Let \(a > 0\) . Evaluate \(\textstyle \int _ { 0 } ^ { \infty } { \frac { \cos x } { ( x ^ { 2 } + a ^ { 2 } ) ^ { 2 } } } d x\)
Solution
Set \[ F(z)=\frac{e^{iz}}{(z^2+a^2)^2}. \] For \(R>a\), integrate \(F\) over the positively oriented upper semicircle with diameter \([-R,R]\).
1. The only pole of \(F\) inside the contour is the double pole at\(z=ia\), and\[ \Res(F;ia) = -\frac{i e^{-a}(a+1)}{4a^3}. \]
Factor \[ (z^2+a^2)^2 = (z-ia)^2(z+ia)^2. \] For the double pole at \(ia\), \[ \Res(F;ia) = \left. \frac{d}{dz} \left( \frac{e^{iz}}{(z+ia)^2} \right) \right|_{z=ia}. \] Differentiating gives \[ \frac{d}{dz} \left( \frac{e^{iz}}{(z+ia)^2} \right) = \frac{i e^{iz}}{(z+ia)^2} -\frac{2e^{iz}}{(z+ia)^3}. \] At \(z=ia\), \[ \begin{aligned} \Res(F;ia) &= \frac{i e^{-a}}{(2ia)^2} -\frac{2e^{-a}}{(2ia)^3}\\ &= -\frac{i e^{-a}}{4a^2} -\frac{i e^{-a}}{4a^3}\\ &= -\frac{i e^{-a}(a+1)}{4a^3}. \end{aligned} \]
2. The integral over the upper semicircular arc tends to zero as\(R\to\infty\).
For \(z\) in the upper half-plane, \[ \abs{e^{iz}}\leq1. \] On \(\abs{z}=R\), \[ \abs{z^2+a^2} \geq R^2-a^2. \] Therefore \[ \abs{F(z)} \leq \frac1{(R^2-a^2)^2}. \] The arc has length \(\pi R\), so the ML estimate gives \[ \abs{ \int_{\text{arc}}F(z)\,dz } \leq \frac{\pi R}{(R^2-a^2)^2} \longrightarrow0. \]
3. One has\[ \int_{-\infty}^{\infty} \frac{e^{ix}}{(x^2+a^2)^2}\,dx = \frac{\pi e^{-a}(a+1)}{2a^3}. \]
By the residue theorem and step 1, \[ \begin{aligned} \int_{C_R}F(z)\,dz &= 2\pi i\Res(F;ia)\\ &= 2\pi i \left( -\frac{i e^{-a}(a+1)}{4a^3} \right)\\ &= \frac{\pi e^{-a}(a+1)}{2a^3}. \end{aligned} \] The real-axis integrand is absolutely integrable because its modulus is \((x^2+a^2)^{-2}\). Split the contour into the real segment and the arc, let \(R\to\infty\), and use step 2.
4. The whole-line cosine integral is\[ \int_{-\infty}^{\infty} \frac{\cos x}{(x^2+a^2)^2}\,dx = \frac{\pi e^{-a}(a+1)}{2a^3}. \]
Take real parts in step 3.
5. The requested value is\[ \boxed{ \int_0^{\infty} \frac{\cos x}{(x^2+a^2)^2}\,dx = \frac{\pi e^{-a}(a+1)}{4a^3}. } \]
The integrand in step 4 is even, so its whole-line integral is twice the half-line integral.
6. Q.E.D.
Step 5 is the requested evaluation.