step function integration

**omeganeu** · April 22nd 2009, 07:49 AM

Hi

can I say that $\int$ s(x-a)dx=a where s is the step function of (x-a) and the integration is from zero to infinity.

**HallsofIvy** · April 22nd 2009, 10:51 AM

No, you can't! s(x-a) is equal to 0 from negative infinity to a, 1 from a to infinity.

$\int_{-\infty}^{\infty}s(x-a)dx= \int_a^\infty dx$
which does not exist.

**omeganeu** · April 22nd 2009, 10:31 PM

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.

**omeganeu** · April 22nd 2009, 10:35 PM

$\int$ s(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).

**HallsofIvy** · April 23rd 2009, 05:27 AM

In that case, it is, of course, $\int_0^a dx= a$ , as you had originally.

**omeganeu** · April 23rd 2009, 05:53 AM

Thanks a lot this solve a great problem to me.

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