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

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 $e^{\frac{1}{4}}$ ??

  • 1
    $\begingroup$ Yes, completing the square is an excellent idea for evaluating your integral. Now, what is your question? $\endgroup$ – J. M. is a poor mathematician Jul 22 '12 at 12:32
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    $\begingroup$ yes: $-(x-\frac 12)^2+\frac 14$ is enough. $\endgroup$ – Raymond Manzoni Jul 22 '12 at 12:33
  • $\begingroup$ Ok,That is my question $\endgroup$ – Frank Jul 22 '12 at 12:34

Let us make certain what the OP and the comments have suggested.

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 $e^{1/4}$.


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