Th following integral (A,B are constants) $$\int_{-\infty}^{\infty}\sqrt{A-bx}e^{-x^2}dx$$ pops up as a curve fitting routine
Attempt to integrate it by parts (setting $u=\sqrt{A-bx},v'=e^{x^2}$) gives
EDIT: In detail
$$\int_{-\infty}^{\infty}\sqrt{A-bx}e^{-x^2}dx=\left[\sqrt{A-bx}\left(\int_{-\infty}^{\infty}e^{-x'^2}dx'\right)\right]^{\infty}_{-\infty}-\int_{-\infty}^{\infty}\frac{-b}{2\sqrt{A-bx}}\left(\int_{-\infty}^{\infty}e^{-x'^2}dx'\right)dx$$
Note that $\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}$. Therefore
$$\int_{-\infty}^{\infty}\sqrt{A-bx}e^{-x^2}dx=\left[\sqrt{A-bx}\sqrt{\pi}\right]^{\infty}_{-\infty}-\int_{-\infty}^{\infty}\frac{-b}{2\sqrt{A-bx}}\sqrt{\pi}dx$$
$$\int_{-\infty}^{\infty}\sqrt{A-bx}e^{-x^2}dx=\left[\sqrt{A-bx}\sqrt{\pi}\right]^{\infty}_{-\infty}+\left[\sqrt{A-bx}\sqrt{\pi}\right]_{-\infty}^{\infty}$$
$$=2\sqrt{\pi}\left[\sqrt{A-bx}\right]^{\infty}_{-\infty}$$
which diverges
A brief read on a related link and the discussion with a user about how there is a branch in the $\sqrt{}$ function in $\mathbb{C}$ caused me to suspect whether it is because of the branch point in the squareroot function that lead to integration by parts to fail
- Why does integration by parts fail, what is the source of the diverging term despite the integral is known to converge?
- What is the correct way to integrate it, should I split it into two parts like so: $\int_{-\infty}^{\infty}=\int_{-\infty}^{0}+\int_{0}^{\infty}$?