Question

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

• A. \( \dfrac{\pi \ln 2}{4} \)
• B. \( \dfrac{\pi \ln 2}{2} \)
• C. \( \dfrac{\pi \ln 2}{1} \)
• D. \( \dfrac{\pi \ln 4}{3} \)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is an improper integral of a logarithmic function over a rational function. A common technique for integrals of the form \( \int_0^\infty \frac{\ln x}{x^2 + a^2} dx \) is to use the substitution \( x = a \tan \theta \) or \( x = \frac{a^2}{t} \).

Step 2: Key Formula or Approach:
For \( a > 0 \): \[ \int_0^\infty \frac{\ln x}{x^2 + a^2} dx = \frac{\pi \ln a}{2a} \]
Step 3: Detailed Explanation:
1. Let \[ I = \int_{0}^{\infty} \frac{\ln x}{x^2 + 4} dx \] 2. Substitute \( x = \frac{4}{t} \), then \( dx = -\frac{4}{t^2} dt \). 3. Change limits: As \( x \to 0, t \to \infty \); as \( x \to \infty, t \to 0 \). \[ I = \int_{\infty}^{0} \frac{\ln(4/t)}{(4/t)^2 + 4} \left( -\frac{4}{t^2} \right) dt = \int_{0}^{\infty} \frac{\ln 4 - \ln t}{\frac{16 + 4t^2}{t^2}} \left( \frac{4}{t^2} \right) dt \] \[ I = \int_{0}^{\infty} \frac{4(\ln 4 - \ln t)}{4(t^2 + 4)} dt = \int_{0}^{\infty} \frac{\ln 4}{t^2 + 4} dt - \int_{0}^{\infty} \frac{\ln t}{t^2 + 4} dt \] 4. Notice the second integral is \( I \). So, \[ I = \int_{0}^{\infty} \frac{\ln 4}{x^2 + 4} dx - I \] \[ 2I = \ln 4 \int_{0}^{\infty} \frac{1}{x^2 + 4} dx \] \[ 2I = \ln 4 \left[ \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) \right]_0^\infty = \ln 4 \left( \frac{1}{2} \cdot \frac{\pi}{2} \right) = \frac{\pi \ln 4}{4} \] 5. Simplify: \[ I = \frac{\pi \ln 4}{8} = \frac{\pi \ln(2^2)}{8} = \frac{2\pi \ln 2}{8} = \frac{\pi \ln 2}{4} \]
Step 4: Final Answer:
The value of the integral is \( \frac{\pi \ln 2}{4} \).