Step 1: Understanding the Concept:
For a function $f(x)$ to be a valid Probability Density Function (p.d.f.), the total area under its curve must equal 1, i.e., $\int_{-\infty}^{\infty} f(x) dx = 1$. We utilize this property to determine the constant $k$. The Cumulative Distribution Function (c.d.f.), denoted $F(x)$, is then calculated by integrating the p.d.f. from $-\infty$ up to $x$.
Step 2: Key Formula or Approach:
1. Normalization property of p.d.f.: $\int_{-\infty}^{\infty} f(x) dx = 1$.
2. Definition of c.d.f.: $F(x) = \int_{-\infty}^{x} f(t) dt$.
3. Standard integration formula: $\int \frac{1}{x^2+1} dx = \tan^{-1} x + C$.
Step 3: Detailed Explanation:
First, find the constant $k$ using the property that the total probability must be 1:
\[ \int_{-\infty}^{\infty} f(x) dx = 1 \]
Given the piecewise definition, $f(x)$ is non-zero only for $x>0$:
\[ \int_{0}^{\infty} \frac{k}{x^2+1} dx = 1 \]
Evaluate the integral:
\[ k \left[ \tan^{-1} x \right]_0^\infty = 1 \]
\[ k \left( \lim_{x \to \infty} \tan^{-1} x - \tan^{-1} 0 \right) = 1 \]
Substitute the known values: $\lim_{x \to \infty} \tan^{-1} x = \frac{\pi}{2}$ and $\tan^{-1} 0 = 0$.
\[ k \left( \frac{\pi}{2} - 0 \right) = 1 \]
\[ k \cdot \frac{\pi}{2} = 1 \implies k = \frac{2}{\pi} \]
So, the proper p.d.f. is $f(x) = \frac{2}{\pi(x^2+1)}$ for $x>0$.
Now, calculate the c.d.f. $F(x)$ for an arbitrary $x>0$:
\[ F(x) = \int_{-\infty}^{x} f(t) dt = \int_{-\infty}^{0} 0 \,dt + \int_{0}^{x} \frac{2}{\pi(t^2+1)} dt \]
\[ F(x) = 0 + \frac{2}{\pi} \left[ \tan^{-1} t \right]_0^x \]
Evaluate at the limits:
\[ F(x) = \frac{2}{\pi} (\tan^{-1} x - \tan^{-1} 0) \]
\[ F(x) = \frac{2}{\pi} \tan^{-1} x \]
Step 4: Final Answer:
The c.d.f. is $\frac{2}{\pi} \tan^{-1} x$.