Improper integral $\int_{-\infty}^{+\infty}\frac{1}{a x^2 + bx + c} dx$, $a > 0,a x^2 + bx + c>0$

Question

I have a question about improper integral. If you can help me , I appreciate that.

If a > 0 and the graph of $y=a x^2 + bx + c$ lies entirely above the x-axis, show that $$ \int_{-\infty}^{+\infty} \frac{dx}{a x^2 + bx + c}=\frac{2 \pi}{\sqrt{4ac - b^2}}. $$

Answer 1

Note that if $ax^2+bx+c$ has real roots then the integral doesn't exists, since we will end up with an integral of the form $$\dfrac1{a}\int_{-\infty}^{\infty} \dfrac{dx}{(x-\alpha)(x-\beta)}$$where $\alpha, \beta \in \mathbb{R}$, which clearly diverges. Hence, we can assume that the discriminant is strictly negative, i.e., $4ac -b^2 > 0$.

Given that $4ac-b^2>0$, we have $ax^2+bx+c = a\left(x+\dfrac{b}{2a}\right)^2 + c - \dfrac{b^2}{4a}$. Hence, letting $y=x+\dfrac{b}{2a}$, we have \begin{align} \int_{-\infty}^{\infty} \dfrac{dx}{ax^2+bx+c} & = \int_{-\infty}^{\infty} \dfrac{dy}{ay^2 + c-\dfrac{b^2}{4a}} = \dfrac1{a} \dfrac1{\sqrt{c/a-b^2/(4a^2)}}\left. \arctan\left(\dfrac{y}{\sqrt{c/a-b^2/(4a^2)}}\right)\right \vert_{-\infty}^{\infty}\\ & = \dfrac{2\pi}{\sqrt{4ac-b^2}} \end{align}