endgroup$
    – Deep North
    Commented Mar 5, 2018 at 4:01
  • $\begingroup$ @DeepNorth Havent you been taught this? $\int f(x) dx = F(x) + K$ so we can in fact use any branch we wish of arctan? $\endgroup$
    – King Tut
    Commented Mar 5, 2018 at 4:04

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