Hello everyone!
I have a question about the derivative of the regularized upper incomplete gamma function. Considering the gamma function and the incomplete gamma function \begin{eqnarray} \Gamma(x)&=&\int_0^\infty{t^{x-1}e^{-t}\mathrm{d}t},\\ \Gamma(x,\lambda)&=&\int_\lambda^\infty{t^{x-1}e^{-t}\mathrm{d}t},\end{eqnarray} where the quotient of $\Gamma(x,\lambda)$ divided by $\Gamma(x)$ is usually denoted as regularized upper incomplete gamma function \begin{equation} Q(x,\lambda)=\frac{\Gamma(x,\lambda)}{\Gamma(x)}, \end{equation} there seems to be a precise estimation for the derivative $\frac{\partial Q(x,\lambda)}{\partial x}$, but I cannot prove it.
With numerical experiment, I found that given $\lambda$ is not small, $\frac{\partial Q(x,\lambda)}{\partial s}$ is almost proportional to \begin{equation} \frac{\lambda^xe^{-\lambda}}{x^xe^{-x}}=\exp\left(-\left(x\log x-x\log\lambda-x+\lambda\right)\right). \end{equation} Denoting \begin{equation} I(\lambda)=\int_0^\infty{\frac{\lambda^xe^{-\lambda}}{x^xe^{-x}}\mathrm{d}x}, \end{equation} the partial derivative $\frac{\partial Q(x,\lambda)}{\partial x}$ seems to have the following approximation \begin{equation} \frac{\partial Q(x,\lambda)}{\partial x} \approx\frac{\lambda^xe^{-\lambda}}{x^xe^{-x}}/I(\lambda). \end{equation}
I give the comparison of these two formulas from the numerical experiments with Matlab in following figures, where the blue solid line and green dotted line represent $\frac{\partial Q(x,\lambda)}{\partial x}$ and $\frac{\lambda^xe^{-\lambda}}{x^xe^{-x}}/I(\lambda)$ respectively.
From these figures, it's obvious that as the increase of $\lambda$, the two lines approach to be equal. However, I cannot prove such a relationship due to the complexity of the derivative $\frac{\partial Q(x,\lambda)}{\partial x}$.
Could you help me with some comments about this question? Any suggestions will be helpful and thank you very much!
