Question

The value of the integral $\int_0^\infty \frac{\log_e (x)}{x^2 + 4} dx$ is:

• A. $\frac{\pi \log_e (2)}{2}$
• B. $\frac{\pi \log_e (2)}{4}$
• C. $1 + \pi \log_e (2)$
• D. $2 + \pi \log_e (2)$

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks to evaluate a definite integral involving the natural logarithm in the numerator and a quadratic expression in the denominator over the interval $[0, \infty)$.
Step 2: Key Formula or Approach:
A standard definite integral property is:
\[ \int_0^\infty \frac{\ln x}{x^2 + a^2} dx = \frac{\pi \ln a}{2a} \]
We can also solve this by substituting $x = 2 \tan \theta$.
Step 3: Detailed Explanation:
Let $I = \int_0^\infty \frac{\ln x}{x^2 + 4} dx$.
Substitute $x = 2 \tan \theta$. Then $dx = 2 \sec^2 \theta d\theta$.
When $x = 0, \theta = 0$. When $x \to \infty, \theta \to \frac{\pi}{2}$.
\[ I = \int_0^{\pi/2} \frac{\ln (2 \tan \theta)}{(2 \tan \theta)^2 + 4} (2 \sec^2 \theta) d\theta \]
\[ I = \int_0^{\pi/2} \frac{\ln (2 \tan \theta)}{4 \tan^2 \theta + 4} (2 \sec^2 \theta) d\theta \]
\[ I = \int_0^{\pi/2} \frac{\ln (2 \tan \theta)}{4 \sec^2 \theta} (2 \sec^2 \theta) d\theta \]
\[ I = \frac{1}{2} \int_0^{\pi/2} [\ln 2 + \ln (\tan \theta)] d\theta \]
\[ I = \frac{1}{2} \left[ \int_0^{\pi/2} \ln 2 d\theta + \int_0^{\pi/2} \ln (\tan \theta) d\theta \right] \]
We know that $\int_0^{\pi/2} \ln (\tan \theta) d\theta = 0$ because:
\[ \int_0^{\pi/2} \ln (\tan \theta) d\theta = \int_0^{\pi/2} \ln (\cot \theta) d\theta = -\int_0^{\pi/2} \ln (\tan \theta) d\theta \]
Thus, the integral becomes:
\[ I = \frac{1}{2} [\ln 2 \cdot (\frac{\pi}{2} - 0) + 0] \]
\[ I = \frac{\pi \ln 2}{4} \]
Step 4: Final Answer:
The value of the integral is $\frac{\pi \log_e (2)}{4}$.