Integration technique for definite integrals involving an infinite series

[chisigma]

Some year ago, in an Italian mathematical community, the following definite integral has been proposed...

$\displaystyle I= \int_{0}^{\infty} \frac{1 - \cos x}{x^{2}}\ e^{-x}\ dx$ (1)

I don't mask the fact that I like to follow 'unconventional' ways and in that situation I was perfecly coherent. Combining the series expansion...

$\displaystyle \frac{1 - \cos x}{x^{2}} = \sum_{n=0}^{\infty} (-1)^{n}\ \frac{x^{2 n}}{(2n+2)!}$ (3)

... and the well known result...

$\displaystyle \int_{0}^{\infty} x^{n}\ e^{-x}\ dx = n!$ (4)

... without serious efforts I obtained...

$\displaystyle I= \sum_{n=0}^{\infty} \frac{(-1)^{n}} {(2n +1) (2n+2)} = \sum_{n=0}^{\infty} (-1)^{n}\ (\frac{1}{2n+1} - \frac{1}{2n + 2}) = 1 - \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + ... = \frac{\pi}{4} - \frac{\ln 2}{2}$ (5)

Avoiding any comment about the 'elegance' of the solution, a spontaneous question is: is that a valid integration technique?... or, in other words, if f(x) is an entire function [i.e. an f(x) the series expansion ofr which converges for all x...], the integral...

$\displaystyle \int_{0}^{\infty} f(x)\ e^{-x}\ dx$ (6)

... if converges, can always be solved with the technique I have described?...

... unfortunately not, as in the following example...

$\displaystyle \cos x = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n}}{(2n)!}$ (7)

It is well known that is...

$\displaystyle \int_{0}^{\infty} \cos x\ e^{-x}\ dx = \frac{1}{2}$ (8)

... but if we try to apply the technique I used to solve (1) we obtain...

$\displaystyle \int_{0}^{\infty} \cos x\ e^{-x}\ dx = \sum_{n=0}^{\infty} (-1)^{n} = 1 - 1 + 1 - 1 +...$ (9)

... and the series don't converge!...

Dear friends of MHB... why don't dedicate a little of our time to discuss about when this integration technique can be applied and when it can't...

Kind regards

$\chi$ $\sigma$

Comments and questions should be posted here:

http://mathhelpboards.com/commentary-threads-53/commentary-not-very-advanced-integration-technique-4215.html

 

[alyafey22]

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 ...

 

[chisigma]

According to series integration theorem, if f(x) can be written in (a,b) as...

$\displaystyle f(x)= \sum_{n=1}^{\infty} u_{n} (x)$ (1)

... and in (a,b) the series (1) uniformely converges, then You can write...

$\displaystyle \int_{a}^{b} f(x)\ dx = \int_{a}^{b} \{\sum_{n=1}^{\infty} u_{n} (x)\}\ dx = \sum_{n=1}^{\infty} \int_{a}^{b} u_{n} (x)\ dx$ (2)

The functions we are analysing in this thread in general don't satisfy uniform convergence in $(0, \infty)$ so that the (2) can be possible or impossible depending from f(x), and the examples I have supplied in the first post confirm that. Scope of this thread is to find a sort of 'feasibility criterion' in applying (2) for series of functions non uniformely convergent in $(0,\infty)$... Kind regards $\chi$ $\sigma$

 

[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

$\chi$ $\sigma$

 

[alyafey22]

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 ?

 

[chisigma]

I think that the best answer to legitimate doubt is to clarify what's the scope of this thread. Let's suppose to have to solve [if it converges...] the definite integral...

$\displaystyle I= \int_{0}^{\infty} f(x)\ e^{-x}\ dx$ (1)

... where f(x) is an entire function [a function which is analytic in the whole complex plane...] so that is for x in $[0,\infty)$...

$\displaystyle f(x)= \sum_{n=0}^{\infty} a_{n}\ x^{n}$ (2)

In particular we intend to verify, remembering that is...

$\displaystyle \int_{0}^{\infty} x^{n}\ e^{-x}\ dx = n!$ (3)

... if and when we can write...

$\displaystyle I= \int_{0}^{\infty} f(x)\ e^{-x}\ dx = \sum_{n=0}^{\infty} a_{n}\ n!$ (4)

In the first post we found that for $\displaystyle f(x)= \frac{1-\cos x}{x^{2}}$ the (4) works and for $\displaystyle f(x)=\cos x$ the (4) fails. In order to get more information let's try with $\displaystyle f(x)= \frac{\sin x}{x}$. Remembering that is...

$\displaystyle \int_{0}^{\infty} \frac{\sin x}{x}\ e^{-x}\ dx = \frac{\pi}{4}$ (5)

... and...

$\displaystyle \frac{\sin x}{x}= \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2 n}}{(2 n + 1)!}$ (6)

... the (4) supplies to us...

$\displaystyle I= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2 n + 1} = \frac{\pi}{4}$ (7)

... which is exact!...

The two functions $\displaystyle f(x)= \frac{\sin x} {x}$ and $\displaystyle f(x)= \frac{1 - \cos x}{x^{2}}$ have one common property: in x=0 there is a removable singularity and if we reflect for a moment soon realize why (4) in this case works. Two simple questions...

a) can we hypothize that if f(x) is entire and has in x=0 a removable singularity, the (4) supplies the exact value of (1)?...

b) is a) the only case in which the (4) supplies the exact value of (1)?...

That's what we discuss in the successive posts...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

... two simple questions...

a) can we hypothize that if f(x) is entire and has in x=0 a removable singularity, the (4) supplies the exact value of (1)?...

b) is a) the only case in which the (4) supplies the exact value of (1)?...

That's what we discuss in the successive posts...

An interesting example permits us to the question b) with a 'no'. An important class of entire functions are the so called 'Bessel functions of the first kind' defined by the series expansion...

$\displaystyle J_{m} (x)= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n! (n+m)!}\ (\frac {x}{2})^{2 n+m}$ (1)

Proceeding 'step by step' we first try to apply the technique under investigation in the case m=0 obtaining...

$\displaystyle I_{0}= \int_{0}^{\infty} J_{0} (x)\ e^{-x}\ dx = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2^{2 n}}\ \frac{(2n)!}{(n!)^{2}} = 1 - \frac{1}{2} + \frac{1 \cdot 2}{2 \cdot 4} - \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} + ...$ (2)

Now we have to remember the binomial expansion...

$\displaystyle (1+x^{2})^{- \frac{1}{2}} = \sum_{n=0}^{\infty} \binom{- \frac{1}{2}}{n} x^{2 n} = 1 - \frac{1}{2} x^{2} + \frac{1 \cdot 2}{2 \cdot 4} x^{4} - \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} x^{6} + ... $ (3)

... and we can conclude that is $\displaystyle I_{0}= \frac{1}{\sqrt{2}}$. Proceeding [with a little of patience...] in the same way we discover that the technique works for the Bessel functions of any order and is...

$\displaystyle I_{m}= \int_{0}^{\infty} J_{m}(x)\ e^{-x}\ dx = \frac{(\sqrt{2}-1)^{m}}{\sqrt{2}}$ (4)

Leaving to the reader the details, it is important to analyse other entire functions suitable to the scope and one of them is the Sine integral function defined as...

$\displaystyle \text{Si} (x)= \int_{0}^{x} \frac{\sin t}{t}\ dt = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2 n + 1}}{(2 n+1)\ (2n+1)!}$ (5)

This case is easier than the previous of course and is...

$\displaystyle \int_{0}^{\infty} \text{Si} (x)\ e^{-x}\ dx = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2 n + 1} = \frac{\pi}{4}$ (6)

Now let's terminate this post with a very interesting case of a non entire function: the function...

$\displaystyle \text{erf} (\sqrt{x}) = \frac{2}{\sqrt{\pi}}\ \int_{0}^{\sqrt{x}} e^{- t^{2}}\ dt = \frac{2}{\sqrt{\pi}} \sum_{n=0}^{\infty} (-1)^{n}\ \frac{x^{n + \frac{1}{2}}}{(2n + 1)\ n!}$ (7)

Remembering that is... $\displaystyle (n+\frac{1}{2})! = \frac{\sqrt{\pi}}{2}\ \frac{(2n+1)!}{2^{n}}$ (8)

... we discover to be in the same case of (2)...

$\displaystyle \int_{0}^{\infty} \text{erf}(\sqrt{x})\ e^{-x}\ dx = 1 - \frac{1}{2} + \frac{1 \cdot 3}{2 \cdot 4} - \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} + ... = \frac{1}{\sqrt{2}}$ (9)

So we have examined different type functions for which the techique explained in the first post can be usefully used... it remain to extablish if a general criterion exists to have convergence in the series expression of the integral $\displaystyle \int_{0}^{\infty} f(x)\ e^{-x}\ dx$ ... Kind regards $\chi$ $\sigma$