Let us consider the incomplete Gamma function to $a = \frac{1}{3}$ $$\gamma(x) = \int_0^x s^{-2/3}\exp(-s) \,ds.$$ I am interested in the function $f(t) = t^4 \gamma(t^3)$. A quick WolframAlpha search from here says that this function is in $L^\infty(0,\infty)$. I am currently investigating if this is really true. As it is $\lim_{t \to \infty} \gamma(t) = c>0$ for some constant $c$. I find it quite hard to believe that $\lim_{t \to \infty} \gamma(t) = 0$.
Asked
Active
Viewed 31 times
0
-
I've edited your post. Please ensure that it still represents what you originally intended. – Clayton Jul 06 '19 at 19:48
-
To use the lower incomplete Gamma function, you need to introduce the additional middle argument as 0: x^4Gamma(1/3, 0, x^3) – Simply Beautiful Art Jul 06 '19 at 19:54
1 Answers
2
Wolfram's definition of the incomplete gamma function is as follows $$\Gamma(a,z)=\int_z^\infty t^{a-1}e^{-t}\mathrm{d}t$$ What you meant to write was $\Gamma(1/3)-\Gamma(1/3,x^3)$ and the limit is indeed unbounded. See their reference here - https://reference.wolfram.com/language/ref/Gamma.html
Peter Foreman
- 19,947