Why is this integral not convergent? Could be a technicality

[CookieSalesman]

Homework Statement

Wolfram alpha and integral calculator say zero, but my school's HW on webworks does say it's divergent. I don't suppose it's wrong either, since the issue is probably a technicality
Arguably, on desmos.com the graph is pretty damn near odd.

Homework Equations

Int(-inf)(inf) of (-5x)/(1+x^2)
Sorry for the lack of Latex, but basically it's simple, just from neg inf to infinity of the equation written.

The Attempt at a Solution

Been trying at this for a few days now, and it comes up to zero each time. Supposedly it's divergent...
Am I doing something wrong when I take the limit with the ln()'s?

Of course now each time, I split the integral in half, just to be safe. (Negative five multiplied by the whole quantity of the two integrals. The first, negative infinity to zero, the second, zero to infinity)

 

[Dick]

Homework Statement

Wolfram alpha and integral calculator say zero, but my school's HW on webworks does say it's divergent. I don't suppose it's wrong either, since the issue is probably a technicality
Arguably, on desmos.com the graph is pretty damn near odd.

Homework Equations

Int(-inf)(inf) of (-5x)/(1+x^2)
Sorry for the lack of Latex, but basically it's simple, just from neg inf to infinity of the equation written.

The Attempt at a Solution

Been trying at this for a few days now, and it comes up to zero each time. Supposedly it's divergent...
Am I doing something wrong when I take the limit with the ln()'s?

Of course now each time, I split the integral in half, just to be safe. (Negative five multiplied by the whole quantity of the two integrals. The first, negative infinity to zero, the second, zero to infinity)

It's integrable only in the principal value sense. I.e. if you take the limits in a special way. Look it up. The integral from 0 to infinity is divergent and the integral from -infinity to zero is divergent. That might be a clue that the integrability is somewhat dodgy.

 

[Ray Vickson]

Homework Statement

Wolfram alpha and integral calculator say zero, but my school's HW on webworks does say it's divergent. I don't suppose it's wrong either, since the issue is probably a technicality
Arguably, on desmos.com the graph is pretty damn near odd.

Homework Equations

Int(-inf)(inf) of (-5x)/(1+x^2)
Sorry for the lack of Latex, but basically it's simple, just from neg inf to infinity of the equation written.

The Attempt at a Solution

Been trying at this for a few days now, and it comes up to zero each time. Supposedly it's divergent...
Am I doing something wrong when I take the limit with the ln()'s?

Of course now each time, I split the integral in half, just to be safe. (Negative five multiplied by the whole quantity of the two integrals. The first, negative infinity to zero, the second, zero to infinity)

The integral
F(a,b) = \int_{-a}^{b} \frac{x}{x^2+1} \, dx
does not have a limit as $(a,b) \to (\infty,\infty)$; what you get will depend on the path through which $(a,b) \to (\infty,\infty)$. Depending on the choice of path you can get any "answer" from $-\infty$ to $+\infty$. Therefore, $F(\infty,\infty)$ does not exist, and that is not just a "technicality".

 

[Nathanael]

Arguably (...) the graph is pretty damn near odd.

The function is indeed exactly odd! (Check by the definition of odd, -y(-x)=y(x)) But this is not enough when it comes to infinity. The other answers were interesting, and I don't have much insight to contribute, but I wanted to share a few sentences from my Calculus book (by Gilbert Strang).

It says, "the (...) integral (...) from a=-∞ to b=∞ (...) is split into two parts, and each part much converge. By definition, the limits at -∞ and ∞ are kept separate."
(The bold and italics are from him not me.)

He goes on to emphasize this point a little later with an odd function like yours:

"\int_{-∞}^{∞}x.dx is not defined even though \int_{-b}^{b}x.dx=0 for every b. The area under y=x is +∞ on one side of zero. The area is -∞ on the other side. We cannot accept ∞-∞=0. The two areas must be separately finite, and in this case they are not."