Okay, found out how I can prove it:
lower-gamma(z,infinity)=gamma(z)
lower-gamma(z,x)+upper-gamma(z,x)=gamma(z)
If x=infinity:
gamma(z)+upper-gamma(z,infinity)=gamma(z)
Thus:
upper-gamma(z,infinity)=0
I noticed that the equation for the incomplete gamma function yields an issue for gamma(z,infinity)
It makes the integral representation have infinity for both limits of integration.
My question for this is, does this have a solution, and if so, would I be correct in assuming that the answer is zero?
No, it's not good. I'm too tired to provide a good answer, but the gamma function is an improper integral but it is defined over . Maybe check out Gamma function - Wikipedia, the free encyclopedia.Okay, found out how I can prove it:
lower-gamma(z,infinity)=gamma(z)
lower-gamma(z,x)+upper-gamma(z,x)=gamma(z)
If x=infinity:
gamma(z)+upper-gamma(z,infinity)=gamma(z)
Thus:
upper-gamma(z,infinity)=0
And forThe limits of the integral make it an improper integral. But the value of the gamma function (so of the improper integral) is always finite for a given z. (I'm not that sure though).I noticed that the equation for the incomplete gamma function yields an issue for gamma(z,infinity)
It makes the integral representation have infinity for both limits of integration.
My question for this is, does this have a solution, and if so, would I be correct in assuming that the answer is zero?
What happens for gamma(z,x) when both z and x tends to infinity?
If we use Matlab and type
gamma(10)*gammainc(1000,10,'upper'), where x = 1000 and z = 10, the answer is 0.
but when we type
gamma(1000)*gammainc(1000,1000,'upper'), where x = 1000 and z = 1000, the answer is infinity.
The below integral is given in
Complementary error function: Integration (formula 06.27.21.0019)
If the limit is from 0 to infinity, the condition whereby k and z (in the incomplete gamma function) tends to infinity will arise. Does that means that the integral below cannot be solve when the limit is from 0 to infinity?