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The integral to compute is $\displaystyle\int_0^\infty \frac{1}{3+x^2} \, \mathrm dx$.

I know how to compute the indefinite integral of this function its gives me

$$\frac{\sqrt{3}}{3} \arctan\left(\frac{x}{\sqrt{3}}\right).$$

When i compute the definite integral it now gives me :

$$\frac{\sqrt{3}}{3} \left(\lim\limits_{x \to \infty }\arctan\left(\frac{x}{\sqrt{3}}\right)-\arctan(0)\right).$$

Then i don't understand why my teacher writes that arctan(0)=0 because it also can be equal to π, and more stranger he found i don't know how that $\lim\limits_{x \to \infty }\arctan\left(\frac{x}{\sqrt{3}}\right)=\pi/2$. Thank you for help !

EDIT : I only need how to compute the limit now.

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    $\begingroup$ the range of $arctan(x)$ is [-pi/2,pi/2].$ the arctan function is hence injective. so arctan(0)=0 $\endgroup$ – dajoker Jun 7 '13 at 14:23
  • $\begingroup$ @VijayRaghavan Thank you for this. What about the limit ? $\endgroup$ – The Answer Jun 7 '13 at 14:25
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    $\begingroup$ 3^0.5 is a finite number.An infinitely large number(x tends to infinity) is divided by a finite real number still results in an infinitely large number. the expression reduces to arctan(infinity) which is pi/2 $\endgroup$ – dajoker Jun 7 '13 at 14:27
  • $\begingroup$ @VijayRaghavan But how can you show that arctan(infinity)=$\pi$/2 $\endgroup$ – The Answer Jun 7 '13 at 14:29
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    $\begingroup$ simple. what is tan(pi/2)? . the arctan function accepts any real number and returns a value in [-pi/2,pi/2] so that tan(the returned value)=the entered real number $\endgroup$ – dajoker Jun 7 '13 at 14:32
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Actually, you don't really have to have $\arctan(0)=0$, you can have $\arctan(0)=\pi$ but your arctan function should be continuous. Hence if you decide to make your range of arctan such that $\arctan(0)=\pi$, you should have that $\lim_{x\rightarrow \infty}\arctan(x)=\frac{3\pi}{2}$, which will give the same answer as your teacher's.

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  • $\begingroup$ @Jonh Where was his teacher's answer even mentioned? $\endgroup$ – Git Gud Jun 7 '13 at 14:40
  • $\begingroup$ In the last paragraph he goes through the values of arctan that his teacher found, which he didn't understand. While the answer isn't explicitly stated anywhere, given that these values were all that remained to be found, the teacher's answer is easily assumed. $\endgroup$ – john Jun 7 '13 at 14:43
  • $\begingroup$ He said that his teacher is considering $\arctan (0)=0$, he didn't say it could be $\arctan (0)=\pi$. And he explicitly says that his teacher found the limit to be $\pi /2$, not $3\pi /2$. $\endgroup$ – Git Gud Jun 7 '13 at 14:46
  • $\begingroup$ Exactly. The OP was confused why the teacher said that $\arctan(0) = 0$ when the OP believed it could also be $\pi$. I'm saying that we could indeed have $\arctan(0) = \pi$ as the OP suggested but this would then also affect the value of arctan we use in the limit as x goes to infinity. If we do this though, we get the same answer, as the two values still differ by $\pi/2$ $\endgroup$ – john Jun 7 '13 at 14:54
  • $\begingroup$ Your comment clarifies things, but that's not what you wrote on your answer. You said that: "$\lim_{x\rightarrow \infty}\arctan(x)=\frac{3\pi}{2}$, which will give the same answer as your teacher's", which isn't true. $\endgroup$ – Git Gud Jun 7 '13 at 14:56
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The way i would approach this problem is . $$ \int_0^\infty \frac{1}{3+x^2}dx $$

I would divide the $$ x^2+3 $$ by 3 to factor out the constants.

$$ \int_0^\infty \frac{1}{3(\frac{x^2}{3})+1}dx $$

then we can substitute our function with

$ u^2=\frac{x^2}{3}$

$u= \frac{x}{\sqrt{3}} $

The integral of

$$ \int_0^\infty \frac{1}{3(\frac{x^2}{3})+1}dx $$

$du= \frac{1}{\sqrt{3}}dx$

When we substitute we get

$$ \frac{1}{\sqrt{3}}\int_0^\infty \frac{1}{{u^2}+1}dx $$

the integral would result to

$$ \frac{1}{\sqrt{3}}\int_0^\infty \frac{1}{{u^2}+1}dx =\frac{1}{\sqrt{3}}arctan(u)|_0^\infty $$

Then we substitute back $u= \frac{x}{\sqrt{3}} $

This would give us the step before the final answer.

$$\frac{1}{\sqrt{3}}arctan(\frac{x}{\sqrt{3}})|_0^\infty $$

then we can evaluate the value and the limit

$$\frac{1}{\sqrt{3}}\left(\lim\limits_{a \to \infty }\arctan\left(\frac{a}{\sqrt{3}}\right)-\arctan(0)\right)= (\frac{pi}{2\sqrt{3}}) $$

because as "a" approaches infinity arctan(a/sqrt(3)) = Pi/2 and arctan(0/sqrt(3)) = 0

$$ \frac{1}{\sqrt{3}}( \frac{pi}{2} - 0 ) = (\frac{pi}{2\sqrt{3}}) $$

Thats the final correct answer and evaluation.

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