Calculate the following improper integral or show that it diverges.


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:

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

  • $\begingroup$ @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

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

  • $\begingroup$ 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

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.