How does one solve the following integral?

$$\int_{-\infty}^{\infty} e^{-x^2-x} dx$$

I've tackled this thing every way I can think of and I'm still lost.


1 Answer 1


Completing the square can put it into the standard form here with the appropriate substitution.

$$\int_{-\infty}^{\infty} e^{-x^2-x} dx = \int_{-\infty}^{\infty} e^{-(x+\frac{1}{2})^2+\frac{1}{4}} dx = e^{\frac{1}{4}}\int_{-\infty}^{\infty} e^{-(x+\frac{1}{2})^2} dx$$

See this question also


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