Understanding the incomplete gamma function

  • #1

Main Question or Discussion Point

I know that the gamma function (from 0 to infinity):
[tex]\int [/tex] e-t tx-1 dt = [tex]\Gamma[/tex](x)

and that the relation exists...

[tex]\int [/tex] e-ut tx-1 dt = 1/ux [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 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 [tex]I[/tex](s,x) = 1/ux P(a,x) [tex]\Gamma[/tex](x) = [tex]\int [/tex] 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?
 
Last edited:

Answers and Replies

  • #2
I know that the gamma function (from 0 to infinity):
[tex]\int [/tex] e-t tx-1 dt = [tex]\Gamma[/tex](x)

and that the relation exists...

[tex]\int [/tex] e-ut tx-1 dt = 1/ux [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 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 [tex]I[/tex](s,x) = 1/ux P(a,x) [tex]\Gamma[/tex](x) = [tex]\int [/tex] 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?
Thus, instead of just a -t in the exponent, there is now a constant with it! So is this legal?
 

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