Improper integral, tarctan(t)/(1+t^2)^2, from 0 to infinity, convergent or divergent?

**electricsparks** · January 25th 2010, 11:28 AM

Hello,

I just had a quiz and I did not know how to do this question even though it looked so familiar.

Is the improper integral, from 0 to infinity,
tarctan(t)/(1+t^2)^2
convergent or divergent?

I tried using comparison theorem but it didn't work out.
And then I just tried solving the integral by substitution as in u=arctan(t) and du=1/1+t^2 dt but it didn't cancel out!

Please help! Thanks in advance!

**Jhevon** · January 25th 2010, 11:39 AM

Originally Posted by electricsparks

Hello,

I just had a quiz and I did not know how to do this question even though it looked so familiar.

Is the improper integral, from 0 to infinity,
tarctan(t)/(1+t^2)^2
convergent or divergent?

I tried using comparison theorem but it didn't work out.
And then I just tried solving the integral by substitution as in u=arctan(t) and du=1/1+t^2 dt but it didn't cancel out!

Please help! Thanks in advance!

comparison didn't work? really?

try again: $\int_0^\infty \frac {t \arctan t}{(1 + t^2)^2}~dt \le \frac {\pi}2 \int_0^\infty \frac t{(1 + t^2)^2}~dt$

**Krizalid** · January 25th 2010, 11:40 AM

Of course, it's a convergent integral. (Even an absolutely one.)

The integral converges as long as $\int_1^\infty\frac{t\arctan t}{(1+t^2)^2}\,dt<\infty,$ which in efect, it's convergent, since for all $t\ge1$ is $\frac{t\arctan (t)}{\left( 1+t^{2} \right)^{2}}<\frac{\pi t}{2t^{4}}=\frac{\pi }{2t^{3}}.$ Thus our integral is bounded by a convergent integral, hence the original integral converges.

(Beaten, ahahah.)

**simplependulum** · January 25th 2010, 11:15 PM

Why don't we compute the exact value of the integral ?

$I = \int_0^{\infty} \frac{x \tan^{-1}(x) }{ ( 1 + x^2 )^2 } ~dx$

Sub. $x = \frac{1}{t} , dx = - \frac{dt}{t^2}$

$I = \int_0^{\infty} \frac{t \tan^{-1}(\frac{1}{t})}{ ( t^2 + 1)^2 }~dt$

but $\tan^{-1}(\frac{1}{t}) = \frac{\pi}{2} - \tan^{-1}(t)$

so we have $I = \frac{\pi}{2}\int_0^\infty \frac t{(1 + t^2)^2}~dt - I<br />$

$2I = \frac{\pi}{2} \frac{1}{2} \implies I = \frac{\pi}{8}$

**Moo** · January 25th 2010, 11:29 PM

simplependulum : theoretically, before using integration techniques (substitution, integration by parts, ...), you have to prove that it converges first !

**Krizalid** · January 26th 2010, 04:56 AM

I really don't care about computing its value, I'm just doing what the problem asks.

I can put $\int_0^\infty\frac{\sin x}x\,dx$ and ask to show its convergence, and I know someone's gonna appear and batantly say $\int_0^\infty\frac{\sin x}x\,dx=\frac\pi2,$ which is far of the answer that I was looking for.

I know how to compute its value, but in this case, we're talking about convergence, besides, we don't know if the OP has covered computations on improper integrals.

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