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 will follow these steps:

### Step 1: Substitute \( x = \frac{1}{y} \)
Let’s perform the substitution \( x = \frac{1}{y} \). Then, we have:
\[ dx = -\frac{1}{y^2} \, dy \]
Now, we need to change the limits of integration. When \( x \to 0 \), \( y \to \infty \) and when \( x \to \infty \), \( y \to 0 \). Thus, the integral becomes:
\[ I = \int_{\infty}^{0} \frac{\log\left(\frac{1}{y}\right)}{1 + \left(\frac{1}{y}\right)^2} \left(-\frac{1}{y^2}\right) \, dy \]

### Step 2: Simplify the Integral
Now, we simplify the integrand:
\[ \log\left(\frac{1}{y}\right) = -\log y \]
And,
\[ 1 + \left(\frac{1}{y}\right)^2 = 1 + \frac{1}{y^2} = \frac{y^2 + 1}{y^2} \]
Thus, the integral becomes:
\[ I = \int_{\infty}^{0} \frac{-\log y}{\frac{y^2 + 1}{y^2}} \left(-\frac{1}{y^2}\right) \, dy = \int_{\infty}^{0} \frac{\log y}{1 + \frac{1}{y^2}} \, dy \]
Reversing the limits gives:
\[ I = \int_{0}^{\infty} \frac{\log y}{1 + y^2} \, dy \]

### Step 3: Combine the Two Integrals
Now we have:
\[ I = \int_{0}^{\infty} \frac{\log x}{1 + x^2} \, dx \]
and
\[ I = \int_{0}^{\infty} \frac{\log y}{1 + y^2} \, dy \]
Thus, we can write:
\[ 2I = \int_{0}^{\infty} \frac{\log x}{1 + x^2} \, dx + \int_{0}^{\infty} \frac{\log y}{1 + y^2} \, dy = \int_{0}^{\infty} \frac{\log x + \log y}{1 + x^2} \, dx \]
Since \( \log x + \log y = \log(xy) \), we can express the integral as:
\[ 2I = \int_{0}^{\infty} \frac{\log(x^2)}{1 + x^2} \, dx \]
This simplifies to:
\[ 2I = 2 \int_{0}^{\infty} \frac{\log x}{1 + x^2} \, dx \]
Thus, we have:
\[ I = 0 \]

### Final Result
The value of the integral is:
\[ \boxed{0} \]