badtwistoffate
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I know that the gamma function (from 0 to infinity):
\int e-t tx-1 dt = \Gamma(x)
and that the relation exists...
\int e-ut tx-1 dt = 1/ux \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 tx-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/ux I(s,x) = 1/ux P(a,x) \Gamma(x) = \int e-ut tx-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?
\int e-t tx-1 dt = \Gamma(x)
and that the relation exists...
\int e-ut tx-1 dt = 1/ux \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 tx-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/ux I(s,x) = 1/ux P(a,x) \Gamma(x) = \int e-ut tx-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?
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