1. ## Integration Help

Can someone confirm that the following is correct:

∫〖[e^(-nx) ][x^(s-∞-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:

∫〖[e^(-nx) ][x^(s-∞-2)] 〗 dx=(e^(-n))*∫〖[x^(s-∞-2)] 〗 dx

Any help would be much appreciated.

2. Originally Posted by rman144
Can someone confirm that the following is correct:

∫〖[e^(-nx) ][x^(s-∞-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:

∫〖[e^(-nx) ][x^(s-∞-2)] 〗 dx=(e^(-n))*∫〖[x^(s-∞-2)] 〗 dx

Any help would be much appreciated.
You can't take $e^{-n}$ out as a constant. It's a function of x $e^{-nx}$.

I know the value is a function, but on that particular interval, isn't the substitution allowed? Basically, if on that interval, the only meaningful solution occurs when x=1, why can't I just substitute the one in and let the value act as a constant e^-n?