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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.

Comparison for $\lambda=1$ Comparison for $\lambda=2$ Comparison for $\lambda=5$ Comparison for $\lambda=10$

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!

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wikipedia has a formula for the derivative wrt the first argument, maybe simplifying that formula gives the approximation that you are observing. – Suvrit Feb 22 '13 at 18:39
@Suvrit Thank you for your attention! Do you mean there is a formula for $\frac{\partial Q(x,λ)}{\partial x}$ in wikipedia? I am sorry that I did not find such a formula but only the formula $\frac{\partial\Gamma(x,λ)}{\partial x}=\ln\lambda\Gamma(x,\lambda)+\lambda T(3,x,\lambda)$ instead. If you found the formula for the derivative of $Q(x,\lambda)$ w.r.t. $x$, could you please show me the url of the wikipedia? Thank you very much! – ppyang Feb 23 '13 at 12:30
there seems to be a typo (transposition) in your definition of $Q(x,\lambda)$....also, some confusion is arising from the non-standard notation: usually, $x$ is the second argument. But I guess, indeed, the formula in Wikipedia was only for the version that you quote --- taking logs and then diff. will allow a use of this formula, however... – Suvrit Feb 24 '13 at 18:58
@Suvrit Sorry to have made confusion by using different variables in the definition of $Q$. Thank you for your attention, but I cannot understand your last sentense clearly. Can you explain how to use this formula to solve my question in more detail? Thank you very much! – ppyang Feb 25 '13 at 0:35

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