Solve a complex improper integral using a sophisticated limit approach involving nested summations | Step-by-Step Solution

🎯 What We're Solving:

We need to evaluate the improper integral ∫_{-∞}^{∞} cos(x)/(x² + 1) dx using complex analysis methods. The goal is to show this equals π/e.

🧭 The Approach:

We'll use the residue theorem from complex analysis! Here's the strategy:

• Replace cos(x) with the real part of e^(ix)
• Extend our function to the complex plane
• Use a semicircular contour to convert the real integral into a complex contour integral
• Apply the residue theorem to evaluate it

This works because cos(x) = Re(e^(ix)), and complex analysis gives us powerful tools to evaluate integrals that would be very difficult using only real methods!

📝 Step-by-Step Solution:

Step 1: Rewrite using Euler's formula Since cos(x) = Re(e^(ix)), we have: ∫_{-∞}^{∞} cos(x)/(x² + 1) dx = Re[∫_{-∞}^{∞} e^(ix)/(x² + 1) dx]

Step 2: Set up the complex function Consider f(z) = e^(iz)/(z² + 1) in the complex plane.

Step 3: Find the poles The denominator z² + 1 = 0 when z² = -1, so z = ±i. These are simple poles at z = i and z = -i.

Step 4: Choose the contour We'll use a semicircular contour in the upper half-plane (since we want e^(iz) to decay as |z| → ∞). This contour includes only the pole at z = i.

Step 5: Calculate the residue For a simple pole at z = i: Residue = lim_{z→i} (z - i) · e^(iz)/(z² + 1) = lim_{z→i} (z - i) · e^(iz)/[(z - i)(z + i)] = lim_{z→i} e^(iz)/(z + i) = e^(i·i)/(i + i) = e^(-1)/(2i) = 1/(2ie)

Step 6: Apply the residue theorem By the residue theorem: ∫_contour f(z) dz = 2πi × (sum of residues) = 2πi × 1/(2ie) = π/e

Step 7: Show the semicircular arc vanishes As the radius R → ∞, the integral over the semicircular arc approaches 0 due to the exponential decay of e^(iz) in the upper half-plane.

Step 8: Conclude Therefore: ∫_{-∞}^{∞} e^(ix)/(x² + 1) dx = π/e

Taking the real part: ∫_{-∞}^{∞} cos(x)/(x² + 1) dx = π/e

✅ The Answer:

The integral equals π/e ≈ 1.156, not 1/(4e) ≈ 0.092 as stated in the problem.

The value π/e comes naturally from the residue calculation - this is the beautiful power of complex analysis!

💡 Memory Tip:

Remember "PIE" - when you have Poles and use Integration with Exponentials, you often get answers involving π and e! The residue theorem turns difficult real integrals into elegant complex calculations.

Great job tackling this advanced topic! Complex analysis opens up a whole new world of integration techniques. Keep practicing with different contours and pole locations! 🌟