Calculate \(\textstyle \int _ { 0 } ^ { \infty } { \frac { d x } { ( 1 + x ^ { 2 } ) ( 1 + 9 x ^ { 2 } ) } }\)
Solution
Set \[ F(z)=\frac1{(1+z^2)(1+9z^2)}. \] For \(R>1\), let \(C_R\) be the positively oriented contour consisting of the interval \([-R,R]\) and the upper semicircle \[ \Gamma_R=\{Re^{i\theta}:0\leq\theta\leq\pi\}. \]
1. The only poles of \(F\) inside \(C_R\) are\[ z=i \qquad\text{and}\qquad z=\frac i3, \]and\[ \Res(F;i) = \frac{i}{16}, \qquad \Res\left(F;\frac i3\right) = -\frac{3i}{16}. \]
The poles are the zeros of \[ (1+z^2)(1+9z^2), \] namely \[ \pm i, \qquad \pm\frac i3. \] The two with positive imaginary part lie inside \(C_R\) for \(R>1\).
At \(z=i\), \[ \begin{aligned} \Res(F;i) &= \frac1{(2i)(1+9i^2)}\\ &= \frac1{-16i} = \frac{i}{16}. \end{aligned} \] At \(z=i/3\), using \[ \frac{d}{dz}(1+9z^2)=18z, \] one gets \[ \begin{aligned} \Res\left(F;\frac i3\right) &= \frac1{ \left(1+(i/3)^2\right) 18(i/3) }\\ &= \frac1{(8/9)(6i)}\\ &= -\frac{3i}{16}. \end{aligned} \]
2. The integral of \(F\) over the semicircular arc tends to zero:\[ \int_{\Gamma_R}F(z)\,dz\longrightarrow0 \]as \(R\to\infty\).
On \(\Gamma_R\), \[ \abs{1+z^2}\geq R^2-1 \] and \[ \abs{1+9z^2}\geq9R^2-1. \] Hence \[ \abs{F(z)} \leq \frac1{(R^2-1)(9R^2-1)}. \] The arc has length \(\pi R\), so the ML estimate gives \[ \abs{ \int_{\Gamma_R}F(z)\,dz } \leq \frac{\pi R}{(R^2-1)(9R^2-1)} \longrightarrow0. \]
3. The integral over the whole real line is\[ \int_{-\infty}^{\infty} \frac{dx}{(1+x^2)(1+9x^2)} = \frac{\pi}{4}. \]
By the residue theorem and step 1, \[ \begin{aligned} \int_{C_R}F(z)\,dz &= 2\pi i \left( \frac{i}{16} -\frac{3i}{16} \right)\\ &= 2\pi i\left(-\frac{i}{8}\right)\\ &= \frac{\pi}{4}. \end{aligned} \] Splitting the contour integral, \[ \int_{-R}^{R}F(x)\,dx + \int_{\Gamma_R}F(z)\,dz = \frac{\pi}{4}. \] Letting \(R\to\infty\) and using step 2 gives the displayed real-line integral.
4. The requested integral is\[ \boxed{ \int_0^{\infty} \frac{dx}{(1+x^2)(1+9x^2)} = \frac{\pi}{8}. } \]
The integrand is even. Therefore step 3 gives \[ 2\int_0^{\infty} \frac{dx}{(1+x^2)(1+9x^2)} = \frac{\pi}{4}. \] Divide by \(2\).
5. Q.E.D.
Step 4 is the requested value.