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Calculate the following improper integral or show that it diverges.

$$\int_{0}^{\infty}\frac{\arctan(x)}{(1+x^2)^{3/2}}dx$$

I'm really lost. Your help would be very appreciated.

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Sub $x=\tan{t}$ and the integral is equal to

$$\int_0^{\pi/2} dt \, t \, \cos{t} $$

Which may easily be evaluated by parts:

$$[ t \sin{t}]_0^{\pi/2} - \int_0^{\pi/2} dt \, \sin{t} = \frac{\pi}{2}-1 $$

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  • $\begingroup$ @DavidH: Thanks. There's more than one way to skin a cat. $\endgroup$ – Ron Gordon Nov 10 '13 at 0:47
  • $\begingroup$ Why is the interval (0, pi/2) instead of (0, ∞)? And where does cos t come from? $\endgroup$ – William Nov 10 '13 at 1:09
  • $\begingroup$ @William: because $\arctan{(\infty)}=\pi/2$. $\cos{t} = \sec^2{t}/\sec^3{t}$ from the substitution. $\endgroup$ – Ron Gordon Nov 10 '13 at 1:13
  • $\begingroup$ @RonGordon then my next question is, how come we get sec^2 t/sec^3 t from the substitution? $\endgroup$ – William Nov 10 '13 at 11:06
  • $\begingroup$ @William: $x=\tan{t} \implies dx = \sec^2{t} dt$ and $(1+x^2)^{3/2}=(1+\tan^2{t})^{3/2}=(\sec^2{t})^{3/2}=\sec^3{t}$. $\endgroup$ – Ron Gordon Nov 10 '13 at 11:15
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Hint: For all $x \in (0, \infty)$, we have

$$\frac{\arctan{x}}{(1 + x^2)^{3/2}} \le \frac{\pi/2}{(1 + x^2)^{3/2}}$$

Now what can you say about this integral?

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


It's clear that $\int_0^1 (1 + x^2)^{-3/2} dx$ is finite, since the function is bounded.

Now on the interval $(1, \infty)$, use the fact that

$$1 + x^2 > x^2 \implies (1+x^2)^{-3/2} < (x^2)^{-3/2} = x^{-3}$$

But what can you say about: $$\int_1^{\infty} x^{-3} dx$$

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  • $\begingroup$ Thanks for the answer. I understand what you did there, but I'm struggling to take the next step. I should try to find a primtive function, right? I can see that the expression we have is similar to the derivative of arctan x, but I think I need another hint to solve this. $\endgroup$ – William Nov 10 '13 at 0:27
  • $\begingroup$ @William Edited. $\endgroup$ – user61527 Nov 10 '13 at 0:31
  • $\begingroup$ Okay so it's a bit easier to find a primitive function to this. $$ \left. \dfrac{-1}{2x^2} \right\vert_1^{∞} = -1/∞ + 1$$ Which gives us the value 1, right? But what does it mean that we also got a finite integral? This whole thing about improper integrals is new to me so I didn't really understand what the consequence of the first conclusion is. $\endgroup$ – William Nov 10 '13 at 0:56
  • $\begingroup$ I must be doing something wrong since the answer should be pi/2 - 1 according the other solutions $\endgroup$ – William Nov 10 '13 at 11:11

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