Question

The value of \( \int_{0}^{\infty} \frac{\ln x}{x^2 + 4} \, dx \) is equal to:

• A. \( \frac{\pi \ln 2}{4} \)
• B. \( \frac{\pi \ln 2}{2} \)
• C. \( \frac{\pi \ln 4}{3} \)
• D. \( \frac{3\pi \ln 2}{4} \)

Hint:

The integral of \( \ln(\tan \theta) \) from 0 to \( \pi/2 \) is a famous zero-integral. Recognizing these "Special Functions" properties saves significant calculation time.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a standard improper integral that can be solved using substitution or complex analysis (residues). A common substitution for integrals of the form \( \frac{\ln x}{x^2 + a^2} \) is \( x = a \tan \theta \).

Step 2: Key Formula or Approach:
Let \( I = \int_{0}^{\infty} \frac{\ln x}{x^2 + a^2} \, dx \). Using the property \( \int_{0}^{\infty} \frac{\ln x}{x^2 + a^2} \, dx = \frac{\pi \ln a}{2a} \).

Step 3: Detailed Explanation:
1. Here \( a^2 = 4 \), so \( a = 2 \). 2. Substitute \( x = 2 \tan \theta \), then \( dx = 2 \sec^2 \theta \, d\theta \). 3. Limits: \( x=0 \to \theta=0 \); \( x \to \infty \to \theta = \pi/2 \). \[ I = \int_{0}^{\pi/2} \frac{\ln(2 \tan \theta)}{4 \sec^2 \theta} (2 \sec^2 \theta) \, d\theta = \frac{1}{2} \int_{0}^{\pi/2} (\ln 2 + \ln \tan \theta) \, d\theta \] 4. Split the integral: \[ I = \frac{1}{2} \left[ \int_{0}^{\pi/2} \ln 2 \, d\theta + \int_{0}^{\pi/2} \ln \tan \theta \, d\theta \right] \] 5. We know \( \int_{0}^{\pi/2} \ln \tan \theta \, d\theta = 0 \). 6. So, \( I = \frac{1}{2} [\ln 2 \cdot \frac{\pi}{2}] = \frac{\pi \ln 2}{4} \).

Step 4: Final Answer:
The value of the integral is \( \frac{\pi \ln 2}{4} \).