0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Do you want an antiderivative, or just to integrate it from $-\infty$ to $\infty$? $\endgroup$
    – Arthur
    Dec 1, 2015 at 20:35
  • 4
    $\begingroup$ Maybe try to complete the square and then do a change of variables. $\endgroup$ Dec 1, 2015 at 20:35
  • $\begingroup$ I need to integrate it from $-\infty$ to $\infty$ $\endgroup$
    – Suslik
    Dec 1, 2015 at 20:38

2 Answers 2

4
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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!

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .