Understanding the incomplete gamma function

[badtwistoffate]

I know that the gamma function (from 0 to infinity):
\int e^-t t^x-1 dt = \Gamma(x)

and that the relation exists...

\int e^-ut t^x-1 dt = 1/u^x \Gamma(x)

Now for the lower bound incomplete gamma function... I see that from
http://people.math.sfu.ca/~cbm/aands/page_260.htm (equation 6.5.2):

I(s,x) = P(a,x) \Gamma(x) = \int e^-t t^x-1 dt (evaluated from 0 to x). Where the far left hand side is the result of the incomplete gamma function (lower bound).

Thus, my question/problem is can we use this relation (second equation) in the evaluation of the lower bound incomplete gamma function.

that:


1/u^x I(s,x) = 1/u^x P(a,x) \Gamma(x) = \int e^-ut t^x-1 d (evaluated from 0 to x)

Since we are just pulling out that factor of 1/u^x anyway. This correct? Can we use that relation this way?

 

[badtwistoffate]

I know that the gamma function (from 0 to infinity):
\int e^-t t^x-1 dt = \Gamma(x)

and that the relation exists...

\int e^-ut t^x-1 dt = 1/u^x \Gamma(x)

Now for the lower bound incomplete gamma function... I see that from
http://people.math.sfu.ca/~cbm/aands/page_260.htm (equation 6.5.2):

I(s,x) = P(a,x) \Gamma(x) = \int e^-t t^x-1 dt (evaluated from 0 to x). Where the far left hand side is the result of the incomplete gamma function (lower bound).

Thus, my question/problem is can we use this relation (second equation) in the evaluation of the lower bound incomplete gamma function.

that:1/u^x I(s,x) = 1/u^x P(a,x) \Gamma(x) = \int e^-ut t^x-1 d (evaluated from 0 to x)

Since we are just pulling out that factor of 1/u^x anyway. This correct? Can we use that relation this way?

Thus, instead of just a -t in the exponent, there is now a constant with it! So is this legal?