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.
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$\begingroup$ As a side comment - quite often integrals of this nature (not her) can be simplified with Glasser's Master Theorem en.wikipedia.org/wiki/Glasser%27s_master_theorem $\endgroup$– user150203Dec 12, 2018 at 12:43
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1 Answer
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Completing the square can put it into the standard form here with the appropriate substitution.
$$\int_{-\infty}^{\infty} e^{-x^2-x} dx = \int_{-\infty}^{\infty} e^{-(x+\frac{1}{2})^2+\frac{1}{4}} dx = e^{\frac{1}{4}}\int_{-\infty}^{\infty} e^{-(x+\frac{1}{2})^2} dx$$
