We want to show the following:
$\int_{0}^{\infty}e^{(-y-y^2)/(2c^2)}dy =\int_{0}^{\infty} c \cdot e^{-y^2 - 2cy}dy$
To the best of my abilities, I have tried the standard tricks including u-substitution, integration by parts, etc., but I wind up either having to perform another substitution or another integration by parts.
Intuition that I am unable to translate/verify tells me that in order to bring the $c$ down, I should somehow derive $2c^2$ with respect to $c$.
Can someone please provide some suggestions?
Completing the squares and assuming $c>0$, you face the gaussian integral making $$\int_0^\infty e^{-\frac{y+y^2}{2 c^2}}\,dy=\sqrt{\frac{\pi }{2}} c\, e^{\frac{1}{8 c^2}} \text{erfc}\left(\frac{1}{2 \sqrt{2} c}\right)$$ $$\int_0^\infty c\, e^{-(2 c y+y^2)}\,dy=\frac{1}{2} \sqrt{\pi }\, c\, e^{c^2} \text{erfc}(c)$$
As said in comments, they are equal only if $c=0.4117$