I am working with a probability distribution and I have to find the derivative of incomplete gamma function as \begin{equation*} \Gamma(\frac{\theta}{\beta}x^{2},\theta) = \int_{0}^{\frac{\theta}{\beta}x^{2}}t^{\theta-1}\cdot e^{-t}dt \end{equation*} How can I find the derivatives of above function with respect to $\theta$ and $\beta$ both??
1 Answer
These are quite complicated, but Mathematica is up to the task: \begin{align*} \frac{\partial}{\partial\theta}\,\Gamma\!\left(\frac{\theta x^2}{\beta},\theta\right) &=\frac{x^2\left(G_{2,3}^{3,0}\!\left(\theta\left|\begin{array}{c}1,1\\0,0,\frac{\theta x^2}{\beta}\\\end{array}\right.\right)+\log(\theta)\,\Gamma\!\left(\frac{\theta x^2}{\beta},\theta\right)\right)}{\beta}-e^{-\theta}\,\theta^{\frac{\theta x^2}{\beta}-1}\\ \frac{\partial}{\partial\beta}\,\Gamma\!\left(\frac{\theta x^2}{\beta},\theta\right) &=-\frac{\theta x^2\left(G_{2,3}^{3,0}\!\left(\theta\left|\begin{array}{c}1,1\\0,0,\frac{\theta x^2}{\beta}\\\end{array}\right.\right)+\log(\theta)\,\Gamma\!\left(\frac{\theta x^2}{\beta},\theta\right)\right)}{\beta^2}. \end{align*} Here the complicated $G$ function thingy is called the Meijer G function. Not sure of how much use this is to you, but I've certainly seen worse.