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:
We are required to evaluate an improper definite integral containing a logarithmic function and a quadratic term in the denominator.
Step 2: Key Formula or Approach:
The denominator $x^2+4$ suggests a trigonometric substitution involving tangent. Let $x = 2\tan\theta$. This will transform the algebraic integrand into a trigonometric one, which may be simpler to evaluate. We will also use the standard integral result $\int_0^{\pi/2} \ln(\tan\theta)d\theta = 0$.
Step 3: Detailed Explanation:
Let the integral be $I = \int_0^\infty \frac{\ln x}{x^2 + 4} dx$.
We use the substitution $x = 2 \tan \theta$.
Then $dx = 2 \sec^2 \theta \, d\theta$.
The limits of integration also change:
- When $x = 0$, $2 \tan \theta = 0 \implies \theta = 0$.
- When $x \to \infty$, $2 \tan \theta \to \infty \implies \theta \to \frac{\pi}{2}$.
Substituting into the integral:
\[ I = \int_0^{\pi/2} \frac{\ln(2 \tan \theta)}{(2 \tan \theta)^2 + 4} (2 \sec^2 \theta) d\theta \] \[ I = \int_0^{\pi/2} \frac{\ln(2 \tan \theta)}{4(\tan^2 \theta + 1)} (2 \sec^2 \theta) d\theta \] Using the identity $1 + \tan^2 \theta = \sec^2 \theta$:
\[ I = \int_0^{\pi/2} \frac{\ln(2 \tan \theta)}{4 \sec^2 \theta} (2 \sec^2 \theta) d\theta = \int_0^{\pi/2} \frac{1}{2} \ln(2 \tan \theta) d\theta \] Using the logarithm property $\ln(ab) = \ln a + \ln b$:
\[ I = \frac{1}{2} \int_0^{\pi/2} (\ln 2 + \ln(\tan \theta)) d\theta \] \[ I = \frac{1}{2} \left[ \int_0^{\pi/2} \ln 2 \, d\theta + \int_0^{\pi/2} \ln(\tan \theta) \, d\theta \right] \] The first part is simple: $\int_0^{\pi/2} \ln 2 \, d\theta = \ln 2 \cdot [\theta]_0^{\pi/2} = \frac{\pi}{2}\ln 2$.
For the second part, let $J = \int_0^{\pi/2} \ln(\tan \theta) \, d\theta$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$:
$J = \int_0^{\pi/2} \ln(\tan(\frac{\pi}{2}-\theta)) d\theta = \int_0^{\pi/2} \ln(\cot \theta) d\theta = \int_0^{\pi/2} -\ln(\tan \theta) d\theta = -J$.
$J = -J \implies 2J = 0 \implies J = 0$.
Substituting back into the expression for $I$:
\[ I = \frac{1}{2} \left[ \frac{\pi}{2} \ln 2 + 0 \right] = \frac{\pi \ln 2}{4} \] Step 4: Final Answer:
The value of the integral is $\frac{\pi \log_e (2)}{4}$.