I have two questions
1) How to get the closed-form expression of
$\int_d^e (x-g)^{-f} x^{-c} (b x)^{-a} \Gamma (a,b x) dx=?$
My thought: I can only get the following closed-form expression [This is from my note. I am really sorry I forget where I found this. I appreciate if someone can help me to understand how to get this. I have tested in Matlab, it is right.]
$\int_d^e x^{-c} (b x)^{-a} \Gamma (a,b x) dx=\frac{x^{1-c} (b x)^{c-1} \Gamma (1-c,b x)}{a+c-1}-\frac{x^{1-c} (b x)^{-a} \Gamma (a,b x)}{a+c-1} \bigg |_d^e$
2) How to get the closed-form expression of
$\int_d^e x^{-c} \log (x+1) (b x)^{-a} \Gamma (a,b x) dx = ?$
My thought: I can only get the following closed-form expression [This is from my note. I am really sorry I forget where I found this. I appreciate if someone can help me to understand how to get this. I have tested in Matlab, it is right.]
$\int_d^e x^{-c} \log (x) (b x)^{-a} \Gamma (a,b x)=\frac{x^{-c} \left(b x (a+c-1) \, _2F_2(1-c,1-c;2-c,2-c;-b x)+(c-1)^2 (b x)^c (\Gamma (1-c,b x)-(a+c-1) \log (x) \Gamma (1-c,0,b x))\right)}{b (c-1)^2 (a+c-1)^2}-\frac{x^{1-c} (b x)^{-a} \Gamma (a,b x) ((a+c-1) \log (x)+1)}{(a+c-1)^2} \bigg |_d^e$
where $\Gamma (a,z)=\int _z^{\infty } t^{a-1} e^{-t} dt$ is the incomplete gamma function