Use residues to compute the integral \[\begin{align*} \int_{0}^{\infty} \dfrac{\cos x}{(x^2+1)^2} \mathrm{d}x \end{align*}\]
Solution
Goal: Compute the definite integral \(I = \int_{0}^{\infty} \frac{\cos x}{(x^2+1)^2} \, dx\) using the calculus of residues.
Step 1: Parity and Complex Extension
1. Symmetry of the integrand over the real line.
1.1. The integrand \(g(x) = \frac{\cos x}{(x^2+1)^2}\) is an even function: \(g(-x) = \frac{\cos(-x)}{((-x)^2+1)^2} = \frac{\cos x}{(x^2+1)^2} = g(x)\).Proof: \(\cos(-x) = \cos x\) and \((-x)^2 = x^2\).
1.2. Therefore, \(I = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\cos x}{(x^2+1)^2} \, dx = \frac{1}{2} \text{Re}\left( \int_{-\infty}^{\infty} \frac{e^{ix}}{(x^2+1)^2} \, dx \right)\).Proof: \(e^{ix} = \cos x + i \sin x\), and \(\frac{\sin x}{(x^2+1)^2}\) is an odd function whose integral over \((-\infty, \infty)\) vanishes.
1.3. Define \(f(z) = \frac{e^{iz}}{(z^2+1)^2} = \frac{e^{iz}}{(z-i)^2(z+i)^2}\).Proof: Factoring \(z^2+1 = (z-i)(z+i)\).
1.4. Q.E.D.
Step 2: Set up Contour Integration in the Upper Half-Plane
2. Let \(\Gamma_R = [-R, R] \cup C_R\) where \(C_R = \{R e^{i\theta} : \theta \in [0, \pi]\}\) with \(R > 1\).
2.1. \(\Gamma_R\) is a simple closed positively oriented contour enclosing the upper half-plane disk region.Proof: Standard semicircular contour.
2.2. The only singularity of \(f(z)\) in the upper half-plane \(\mathbb{H}\) is at \(z = i\), which is a pole of order 2. Proof: The zeros of \((z^2+1)^2\) are \(z = \pm i\).Only \(z = i\) has \(\text{Im}(z) > 0\).
2.3. By the Cauchy Residue Theorem: \[\oint_{\Gamma_R} f(z) \, dz = \int_{-R}^R \frac{e^{ix}}{(x^2+1)^2} \, dx + \int_{C_R} f(z) \, dz = 2\pi i \, \text{Res}(f, i).\] Proof: Application of the residue theorem to the single enclosed pole \(z = i\).
2.4. Q.E.D.
Step 3: Compute the Residue at \(z = i\)
3. \(\text{Res}(f, i) = \frac{1}{4ie}\).
3.1. For a pole of order 2, the residue formula is: \[\text{Res}(f, i) = \lim_{z \to i} \frac{d}{dz} \left[ (z-i)^2 f(z) \right] = \lim_{z \to i} \frac{d}{dz} \left[ \frac{e^{iz}}{(z+i)^2} \right].\] Proof: Standard residue formula for order 2 poles.
3.2. Compute the derivative using the quotient rule: \[\frac{d}{dz}\left( \frac{e^{iz}}{(z+i)^2} \right) = \frac{i e^{iz}(z+i)^2 - e^{iz} \cdot 2(z+i)}{(z+i)^4} = \frac{e^{iz} \big( i(z+i) - 2 \big)}{(z+i)^3}.\] Proof: Direct differentiation.
3.3. Evaluate at \(z = i\): \[\text{Res}(f, i) = \frac{e^{i(i)} \big( i(2i) - 2 \big)}{(2i)^3} = \frac{e^{-1} (-2 - 2)}{-8i} = \frac{-4 e^{-1}}{-8i} = \frac{1}{2i e} = -\frac{i}{2e}.\] Proof: Arithmetic evaluation: \(i^2 = -1\), \((2i)^3 = -8i\).
3.4. Therefore, \(2\pi i \, \text{Res}(f, i) = 2\pi i \left( -\frac{i}{2e} \right) = \frac{2\pi}{2e} = \frac{\pi}{e}\).Proof: \(i(-i) = 1\).
3.5. Q.E.D.
Step 4: Vanishing of the Arc Integral
4. \(\lim_{R \to \infty} \int_{C_R} f(z) \, dz = 0\).
4.1. On \(C_R\), \(z = R e^{i\theta}\) with \(\theta \in [0, \pi]\), so \(\text{Im}(z) = R \sin\theta \geq 0\).Proof: \(\sin\theta \geq 0\) for \(\theta \in [0, \pi]\).
4.2. Thus \(|e^{iz}| = e^{-\text{Im}(z)} = e^{-R\sin\theta} \leq 1\).Proof: Exponential of non-positive real number.
4.3. By the reverse triangle inequality, \(|z^2+1| \geq |z|^2 - 1 = R^2 - 1\), so \(|(z^2+1)^2| \geq (R^2-1)^2\).Proof: Reverse triangle inequality for \(R > 1\).
4.4. The integrand is bounded by \(|f(z)| \leq \frac{1}{(R^2-1)^2}\) on \(C_R\), and the length of \(C_R\) is \(\pi R\).Proof: Definition of arc length and supremum bound.
4.5. By the \(ML\)-inequality: \[\left| \int_{C_R} f(z) \, dz \right| \leq \frac{\pi R}{(R^2-1)^2} \to 0 \quad \text{as } R \to \infty.\] Proof: The degree of the denominator (\(R^4\)) exceeds the numerator (\(R\)) by 3.
4.6. Q.E.D.
Step 5: Final Evaluation
5. Compute the real integral \(I\).
5.1. Taking the limit as \(R \to \infty\) in <1>2.<2>3: \[\int_{-\infty}^\infty \frac{e^{ix}}{(x^2+1)^2} \, dx = 2\pi i \, \text{Res}(f, i) - \lim_{R\to\infty} \int_{C_R} f(z)\,dz = \frac{\pi}{e} - 0 = \frac{\pi}{e}.\] Proof: Follows from <1>3.<2>4 and <1>4.<2>5.
5.2. Therefore, from <1>1.<2>2: \[I = \int_0^\infty \frac{\cos x}{(x^2+1)^2} \, dx = \frac{1}{2} \text{Re}\left( \frac{\pi}{e} \right) = \frac{\pi}{2e}.\] Proof: Halving the bilateral integral.
5.3. Q.E.D.