Memorizing the result for Gaussian integrals
I too would like to write down the result of a Gaussian integral by inspection…so this is an attempt to make it easier to do that.
Start with the following:
\[\begin{aligned} a x^2 + bx +c &= a\left[\frac{ax^2 + bx + c}{a}\right] \\ &= a\left[\frac{4a^2x^2 +4abx + 4ac}{4a^2}\right] \\ &= a\left[\frac{(b+2ax)^2}{4a^2}-\frac{b^2-4ac}{4a^2}\right] \\ &= a\left[(x+\frac{b}{2a})^2-\frac{b^2-4ac}{4a^2}\right] \\ &= a(x+\frac{b}{2a}-\frac{\sqrt{b^2-4ac}}{2a})(x+\frac{b}{2a}+\frac{\sqrt{b^2-4ac}}{2a}) \end{aligned}\]Therefore, setting \(c=0\),
\[\begin{aligned} a x^2 + bx &= a\left[(x+\frac{b}{2a})^2-\frac{b^2}{4a^2}\right] \\ &= a(x+\frac{b}{2a})^2-\frac{b^2}{4a} \end{aligned}\]So the key thing to remember is that the quadratic term has coefficient
\[\boxed{a}\]and the constant term is
\[\boxed{-\frac{b^2}{4a}}\]And the rationale is that factoring out $a$ gives a monic quadratic, which can be factored into the product of $x$ minus the two roots, and of course the coefficient of $x^2$ remains $a$. Since the two roots are something plus/minus the discriminant over $2a$, the product must contain the negative discriminant squared over $4a^2$, which when multiplied back with $a$ gives the negative discriminant squared over $4a$. And the discriminant squared is just $b^2$ because $c=0$.
We know that
\[\begin{aligned} \int_{-\infty}^{\infty} d^2x\, e^{-x^2} &= \int_0^{2\pi}d\phi \int_0^{\infty}dr\,r \exp(-r^2)\\ &= \frac{-2\pi}{2}\exp(-r^2)|_0^{\infty} \\ &= \pi \\ &= \int_{-\infty}^{\infty} dx\, e^{-x^2}\int_{-\infty}^{\infty} dy\, e^{-y^2} \\ &= \left(\int_{-\infty}^{\infty} dx\, e^{-x^2}\right)^2 \end{aligned}\]Therefore, \(\int_{-\infty}^{\infty} dx\, e^{-x^2} = \sqrt{\pi}\)
and
\[\int_{-\infty}^{\infty} dx\, e^{-ax^2} = \sqrt{\frac{\pi}{a}} .\]Finally,
\[\begin{aligned} \int_{-\infty}^{\infty} dx\, e^{-(a x^2 + bx)} &= \int_{-\infty}^{\infty} dx\, \exp\left[-a(x+\frac{b}{2a})^2+\frac{b^2}{4a}\right] \\ &= \sqrt{\frac{\pi}{a}} \exp\left(\frac{b^2}{4a}\right) \end{aligned}\]So the formula to memorize is
\[\boxed{ \int_{-\infty}^{\infty} dx\, e^{-(a x^2 + bx)} = \sqrt{\frac{\pi}{a}} \exp\left(\frac{b^2}{4a}\right) }\]And just remember that the exponent on the right side (the answer) must be positive because otherwise in the limit $a \to 0+$, the integral should diverge, but a negative exponent would give cause the right side to vanish.