Today, we will be evaluating the integral $$ \int_{-\infty}^{\infty} e^{ax^2+bx+c} dx \hspace{1cm} a \in \mathbb{R^{-}}, b \in \mathbb{R}, c \in \mathbb{R} $$ The restriction of \(a\) to \(\mathbb{R^{-}}\) is required for the convergence of this integral. My thought process for this integral was initially to minipulate the integrand in such a way that we can employ the result of the Gaussian integral, namely, \(\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}\) . To do this, we complete the square on \(f(x)=ax^2+bx+c\) : $$ y=ax^2+bx+c \implies y-c=ax^2+bx=a(x^2+\frac{b}{a}x) \implies \frac{y-c}{a}=x^2+\frac{b}{a}x $$ Further, $$ \frac{y-c}{a}+\frac{b^2}{4a^2}=x^2+\frac{b}{a}x+\frac{b^2}{4a^2} \implies \frac{y-c}{a}=(x+\frac{b}{2a})^2-\frac{b^2}{4a^2} $$ Lastly, $$ y=a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c $$ Thus we have that $$ I=\int_{-\infty}^{\infty} e^{ax^2+bx+c} dx=\int_{-\infty}^{\infty} e^{a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c} dx=e^{-\frac{b^2}{4a}+c} \int_{-\infty}^{\infty} e^{a(x+\frac{b}{2a})^2} dx $$ Given that this integrand spans all of \(\mathbb{R}\) we can employ the substitution \(t=x+\frac{b}{2a}\) and our bounds will remain the same. Thus, $$ I=e^{-\frac{b^2}{4a}+c} \int_{-\infty}^{\infty} e^{at^2} dt $$ and since \(a\) is negative, we can write $$ I=e^{-\frac{b^2}{4a}+c} \int_{-\infty}^{\infty} e^{-(\sqrt{-a}t)^2} dt $$ Now letting \(v=\sqrt{-a}t\) we have that \(dv=\sqrt{-a} dt \implies \frac{dv}{\sqrt{-a}}=dt\) and thus $$ I=e^{-\frac{b^2}{4a}+c} \int_{-\infty}^{\infty} e^{-(v)^2} \frac{dv}{\sqrt{-a}} \implies I=\frac{e^{-\frac{b^2}{4a}+c}}{\sqrt{-a}} \sqrt{\pi} = e^{-\frac{b^2}{4a}+c} \sqrt{-\frac{\pi}{a}} $$ and we’re done.