Convergence of the Arctangent Integral

[inutard]

Homework Statement

Is $\intop_{-\infty}^{\infty}\arctan(x)\, dx$ convergent?

What about $\lim_{t\rightarrow\infty}\intop_{-t}^{t}\arctan(x)\, dx$?

Homework Equations

The Attempt at a Solution

I think the first integral may actually be divergent the way its written and the second one convergent. This is just a hunch though. Can anyone shed some light on this?

 

[hunt_mat]

Can you integrate arctan first?

 

[eczeno]

they are the same integral, the first is just shorthand for the second, which is shorthand for a more correct expression with two limits.

the divergence of the integral is clear since arctan(x) -> pi/2 as x->inf.

 

[eczeno]

the integral can be evaluated by parts if you want to show that it diverges explicitly.

 

[inutard]

Thanks for the replies guys.

I thought about the question a bit more. I believe that the two integrals are not the same. The first one can be split into the integral from -Infty to 0 + the integral from 0 to +Infty. These two integrals are both divergent and so the sum is divergent (there is no cancelling out effect as the integral is not approaching infinity symmetrically as in the second integral <--- Is this right?). The second one can be shown to converge to zero by just integrating it.

 

[eczeno]

sorry, my post above is incorrect, i was only looking at the integral from 0 to infinity, which does diverge, but your integrals both (and they are the same) converge to zero.

 

[inutard]

Hmm. Can you explain to me this then?
http://www.wolframalpha.com/input/?i=integral+of+arctan(x)+from+negative+infinity+to+infinity

The limit (2nd integral) converges to zero when I test it on mathematica tho.

Also, what is the Cauchy Principal Value? That seems to be zero.

 

[eczeno]

Is $\intop_{-\infty}^{\infty}\arctan(x)\, dx$ convergent?

What about $\lim_{t\rightarrow\infty}\intop_{-t}^{t}\arctan(x)\, dx$?

the first integral does not, technically, mean anything because \infty is not a number. the second integral is what alpha evaluated to get 0. putting the limit in there as you have makes it the Cauchy principal value. It is also how just about anybody would interpret the first integral.

as for approaching \infty symmetrically, i am not sure i follow you. each limit is basically just walking down the x-axis toward infinity, once to the left and once to the right. I don't see how it could be more symmetric than that.now that i think about it some more, i understand your symmetry argument. the first integral my be approaching infinity in a 'non-symmetric' way, that is why we really can't give it a meaning. but the second is well defined and symmetric. another, perfectly reasonable interpretation of the first integral:

$\lim_{t\rightarrow\infty}\intop_{-2t}^{t}\arctan(x)\, dx$

would probably (haven't worked it out) have a value other than zero, and may even diverge.

 

[inutard]

When I say the second integral is symmetric, I guess what i mean is that t is approaching infinity on both sides. But for the first integral, you can interpret it as an integral from a to 0 plus an integral from 0 to b, with 'a' going to negative infinity while 'b' goes to infinity. In that sense, you'll have to add two divergent sums together, making the first integral divergent.

 

[eczeno]

i think we are on the same page now, though i would say the first integral is too ambiguous to evaluate, rather than say it definitely diverges. If you try to give a meaning to the first integral, you pretty much have an infinity of reasonable choices which may or may not converge. one of those choices is the second integral, which converges to zero. another choice is splitting it up int integrals fro a to 0 and 0 to b as you suggest, and, as you correctly point out, we cannot conclude that the sum of those integrals converges.

 

[inutard]

Yup! Thanks for all the help! This has cleared things up for me.

 

[eczeno]

it has cleared it up for me too! I'd say we can call this discussion a success.

cheers.