The value of `int_(0)^(oo) (logx)/(1+x^(2))dx`, is

Options

A. `(pi)/(4)`
B. `(pi)/(2)`
C. 0
D. none of these

Explanation

To solve the integral

\[ I = \int_{0}^{\infty} \frac{\log x}{1 + x^2} \, dx, \]

we can use a substitution and properties of definite integrals. Here’s a step-by-step solution:

### Step 1: Substitution
We will use the substitution \( x = \tan \theta \). This gives us:

\[ dx = \sec^2 \theta \, d\theta. \]

### Step 2: Change of Limits
When \( x = 0 \), \( \theta = 0 \). When \( x \to \infty \), \( \theta \to \frac{\pi}{2} \). Thus, the limits change from \( 0 \) to \( \frac{\pi}{2} \).

### Step 3: Rewrite the Integral
Substituting \( x = \tan \theta \) into the integral, we have:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\log(\tan \theta)}{1 + \tan^2 \theta} \sec^2 \theta \, d\theta. \]

Since \( 1 + \tan^2 \theta = \sec^2 \theta \), we can simplify the integral:

\[ I = \int_{0}^{\frac{\pi}{2}} \log(\tan \theta) \, d\theta. \]

### Step 4: Use Symmetry
We can use the property of integrals:

\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx. \]

Let’s consider:

\[ I = \int_{0}^{\frac{\pi}{2}} \log(\tan \theta) \, d\theta. \]

Now, we can change the variable in the integral:

\[ J = \int_{0}^{\frac{\pi}{2}} \log(\cot \theta) \, d\theta. \]

Using the identity \( \cot \theta = \frac{1}{\tan \theta} \):

\[ J = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1}{\tan \theta}\right) \, d\theta = \int_{0}^{\frac{\pi}{2}} -\log(\tan \theta) \, d\theta = -I. \]

### Step 5: Combine Integrals
Now, we can add \( I \) and \( J \):

\[ I + J = \int_{0}^{\frac{\pi}{2}} \log(\tan \theta) \, d\theta + \int_{0}^{\frac{\pi}{2}} \log(\cot \theta) \, d\theta = \int_{0}^{\frac{\pi}{2}} \log(\tan \theta \cdot \cot \theta) \, d\theta. \]

Since \( \tan \theta \cdot \cot \theta = 1 \):

\[ I + J = \int_{0}^{\frac{\pi}{2}} \log(1) \, d\theta = 0. \]

### Step 6: Solve for \( I \)
Since \( J = -I \), we have:

\[ I - I = 0 \implies 2I = 0 \implies I = 0. \]

### Final Answer
Thus, the value of the integral is

\[ \int_{0}^{\infty} \frac{\log x}{1 + x^2} \, dx = 0. \]

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