1. ## Incomplete gamma function

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?

2. ## Proof

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

3. 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
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 $\displaystyle \mathbb{R}$. Maybe check out Gamma function - Wikipedia, the free encyclopedia.
And for
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?
The 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).

5. Originally Posted by rman144
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
That looks OK, but would have been better if you had used limits in the argument.

That is:

$\displaystyle \Gamma(z,x) =\Gamma(z)-\gamma(z,x)$

Then:

$\displaystyle \Gamma(z,\infty) = \lim_{x \to \infty}\Gamma(z,x)=\lim_{x \to \infty} \left(\Gamma(z)-\gamma(z,x) \right)$

........... $\displaystyle =\Gamma(z)- \lim_{x \to \infty} \left(\gamma(z,x) \right)$

........... $\displaystyle =\Gamma(z)- \Gamma(z)=0$

In fact $\displaystyle \Gamma(z,\infty)$ is just short hand for $\displaystyle \lim_{x \to \infty}\Gamma(z,x)$

RonL

6. 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.

7. Originally Posted by shoryuke
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.
I think this limit is undefined.

CB

8. 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?

9. Originally Posted by shoryuke
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?

I will assume that this makes sense of all $\displaystyle z \in \mathbb{C}$ Now if we want to know what happens when $\displaystyle z \to \infty$ along some curve in $\displaystyle \mathbb{C}$ it is clear that the limiting process in the incomplete gammas takes place in a particular order, first we sum over $\displaystyle k$ then we take the limit over $\displaystyle z$.

CB

10. When we sum over k (k small), the incomplete gamma function will be zero as z tends to infinity. But what happens when k is large, since the summation is from k = 0 to infinity?

11. Originally Posted by shoryuke
When we sum over k (k small), the incomplete gamma function will be zero as z tends to infinity. But what happens when k is large, since the summation is from k = 0 to infinity?
For fixed z as k goes to infinity gamma((a/2)+k,-bz^2) goes to infinity.

CB