Commentary for "A not very advanced integration technique..."

[ZaidAlyafey]

Moderator edit: The following topic is for commentary pertaining to the tutorial:

http://mathhelpboards.com/math-notes-49/not-very-advanced-integration-technique-3297.html

It is not always true that you can interchange an infinite series with an improper integral .
It will work perfectly if the series is always positive ...

 

[Fernando Revilla]

Re: A not very advanced integration technique...

It is not always true that you can interchange an infinite series with an improper integral .
It will work perfectly if the series is always positive ...

Certainly, the corresponding theorem is:

Let $f_n:[a,+\infty)\to \mathbb{R}\;(n\geq 1,\; f_n\geq 0)$ be a sequence of functions such that:

$(i)\quad f_n$ is Riemann integrable on $[a,b]$ for all $b\geq a$ and for all $n\geq 1$.
$(ii)\;\displaystyle\int_a^b\sum_{n=1}^{\infty}f_n(x)\;dx=\sum_{n=1}^{\infty}\int_a^bf_n(x)\;dx$ for all $b\geq a$,

Then,

$ \displaystyle\int_a^{\infty}\sum_{n=1}^{\infty}f_n(x)\;dx=\sum_{n=1}^{\infty}\int_a^{\infty}f_n(x)\;dx$

 

[chisigma]

Re: A not very advanced integration technique...

Certainly, the corresponding theorem is:

Let $f_n:[a,+\infty)\to \mathbb{R}\;(n\geq 1,\; f_n\geq 0)$ be a sequence of functions such that:

$(i)\quad f_n$ is Riemann integrable on $[a,b]$ for all $b\geq a$ and for all $n\geq 1$.
$(ii)\;\displaystyle\int_a^b\sum_{n=1}^{\infty}f_n(x)\;dx=\sum_{n=1}^{\infty}\int_a^bf_n(x)\;dx$ for all $b\geq a$,

Then,

$ \displaystyle\int_a^{\infty}\sum_{n=1}^{\infty}f_n(x)\;dx=\sum_{n=1}^{\infty}\int_a^{\infty}f_n(x)\;dx$

In the first post an obvious counterexample of what You have written has been described. Let's suppose to have the function...

$\displaystyle f(x) = \cos x\ e^{-x} = \sum_{n=0}^{\infty} f_{n} (x) = \sum_{n=0}^{\infty} (-1)^{n}\ \frac{x^{2 n}\ e^{-x}}{(2n)!}$ (1)

It is well known that...

$\displaystyle \int_{0}^{\infty} f(x)\ dx = \frac{1}{2}$ (2)

... but if You 'mechanically' swap between integral and summation You obtain...

$\displaystyle \int_{0}^{\infty} \cos x\ e^{-x}\ dx = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)!}\ \int_{0}^{\infty} x^{2 n}\ e^{-x}\ dx = 1 - 1 + 1 - 1 +...$

... and the series doesn't converge. In general swapping between integral and summation is possible if the series $\displaystyle \sum_{n=0}^{\infty} f_{n} (x)$ uniformely converges in (a,b). For more details see...


Uniform Convergence -- from Wolfram MathWorld

Kind regards

$\chi$ $\sigma$

 

[Fernando Revilla]

Re: A not very advanced integration technique...

In the first post an obvious counterexample of what You have written has been described. Let's suppose to have the function...

$\displaystyle f(x) = \cos x\ e^{-x} = \sum_{n=0}^{\infty} f_{n} (x) = \sum_{n=0}^{\infty} (-1)^{n}\ \frac{x^{2 n}\ e^{-x}}{(2n)!}$ (1)

I've only written a well-known theorem. Your counterexample is not valid because $f_n\not\geq 0$ if $n$ odd.

 

[chisigma]

Re: A not very advanced integration technique...

Let $f_n:[a,+\infty)\to \mathbb{R}\;(n\geq 1,\; f_n\geq 0)$...

I apologize not to be expert in 'hieroglyphics' so that escaped to me that $f_{n} (x) \ge 0$ forall x in (a,b). In that it is 'all right' because if $\displaystyle \sum_{n=0}^{\infty} |f_{n} (x)|$ converges for all x in (a,b) then $\displaystyle \sum_{n=0}^{\infty} f_{n} (x) $ uniformely converges in (a,b) and You can swap integration and summation...

... the scope of this thread however is to analyse what happens when $\displaystyle \sum_{n=0}^{\infty} |f_{n} (x)|$ doesn't converges in (a,b) in the very rescricted case...

$\displaystyle f_{n}(x) = a_{n}\ x^{n}\ e^{-x}$ (1)

... all right?...

... that is all for the moment!...

Kind regards

$\chi$ $\sigma$

 

[ZaidAlyafey]

Re: A not very advanced integration technique...

I didn't realize until recently that we can't swap the integration sign and sigma , I was actually feeling comfortable interchanging whenever needed , this worked for the most time .

But here is a question can you give me an example for a situation where the answer before swaping is a finite real number but if we swap we get a different finite real number ?