Let's compute the integral using the **Residue Theorem**: \[ I = \int_{-\infty}^{\infty} \frac{\cos(4x)}{x^2 + 2x + 2} \, dx \] --- ### **Step 1: Express Cosine as Exponential** Recall: \[ \cos(4x) = \text{Re}(e^{i4x}) \] So, \[ I = \text{Re} \left\{ \int_{-\infty}^{\infty} \frac{e^{i4x}}{x^2 + 2x + 2} \, dx \right\} \] --- ### **Step 2: Identify the Poles** Factor denominator: \[ x^2 + 2x + 2 = (x+1)^2 + 1 \] Set denominator to zero: \[ (x+1)^2 + 1 = 0 \implies (x+1)^2 = -1 \implies x+1 = \pm i \implies x = -1 \pm i \] Poles: **\( x_1 = -1 + i \), \( x_2 = -1 - i \)** --- ### **Step 3: Choose a Contour** Since \(e^{i4x}\) decays in the upper half-plane for large \(x\), close the contour in the **upper half-plane**. Among the poles, only \(x_1 = -1 + i\) is in the upper half-plane. --- ### **Step 4: Compute the Residue at \(x_1\)** For a simple pole: \[ \text{Res}_{x = x_1} \, f(x) = \lim_{x \to x_1} (x - x_1) \frac{e^{i4x}}{(x - x_1)(x - x_2)} = \frac{e^{i4x_1}}{x_1 - x_2} \] Compute \(x_1 - x_2\): \[ x_1 - x_2 = (-1 + i) - (-1 - i) = 2i \] So, \[ \text{Res}_{x = x_1} \, f(x) = \frac{e^{i4x_1}}{2i} \] --- ### **Step 5: Apply the Residue Theorem** \[ \int_{-\infty}^\infty \frac{e^{i4x}}{x^2 + 2x + 2} dx = 2\pi i \cdot \left(\text{Residue at } x_1\right) = 2\pi i \cdot \frac{e^{i4x_1}}{2i} = \pi e^{i4x_1} \] --- ### **Step 6: Take the Real Part** Recall \(I = \text{Re}(\pi e^{i4x_1})\). Compute \(e^{i4x_1}\): \[ x_1 = -1 + i \implies e^{i4x_1} = e^{i4(-1 + i)} = e^{-i4} \cdot e^{-4} \] So, \[ I = \text{Re}(\pi e^{-i4} e^{-4}) = \pi e^{-4} \cos(4) \] --- ## **Final Answer** \[ \boxed{ \int_{-\infty}^{\infty} \frac{\cos(4x)}{x^2 + 2x + 2} \, dx = \pi e^{-4} \cos(4) } \]

Let's carefully compute the integral: \[ I = \int_{-\infty}^{\infty} \frac{\cos(4x)}{x^2 + 2x + 2} \, dx \] using the **Residue Theorem**. Here's a detailed step-by-step process: --- ### **Step 1: Express \(\cos(4x)\) as the real part of an exponential** Recall Euler's formula: \[ \cos(4x) = \Re \left( e^{i4x} \right) \] Thus, \[ I = \Re \left( \int_{-\infty}^{\infty} \frac{e^{i4x}}{x^2 + 2x + 2} \, dx \right) \] Define: \[ J = \int_{-\infty}^{\infty} \frac{e^{i4x}}{x^2 + 2x + 2} \, dx \] so that \[ I = \Re(J) \] --- ### **Step 2: Simplify the denominator** Complete the square: \[ x^2 + 2x + 2 = (x + 1)^2 + 1 \] The poles are where the denominator is zero: \[ (x + 1)^2 + 1 = 0 \implies (x + 1)^2 = -1 \implies x + 1 = \pm i \] So, \[ x = -1 \pm i \] The **poles** are at: \[ x_1 = -1 + i \quad \text{(upper half-plane)} \\ x_2 = -1 - i \quad \text{(lower half-plane)} \] --- ### **Step 3: Choose the contour** Because the integrand involves \(e^{i4x}\), which decays in the **upper half-plane** for large \(\Im(x)\), we close the contour in the **upper half-plane**. Only the pole at: \[ x_1 = -1 + i \] lies inside this contour. --- ### **Step 4: Compute the residue at \(x_1 = -1 + i\)** The integrand: \[ f(x) = \frac{e^{i4x}}{(x+1)^2 + 1} \] has simple poles at \(x = -1 \pm i\). The residue at \(x_1\) is: \[ \operatorname{Res}_{x = x_1} f(x) = \lim_{x \to x_1} (x - x_1) f(x) \] Since the denominator factors as: \[ (x+1)^2 + 1 = (x - x_1)(x - x_2) \] the residue simplifies to: \[ \operatorname{Res}_{x = x_1} f(x) = \frac{e^{i4x_1}}{(x_1 - x_2)} \] because at \(x = x_1\): \[ \lim_{x \to x_1} (x - x_1) \frac{e^{i4x}}{(x - x_1)(x - x_2)} = \frac{e^{i4x_1}}{x_1 - x_2} \] Now, compute \(x_1 - x_2\): \[ x_1 - x_2 = (-1 + i) - (-1 - i) = 2i \] Thus, \[ \operatorname{Res}_{x = x_1} f(x) = \frac{e^{i4x_1}}{2i} \] --- ### **Step 5: Use the Residue Theorem** The integral over the real axis (with the chosen contour) equals: \[ J = 2\pi i \times \text{sum of residues inside the contour} \] since only the pole at \(x_1\) is enclosed, \[ J = 2 \pi i \times \frac{e^{i4x_1}}{2i} = \pi e^{i4x_1} \] --- ### **Step 6: Re-express \(e^{i4x_1}\)** Recall: \[ x_1 = -1 + i \] Then, \[ e^{i4x_1} = e^{i4(-1 + i)} = e^{-i4} \times e^{i4 \times i} = e^{-i4} \times e^{-4} \] because: \[ i \times i = -1 \quad \Rightarrow \quad i4 \times i = -4 \] So, \[ e^{i4x_1} = e^{-i4} \times e^{-4} \] --- ### **Step 7: Write the integral \(J\) in terms of real functions** \[ J = \pi e^{i4x_1} = \pi e^{-4} e^{-i4} \] Recall that: \[ e^{-i4} = \cos(4) - i \sin(4) \] Therefore, \[ J = \pi e^{-4} (\cos 4 - i \sin 4) \] The original integral \(I\) is the **real part** of \(J\): \[ I = \Re(J) = \pi e^{-4} \cos 4 \] --- ### **### **Final Answer:** \[ \boxed{ \int_{-\infty}^{\infty} \frac{\cos(4x)}{x^2 + 2x + 2} \, dx = \pi e^{-4} \cos 4 } \] This completes the detailed step-by-step solution!