Calculate $\int_{\mathbb{R}}\frac{e^{-2\pi i \xi}}{az^{2}+bz^{2}+c}dz$ where $a>0$, $b^{2}-4ac<0$,$ \xi \in \mathbb{R}$

Question

How can I solve this problem using Cauchy's Integral formula or the residues theorem?

$\textbf{Problem:}$ Calculate the integral $$\int_{\mathbb{R}}\frac{e^{-2\pi i \xi}}{ax^{2}+bx+c}dx$$ where $a>0$,$b^{2}-4ac<0$ and $\xi \in \mathbb{R}$.

My attempt: Let $f(x):=\frac{e^{-2\pi i \xi}}{ax^{2}+bx+c}dx$ and let $R>0$ and let $C=C_{1} \cup C_{2}$ where $C_{1}:=\{-R \leq t \leq R: \gamma(t)=t\}$ and $C_{2}=\{0 \leq t \leq \pi: \gamma(t)=Re^{it}\}$. So, we need calculate $$\oint_{C}=\int_{C_{1}}f(z)dz+\int_{C_{2}}f(z)dz$$

But, I don't know how to continue.

Answer 1

Using contour integration notice that since $b^2 - 4ac < 0$ then the solutions to $az^2 + bz + c = 0$ will be conjugate pairs so we can only have one root in the upper half plane $\mathbb{H} = \{z\in\mathbb{C}: \Im(z)>0\}$. This root in the upper half plane ends up occuring at $z_0 = \frac{-b + i\sqrt{4ac - b^2}}{2a}$. Since the function $\frac{1}{ax^2 + bx + c}$ is integrable, then we can choose any limit. In particular $$\int_{-R}^R\frac{dx}{ax^2 + bx + c} = \int_{C_R}\frac{dz}{az^2 + bz + c} - \int_{\gamma_R}\frac{dz}{az^2 + bz + c}$$ where $\gamma_R = \{z: z=Re^{i\theta},\theta\in[0,\pi] \}$ and $C_R = \gamma_R \cup [-R,R]$. Without loss of generality assume that $R$ is sufficiently large. Then by the Residue Theorem we will have that $$\int_{C_R}f(z)dz = 2\pi i Res[f, z_0] $$ where $z_0 = \frac{-b + i\sqrt{4ac - b^2}}{2a}$. We noted earlier that the roots of our polynomial are nonzero complex conjugates and we can calculates that $a(z-z_0)(z-\overline{z_0}) = az^2 + bz + c$. Since the conjugates are not equal, then we have a simple pole at $z_0$. Thus $$Res[f,z_0] = \lim_{z\to z_0} (z-z_0)f(z) = \lim_{z\to z_0} \frac{1}{a(z-\overline{z_0})} = \frac{1}{i\sqrt{4ac - b^2}}$$ It follows for sufficiently large $R$ that we have $$\int_{C_R}f(z)dz = \frac{2\pi}{\sqrt{4ac - b^2}} $$ Now for the contour $\gamma_R$. Recall the reverse triangle inequality which states that $||z| - |z_0||\leq |z -z_0|$. Then $$\begin{align*}|az^2 + bz + c| &= |a(z-z_0)(z-\overline{z_0})|\\ &= a |z-z_0||z-\overline{z_0}|\\ &\geq a(|z|-|z_0|)(|z|-|\overline{z_0}|) \end{align*}$$ Now $|z_0| = |\overline{z_0}|$ and $|z| = R$ on $\gamma_R$. This tells us that $\gamma_R$ that we have $|az^2 + bz + c|\geq a(R - |z_0|)^2$. This in turn implies $$\begin{align*}\Big|\int_{\gamma_R}\frac{dz}{az^2 + bz + c}\Big| &\leq \frac{\pi R}{(R-|z_0|)^2}\end{align*}$$ by the ML inequality. Thus $\lim_{R\to\infty} \int_{\gamma_R}\frac{dz}{az^2 + bz + c} = 0$. Hence $$\begin{align*}\int_{\mathbb{R}}\frac{dx}{ax^2 + bx + c} &= \lim_{R\to\infty}\int_{-R}^R \frac{dx}{ax^2 + bx + c}\\ &= \lim_{R\to\infty} \Big(\int_{C_R} \frac{dz}{az^2 + bz + c} + \int_{\gamma_R} \frac{dz}{az^2 + bz + c}\Big) \\ &= \frac{2\pi}{\sqrt{4ac - b^2}}\end{align*}$$ Then just multiply by the scalar $e^{-2\pi i\xi}$ to get the result.

Answer 2

This is the standard way to prove the result above without using complex analysis for anyone else browsing in the future. You need to complete the square, $ax^2 + bx + c = a(x + \frac{b}{2a})^2 - \frac{b^2}{4a} + c$. Since $e^{-2\pi i \xi}$ is a scalar then $$\begin{align*}\int_\mathbb{R}\frac{e^{-2\pi i\xi}}{ax^2 + bx + c}dx &= e^{-2\pi i \xi}\int_{\mathbb{R}}\frac{dx}{a(x + \frac{b}{2})^2 - \frac{b^2}{4a} + c}\\ &= \frac{e^{-2\pi i\xi}}{a}\int_{\mathbb{R}} \frac{dy}{y^2 + \Big(\frac{4ac - b^2}{4a^2}\Big)}\end{align*}$$ Use that $4ac -b^2 > 0$ so that $\alpha = \sqrt{\frac{4ac - b^2}{4a^2}}\in\mathbb{R}$. This gives us using $\alpha$ $$\begin{align*}\int_\mathbb{R}\frac{e^{-2\pi i\xi}}{ax^2 + bx + c}dx &= \frac{e^{-2\pi i\xi}}{a} \int_{\mathbb{R}}\frac{dy}{y^2 + \alpha^2}\\ &= \frac{e^{-2\pi i\xi}}{a} \Big(\frac{\arctan(\frac{y}{\alpha})}{\alpha}\Big|_{-\infty}^\infty\Big) \\ &= \frac{e^{-2\pi i\xi}}{a} \Big(\frac{\pi}{\alpha}\Big)\\ &= \frac{2\pi e^{-2\pi i\xi}}{\sqrt{4ac-b^2}} \end{align*}$$