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