Hint for integrating exp(x-x^2)

Question

The function $e^{x-x^2}$ is zero if $x \to \infty$ or $x \to -\infty$ it looks like a normal-distribution-curve with the max. value at $x=0.5$.

Has somebody a hint for integrating it from $-\infty$ to $\infty$ ? Thank you.

Answer 1

I assume you need to solve this integral: $$I=\int_{-\infty}^{+\infty}e^{x-x^2}dx.$$ Try this: $x-x^2=-(x-1/2)^2+1/4$. The integral then becomes $$I=e^{1/4}\int_{-\infty}^{+\infty}e^{-(x-1/2)^2}dx.$$ Do a change of variables, recognize the gaussian integral and you're done!

Answer 2

$$x-x^2=-\left(x-\frac{1}{2}\right)^2 +\frac{1}{4}$$

So you are going to get the usual normal $e^{-t^2}$ times a constant ($e^{1/4}$) after substituting $t=x-\frac{1}{2}$.