Let \(a > 0\) . Evaluate \(\begin{array} { r } { \int _ { 0 } ^ { \infty } \frac { x ^ { 2 } } { ( x ^ { 2 } + a ^ { 2 } ) ^ { 2 } } d x } \end{array}\)
Solution
Set \[ F(z)=\frac{z^2}{(z^2+a^2)^2}. \] For \(R>a\), let \(C_R\) be the positively oriented contour consisting of \([-R,R]\) and the upper semicircle of radius \(R\).
1. The only pole of \(F\) inside \(C_R\) is the double pole at \(z=ia\), and\[ \Res(F;ia) = -\frac{i}{4a}. \]
Factor \[ (z^2+a^2)^2 = (z-ia)^2(z+ia)^2. \] Thus \[ (z-ia)^2F(z) = \frac{z^2}{(z+ia)^2}. \] For a double pole, \[ \Res(F;ia) = \left. \frac{d}{dz} \left( \frac{z^2}{(z+ia)^2} \right) \right|_{z=ia}. \] Differentiating, \[ \frac{d}{dz} \left( \frac{z^2}{(z+ia)^2} \right) = \frac{2z}{(z+ia)^2} -\frac{2z^2}{(z+ia)^3}. \] At \(z=ia\) this becomes \[ \begin{aligned} \Res(F;ia) &= \frac{2ia}{(2ia)^2} -\frac{2(ia)^2}{(2ia)^3}\\ &= -\frac{i}{2a} +\frac{i}{4a}\\ &= -\frac{i}{4a}. \end{aligned} \]
2. The integral over the upper semicircular arc tends to zero as\(R\to\infty\).
On \(\abs{z}=R\), \[ \abs{z^2+a^2} \geq R^2-a^2. \] Hence \[ \abs{F(z)} \leq \frac{R^2}{(R^2-a^2)^2}. \] The arc has length \(\pi R\), so \[ \abs{ \int_{\text{arc}}F(z)\,dz } \leq \frac{\pi R^3}{(R^2-a^2)^2} \longrightarrow0. \]
3. The whole-line integral is\[ \int_{-\infty}^{\infty} \frac{x^2}{(x^2+a^2)^2}\,dx = \frac{\pi}{2a}. \]
By the residue theorem and step 1, \[ \int_{C_R}F(z)\,dz = 2\pi i \left(-\frac{i}{4a}\right) = \frac{\pi}{2a}. \] Split the contour into the real segment and arc, let \(R\to\infty\), and use step 2.
4. The requested value is\[ \boxed{ \int_0^{\infty} \frac{x^2}{(x^2+a^2)^2}\,dx = \frac{\pi}{4a}. } \]
The integrand is even, so step 3 is twice the half-line integral.
5. Q.E.D.
Step 4 is the requested evaluation.