Question

Why are these 2 integrals different? \[ \int_{-\infty}^{\infty}\frac{x}{\pi(1+x^2)}dx\ne\lim_{c \rightarrow \infty}\int_{-c}^{+c}\frac{x}{\pi(1+x^2)}dx\]

Answer 1

I will call the left one \(I_1\) and the right one \(I_2\).
I am told that the \(I_1\) is undefined as it gives \(\infty-\infty\). I have shown this and it's ok. But I don't know how it differs from the \(I_2\) integral.

The question hints to show that \( \large \int\limits_{-c}^{+c}\frac{x}{\pi(1+x^2)}dx=0\) for some \(c \ne \pm \infty\) , which again it makes sense to me. But I still don't understand why \(I_1\) is not the same as \(I_2\).

Answer 2

So you're asking why

\[\Large \infty-\infty = 0\]

isn't true?

Answer 3

No. I know \(\infty - \infty\) is undefined. But I'm not not sure why \(I_2\) is not considered to be the same integral as \(I_1\) @jim_thompson5910 It seemed to me from calculus that when you had bounds at infinity, the proper way to define the integral was to relabel the bounds of infinity as constants, and then let the constants approach infinity. I'm not sure what I am missing here :(

Answer 4

well if you focus on the indefinite integral, you get ln(1+x^2)/(2pi) + C

Let's say the limits of integration are p and q (p < q)

so the definite integral would be

[ln(1+q^2)/(2pi) + C] - [ln(1+p^2)/(2pi) + C]

ln(1+q^2)/(2pi) + C - ln(1+p^2)/(2pi) - C

ln(1+q^2)/(2pi) - ln(1+p^2)/(2pi)

Answer 5

if p were to approach -infinity and q were to approach infinity, then we have infinity - infinity, which is an indeterminate form

I'm sure you have something similar to this in your work

Answer 6

however, on the right side, you'll have

ln(1+c^2)/(2pi) - ln(1+c^2)/(2pi)

which turns into 0 no matter what the value of c is

Answer 7

on the right side, c doesn't even have to approach infinity, c could be something small like c = 1

Answer 8

So are you saying that \[\lim_{p\rightarrow -\infty}\lim_{q \rightarrow +\infty}\int_p^q\frac{x}{\pi(1+x^2)}dx \ne\lim_{c\rightarrow \infty}\int_{-c}^{+c}\frac{x}{\pi(1+x^2)}dx \] ? @jim_thompson5910

I just find it strange how the look similar but give different results! I guess it's because of the fact that you evaluate the integral first, then apply the limit, and it just happens that the right integral gives 0 already before applying the limits? (I am so sorry to bother you so much!! )

Answer 9

well when you boil it all down, the left side is a difference of two logs while the right side simplifies to 0

the left side, the difference of two logs, turns into infinity - infinity when you apply the limit to the left side

the right side stays at 0 even when you apply the limit

Answer 10

this is probably why integrals with infinite bounds are evaluated like the right side (instead of the left)

the true answer is indeed 0 and you can see this visually (due to the symmetry) and the answer isn't "undefined" or "indeterminate"

so the right side is the valid answer

Answer 11

Hm I guess I kind of understand, but actually I am doing a statistics course and we are doing expectations. I found this integral as a result of the expectation of the standard Cauchy distribution, \(f(x)= \frac{1}{\pi(1+x^2)}, -\infty<x<\infty\)(which gives the left integral), and it is known that the standard Cauchy distribution has no expectation

Answer 12

so the expected value of the Cauchy distribution is 0?

Answer 13

that makes sense in a way because the center is 0 (where you'd expect most of the values to be clumped around)

Answer 14

No, they says the expected value doesn't exist.. I just find this integration very mysterious..

Answer 15

oh I see

Answer 16

(I appreciate all your help!! I am again sorry for maybe confusing you or delaying you.. I will still give you a medal for your work :) ) If you get more info though about this, feel free to post :)

Answer 17

this page may help

http://www.math.uah.edu/stat/special/Cauchy.html

unfortunately I didn't study this in stats but that's on the list of things to get to

Answer 18

in their proof (see #6) they break up the graph into two pieces (-infinity to 0, then 0 to infinity) and they essentially get to infinity - infinity