Why is this integral not convergent? Could be a technicality

In summary, according to the author of the book, the limits at -∞ and ∞ are not necessarily kept separate, which means that the integral may not exist.
  • #1
CookieSalesman
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5

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)
 
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  • #2
CookieSalesman said:

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.
 
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  • #3
CookieSalesman said:

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
[tex] F(a,b) = \int_{-a}^{b} \frac{x}{x^2+1} \, dx [/tex]
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".
 
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  • #4
CookieSalesman said:
Arguably (...) the graph is pretty damn near odd.
The function is indeed exactly odd! (Check by the definition of odd, [itex]-y(-x)=y(x)[/itex]) 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 [itex]a=-∞[/itex] to [itex]b=∞[/itex] (...) 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:

"[itex]\int_{-∞}^{∞}x.dx[/itex] is not defined even though [itex]\int_{-b}^{b}x.dx=0[/itex] for every b. The area under y=x is +∞ on one side of zero. The area is -∞ on the other side. We cannot accept [itex]∞-∞=0[/itex]. The two areas must be separately finite, and in this case they are not."
 
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FAQ: Why is this integral not convergent? Could be a technicality

Why is this integral not convergent?

There could be several reasons for an integral to not be convergent. Some possible reasons include the function being integrated being discontinuous, the integral having infinite limits, or the function not satisfying the necessary conditions for convergence.

What is a technicality in regards to integrals?

A technicality in regards to integrals refers to a specific condition or rule that must be met in order for the integral to be considered convergent. This could include things like the function being continuous, having finite limits, or satisfying certain convergence tests.

How can I determine if an integral is convergent or not?

To determine if an integral is convergent, you can use various methods such as the comparison test, the ratio test, or the integral test. These tests involve evaluating the function and limits of the integral to see if they meet the necessary conditions for convergence.

Can a seemingly simple integral still be not convergent?

Yes, a seemingly simple integral can still be not convergent. This is because even if a function is simple, it may not satisfy the necessary conditions for convergence. It is important to carefully evaluate the function and limits of the integral to determine if it is convergent.

Are there any common mistakes that can lead to a non-convergent integral?

Yes, there are common mistakes that can lead to a non-convergent integral. These can include forgetting to check for convergence conditions, using an incorrect method or formula for evaluating the integral, or making errors in calculations. It is important to double check your work and make sure all necessary conditions are met when evaluating an integral for convergence.

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