0
$\begingroup$

Let $\Gamma (s)=\int_0^\infty t^{s-1}e^{-t} dt$ - classical gamma function - and $\gamma(s,x)=\int_0^x t^{s-1}e^{-t} dt$ - lower incomplete gamma function. Then is the following true

$$ \lim_{\alpha \to 1} \frac{1}{\Gamma(1-\alpha)} \gamma\left( 1-\alpha, [\Gamma(1-\alpha)]^{-1/\alpha} \right)=1 $$ or false?

Intuitively, looking at the expression of $\gamma$ and taking into account that $\lim_{\alpha\to1}\Gamma(1-\alpha)=\infty$, I would say the limit is $0$. Yet, if we can use the fact that $\gamma(s,x)$ near $x=0$ behaves like $x^s/s$, this would lead us to

\begin{equation} \begin{split} &\lim_{\alpha \to 1} \frac{1}{\Gamma(1-\alpha)} \gamma\left( 1-\alpha, [\Gamma(1-\alpha)]^{-1/\alpha} \right)\\ & \quad=\lim_{\alpha \to 1} \frac{[\Gamma(1-\alpha)]^{-1/\alpha}}{1-\alpha} \\ & \quad=\lim_{\alpha \to 1} \frac{[\Gamma(\alpha)\sin(\pi\alpha)/\pi]^{1/\alpha}}{1-\alpha}\\ & \quad=\lim_{\alpha \to 1} \frac{[\sin(\pi\alpha)/\pi]^{1/\alpha}}{1-\alpha}. \end{split} \end{equation} Then, using the fact that when $x$ is close to $\pi$, $\sin(x) \approx \pi -x$, I would go on with \begin{equation} \hspace{-4em}=\lim_{\alpha \to 1} \frac{[(\pi -\pi \alpha)/\pi]^{1/\alpha}}{1-\alpha}\\ \hspace{-6.4em}=\lim_{\alpha \to 1} \frac{(1-\alpha)^{1/\alpha}}{1-\alpha}\\ \hspace{.7em}=\lim_{\alpha \to 1} \exp \left\lbrace \left(\frac{1}{\alpha}-1\right)\log(1-\alpha) \right\rbrace \end{equation} which eventually equals $1$, since the exponent in the last line converges to $0$. Am I loosing something/making some mistake?

$\endgroup$
  • $\begingroup$ Here's a hint:$$\frac{\Gamma(x)}x=\frac{\Gamma(x+1)}{x^2}\sim_0\frac{\Gamma(1)}{x^2}$$ $\endgroup$ – Simply Beautiful Art Oct 5 '17 at 14:39
0
$\begingroup$

Hint:

$$\frac1{x[\Gamma(x)]^{1/(1-x)}}=\frac{x^{x/(1-x)}}{[x\Gamma(x)]^{1/(1-x)}}=\frac{x^{x/(1-x)}}{[\Gamma(x+1)]^{1/(1-x)}}\to_0\frac1{[\Gamma(1)]^1}=1$$

Since

$$\lim_{x\to0}x^{x/(1-x)}=1$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.