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For the following upper incomplete Gamma function:

\begin{equation} \Gamma(1+d,A-c \ln x)=\int_{A-c\ln x}^{\infty}t^{(1+d)-1}e^{-t}dt \end{equation} I am trying to calculate the derivative of $Γ$ with respect to $x$. In general it holds that: \begin{equation} \frac{\mathrm{d} }{\mathrm{d} x} \Gamma(s,x)=-x^{s-1}e^{-x} \end{equation} After my calculations I ended up with: \begin{equation} \frac{\mathrm{d} }{\mathrm{d} x} \Gamma(1+d,A-c \ln x)=c e^{-A}x^{c-1}(A-c\ln x)^d \end{equation} but the author says that the correct answer is rather: \begin{equation} \frac{\mathrm{d} }{\mathrm{d} x} \Gamma(1+d,A-c \ln x)=-e^{-A}x^{c-1}(A-c\ln x)^d \end{equation} But how is this correct? By applying the chain rule it should be: \begin{equation} \frac{\mathrm{d} }{\mathrm{d} x} \Gamma(1+d,A-c \ln x)=\left[ -(A-c\ln x)^de^{A-c \ln x} \right]\frac{\partial{}}{\partial{x}}(A-c \ln x) \end{equation} right?

Thanks!

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Hint. We have $$ \frac{\partial}{\partial x} Γ(s,x)=-x^{s-1}e^{-x} $$ then, by the chain rule, we get $$ \begin{align} \frac{d}{d x}\left(Γ(1+d,A-c \ln x)\right)&=\frac{-c}{x}\cdot\left.\frac{\partial}{\partial t} Γ(s,t)\right|_{(s,t) =(1+d,A-c \ln x)} \\\\&=\frac{c}{x}\cdot(A-c \ln x)^de^{-(A-c \ln x)} \\\\&=c\:x^{c-1}e^{-A}(A-c \ln x)^d. \end{align} $$

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  • $\begingroup$ Thank you for your answer. Are you sure that there it is $x^c$ and not $x^{c-1}$? It would be of great assistance if you could be a bit more detailed on why this is the result. I would like to understand my fault on applying the chain rule :) $\endgroup$ – Mitscaype Jun 25 '16 at 17:25
  • $\begingroup$ I have a question. Is $Γ_x(1+d,A-c \ln x)$ meaning $\frac{d}{dx}\left(Γ(1+d,A-c \ln x)\right)$ or is it meaning $\left.\left(\frac{\partial}{\partial t}Γ(1+d,t)\right)\right|_{t=A-c \ln x}$? $\endgroup$ – Olivier Oloa Jun 25 '16 at 17:30
  • $\begingroup$ It is $\frac{d}{dx} \left( \Gamma(1+d,A-c\ln x) \right)$. $\endgroup$ – Mitscaype Jun 25 '16 at 17:37
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    $\begingroup$ Ok. I will edit my answer accordingly. Thanks. $\endgroup$ – Olivier Oloa Jun 25 '16 at 17:40
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    $\begingroup$ Ok I see, yes it is. $\endgroup$ – Olivier Oloa Jun 25 '16 at 18:31

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