Why Don't Odd Integrals Necessarily Converge Over Infinite Symmetric Domains?

[member 428835]

hey pf!

can someone shed some light on why odd integrals don't necessarily converge over infinite symmetric domains?

thanks!

 

[HallsofIvy]

I don't see why you would think they should. Perhaps it is because you are thinking of infinite intervals as "symmetric" when they aren't. The integral $\int_{-\infty}^\infty f(x) dx$ is defined as
"$\lim_{A\to\infty}\int_{-A}^a f(x)dx+ \lim_{B\to\infty} \int_a^B f(x)dx$" where the two limits must be taken independently.

(The "Cauchy Principle Value" is defined as $\lim_{A\to\infty} \int_{-A}^A f(x)dx$. If the original integral exists then the Cauchy Principle Value is equal to it. The Cauchy Principle Value may exist when the integral does not, such as for an odd function where the Cauchy Principle Value is 0, but that is NOT the integral and does not have a great deal of use.)

 

[member 428835]

I don't see why you would think they should. Perhaps it is because you are thinking of infinite intervals as "symmetric" when they aren't. The integral $\int_{-\infty}^\infty f(x) dx$ is defined as
"$\lim_{A\to\infty}\int_{-A}^a f(x)dx+ \lim_{B\to\infty} \int_a^B f(x)dx$" where the two limits must be taken independently.

i was never aware of this definition, but now it makes perfect sense. why is it we have defined the reimann integral over $(-\infty,\infty)$ in this way? why not define it as the Cauchy Principle Value suggests (i've studied real analysis, so if you want to get rigorous in your response, i may be able to keep up for a little while)?

also, does this mean that $\int_{-\infty}^{\infty}\sin (x) dx = 0$ when invoking the Cauchy Principle Value?

 

[gopher_p]

i was never aware of this definition, but now it makes perfect sense. why is it we have defined the reimann integral over $(-\infty,\infty)$ in this way? why not define it as the Cauchy Principle Value suggests (i've studied real analysis, so if you want to get rigorous in your response, i may be able to keep up for a little while)?

also, does this mean that $\int_{-\infty}^{\infty}\sin (x) dx = 0$ when invoking the Cauchy Principle Value?

It's technically not a Riemann integral. Riemann integrals are only defined on closed intervals.

There are likely many reasons why improper integrals are defined the way they are as opposed to accepting the Cauchy Principle Value as the definition. I'll just give two.

1) If one were to define the Riemann integral over an interval like (-∞,∞) in a manner similar to the way it is defined over closed intervals, one would need to ensure that the Riemann sums over non-symmetric partitions converged to the same value (and at all) as the Riemann sums over symmetric partitions.

2) When you define convergent improper integrals the way that we do, a lot of the old theorems and properties for Riemann integrals can be extended to the new kind of integral. For instance, $\int_a^bf=\int_a^cf+\int_c^bf$ still works. Also $f$ integrable on $I$ $\Rightarrow$ $f$ integrable on all $J\subset I$ is still true.