# the integral :$\int^\infty_{-\infty}e^{-x^{2}+x}dx$

$$\int^\infty_{-\infty}e^{-x^{2}+x}dx$$

May be completing the square on the $-x^{2}+x$. Then, make a sub.

then end up with the Gaussian integral with a constant multiple of $e^{\frac{1}{4}}$ ??

-
Yes, completing the square is an excellent idea for evaluating your integral. Now, what is your question? –  Guess who it is. Jul 22 '12 at 12:32
yes: $-(x-\frac 12)^2+\frac 14$ is enough. –  Raymond Manzoni Jul 22 '12 at 12:33
Ok,That is my question –  Frank Jul 22 '12 at 12:34

We are doing $\displaystyle \int^\infty_{-\infty}e^{-x^{2}+x}dx$
First, we note that $-x^2 + x = -(x - \frac{1}{2})^2 + \frac{1}{4}$, which we could get from completing the square. Then $e^{-x^2 + x} = e^{1/4}e^{-(x + 1/2)^2}$.
So we are left to compute $\displaystyle \int_{-\infty}^\infty e^{1/4}e^{-(x + 1/2)^2} dx = e^{1/4}\int_{-\infty}^\infty e^{-(x + 1/2)^2} dx$
On the one hand, we could use the substitution $u = x + 1/2$ to finish. Or we could intuit that $\displaystyle \int_\mathbb{R} e^{-(x + 1/2)^2} dx$, the area under $e^{-(x + 1/2)^2}$, is the same as the area under $e^{-x^2}$, as they're the same curve except for a horizontal shift by $1/2$. In either case, we are left with the standard Gaussian integral multiplied by $e^{1/4}$.