- #1

- 81

- 0

## Main Question or Discussion Point

I know that the gamma function (from 0 to infinity):

[tex]\int [/tex] e

and that the relation exists...

[tex]\int [/tex] e

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):

[tex]I[/tex](s,x) = P(a,x) [tex]\Gamma[/tex](x) = [tex]\int [/tex] e

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

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

[tex]\int [/tex] e

^{-t}t^{x-1}dt = [tex]\Gamma[/tex](x)and that the relation exists...

[tex]\int [/tex] e

^{-ut}t^{x-1}dt = 1/u^{x}[tex]\Gamma[/tex](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):

[tex]I[/tex](s,x) = P(a,x) [tex]\Gamma[/tex](x) = [tex]\int [/tex] 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}[tex]I[/tex](s,x) = 1/u^{x}P(a,x) [tex]\Gamma[/tex](x) = [tex]\int [/tex] 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?

Last edited: