# Integration Work Check

1. Dec 23, 2008

### rman144

Can someone confirm that the following is correct:

S [e^(-nx) ][x^(s-oo-2)] dx , limits from x=1 to x=infinity, n is some real integer>=1, and 0<Re(s)<1

What I want to know is, because the equation vanishes for all "x" in the interval x=1 to x=infinity, excluding x=1, can I just bind (e^(-nx)) to e^(-n(1)), leaving:

S [e^(-nx) ][x^(s-oo-2)] dx=(e^(-n))*S [x^(s-oo-2)] dx

Any help would be much appreciated.

Last edited: Dec 23, 2008
2. Dec 23, 2008

### HallsofIvy

Staff Emeritus
2. I don't know what "s-∞-2" could possibly mean.
3. I don't know what "bind (e^(-nx)) to e^(-n(1))" means.

3. Dec 23, 2008

### rman144

First, the s-oo-2 is read as "s" (the variable s) minus infinity minus 2.

Next, when I say bind it to a constant, I mean:

Say you have:

S f(x)*g(x) dx, limits from x=x1 to x=x2

Now if you know that the equation f(x) on the interval x1 to x2 is at most f(k1) and at least f(k2), then the integral:

S f(x)*g(x) dx

Will have a maximum absolute value of:

abs( S f(k1)*g(x) dx )

And a minimum of:

abs( S f(k2)*g(x) dx)

Now what I'm asking is, because within my original equation, the only value in the interval that doesn't go to zero is when x=1, as a result, can I repeat the above process for e^(-nx) and "bind" the value to the interval:

e^-n<=e^(-nx)<=e^-n

Which implies:

e^(-nx)=e^(-n)

I'm not trying to solve the integral; I'm merely attempting to establish whether or not it converges, and if so, how small an interval can I confine the possible solution of the actual integral to.

Last edited: Dec 23, 2008
4. Dec 23, 2008

### HallsofIvy

Staff Emeritus
Yes, even I could read[/i] it that way but what does it MEAN? Since "infinity" is not a number, that makes no sense to me. Is it intended as a limit? And if so what could "infinity minus 2" mean?