I'm insterested in calculating
$$
\frac{\partial}{\partial \alpha} \gamma \left(p, \left( \frac{x-\mu}{\sigma} \right)^2/\alpha^2 \right)
$$
where $\gamma(p,u)=\int_{0}^{u}w^{p-1}e^{-w}dw$ is the incomplete Gamma function. Using this definition, you'd get
$$
\frac{\partial}{\partial \alpha} \int_{0}^{\left( \frac{x-\mu}{\sigma} \right)^2/\alpha^2} w^{p-1}e^{-w}dw.
$$
I thought, well, this looks just like the Fundamental Theorem of Calculus, except for the slight complication that there's a partial derivative and the upper limit of integration isn't just the variable $\alpha$. Anyway, after thinking for a while, I thought that the answer might simply be
$$
\frac{\partial}{\partial \alpha} \int_{0}^{\left( \frac{x-\mu}{\sigma} \right)^2/\alpha^2} w^{p-1}e^{-w}dw = w^{p-1}e^{-w} \cdot \frac{\partial}{\partial \alpha} \left[\left( \frac{x-\mu}{\sigma} \right)^2/\alpha^2\right].
$$
Is this reasoning correct or am I missing something?
Best Answer
You also need to evaluate $w$ at $\left(\frac{x-\mu}{\sigma}\right)^2\frac{1}{\alpha^2},$ then you will obtain the correct answer. Note that after integrating, and evaluating $w$ at the limits, there will be no $w$ terms left. This is the Leibniz integral rule.