# Calculate $\int_{0}^{\infty}\frac{\arctan(x)}{(1+x^2)^{3/2}}dx$ or show that it diverges

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.

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:

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

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

Hint: For all $x \in (0, \infty)$, we have

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

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

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: $$\int_1^{\infty} x^{-3} dx$$

• 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. – William Nov 10 '13 at 0:27
• @William Edited. – user61527 Nov 10 '13 at 0:31
• 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. – William Nov 10 '13 at 0:56
• I must be doing something wrong since the answer should be pi/2 - 1 according the other solutions – William Nov 10 '13 at 11:11