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#### chisigma

##### Well-known member

- Feb 13, 2012

- 1,704

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/commentar...very-advanced-integration-technique-4215.html

$\displaystyle I= \int_{0}^{\infty} \frac{1 - \cos x}{x^{2}}\ e^{-x}\ dx$ (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/commentar...very-advanced-integration-technique-4215.html

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