Nice integral

[MaMo2]

I propose this problem:

http://faculty.missouristate.edu/l/lesreid/Adv98.html

[needmathhelp]

thanks.

[_luca.barletta]

For the last serie:
http://www.matematicamente.it/f/viewtopic.php?t=12658

[needmathhelp]

"Kroldar":[quote="blackdie"]proof?

Ye...

$int_1^(+oo) (((x)))/x^3 dx = sum_(n=1)^(+oo) int_n^(n+1) (x-n)/x^3 dx$

But this is a very simple integral, in fact

$int (x-n)/x^3 dx = int 1/x^2 - n/x^3 dx = n/(2x^2) - 1/x$

So

$int_1^(+oo) (((x)))/x^3 dx = sum_(n=1)^(+oo) int_n^(n+1) (x-n)/x^3 dx = sum_(n=1)^(+oo) 1/(2n(n+1)^2)$

Now we have to find the sum of series... let's try

$1/(2n(n+1)^2) = 1/2 (1/n - (n+2)/(n+1)^2) = 1/2 (1/n - (n+1+1)/(n+1)^2) = 1/2 (1/n - 1/(n+1) - 1/(n+1)^2)$

Then

$sum_(n=1)^(+oo) 1/(2n(n+1)^2) = 1/2 sum_(n=1)^(+oo) (1/n - 1/(n+1) - 1/(n+1)^2) = 1/2 sum_(n=1)^(+oo) (1/n - 1/(n+1)) - 1/2 sum_(n=1)^(+oo) 1/(n+1)^2 = 1/2 (1-((pi^2)/6 - 1)) = 1-(pi^2)/12$

Finally

$int_1^(+oo) (((x)))/x^3 dx = 1-(pi^2)/12$[/quote]

Sorry for the banal question, but I do not study math.hence, would you mind explaining me the first passage? how can you write the integral in that way and then pass to series analysis? thank u very much

[Kroldar]

"Reynolds":That number in the result... $pi^2/12$... It seems that it hints the use of Fourier...

Fourier? Uhmm... I dunno, but I think it would be a great result if someone found a link with Fourier... The problem is that there's no periodic replication of the whole function

[Kroldar]

"blackdie":proof?

Ye...

$int_1^(+oo) (((x)))/x^3 dx = sum_(n=1)^(+oo) int_n^(n+1) (x-n)/x^3 dx$

But this is a very simple integral, in fact

$int (x-n)/x^3 dx = int 1/x^2 - n/x^3 dx = n/(2x^2) - 1/x$

So

$int_1^(+oo) (((x)))/x^3 dx = sum_(n=1)^(+oo) int_n^(n+1) (x-n)/x^3 dx = sum_(n=1)^(+oo) 1/(2n(n+1)^2)$

Now we have to find the sum of series... let's try

$1/(2n(n+1)^2) = 1/2 (1/n - (n+2)/(n+1)^2) = 1/2 (1/n - (n+1+1)/(n+1)^2) = 1/2 (1/n - 1/(n+1) - 1/(n+1)^2)$

Then

$sum_(n=1)^(+oo) 1/(2n(n+1)^2) = 1/2 sum_(n=1)^(+oo) (1/n - 1/(n+1) - 1/(n+1)^2) = 1/2 sum_(n=1)^(+oo) (1/n - 1/(n+1)) - 1/2 sum_(n=1)^(+oo) 1/(n+1)^2 = 1/2 (1-((pi^2)/6 - 1)) = 1-(pi^2)/12$

Finally

$int_1^(+oo) (((x)))/x^3 dx = 1-(pi^2)/12$

[blackdie]

proof?

[needmathhelp]

Jeez, you guys speak such an "Italian English" !!!

[Fioravante Patrone1]

"luca.barletta":I've got yuor same result

me too

[Thomas16]

"Reynolds":That number in the result... $pi^2/12$... It seems that it hints the use of Fourier...

Yeah... but I guess it is sufficient the known formula for the sum of the inverse squares of natural numbers...

[fireball1]

That number in the result... $pi^2/12$... It seems that it hints the use of Fourier...

[_luca.barletta]

I've got yuor same result

[Camillo]

It is really very nice!
I will just give the final result without details for the moment since others will try .
The result is $ 1-pi^2/12 $.