Hi

can I say that http://www.advancedphysics.org/cgi-bin/mimetex.cgi?\ints(x-a)dx=a where s is the step function of (x-a) and the integration is from zero to infinity.

Printable View

- Apr 22nd 2009, 07:49 AMomeganeustep function integration
Hi

can I say that http://www.advancedphysics.org/cgi-bin/mimetex.cgi?\ints(x-a)dx=a where s is the step function of (x-a) and the integration is from zero to infinity. - Apr 22nd 2009, 10:51 AMHallsofIvy
No, you can't! s(x-a) is equal to 0 from negative infinity to a, 1 from a to infinity.

$\displaystyle \int_{-\infty}^{\infty}s(x-a)dx= \int_a^\infty dx$

which does not exist. - Apr 22nd 2009, 10:31 PMomeganeu
but I mean by this step function that it is equall one in the region from zero to a, and it is equall zero in all the regions less than zero and greater than a, so I quessed that its integration from zero to infinity will be reduced to the interval where it is equall one (from zero to a). I thought of it as a multiplication by zero (which gives zero) in all regions except when x goes from zero to a.

- Apr 22nd 2009, 10:35 PMomeganeu
http://www.advancedphysics.org/cgi-bin/mimetex.cgi?\ints(x-a)dx=a

actually I need to integrate (from zero to infinity) a function which is equall one in the region from 0 th a and equalls zero otherwise (when x less than zero or greater than a). - Apr 23rd 2009, 05:27 AMHallsofIvy
In that case, it is, of course, $\displaystyle \int_0^a dx= a$, as you had originally.

- Apr 23rd 2009, 05:53 AMomeganeu
Thanks a lot this solve a great problem to me.