Improper Integrals — Question 9

Question 9

Find $c$ so that $$f(x)=\frac{c}{1+x^2}$$ is a probability density on $(-\infty,\infty)$.

Question 9 – Solution

Step 1: Use the normalization requirement. A probability density must satisfy $$\int_{-\infty}^{\infty}f(x)\,dx=1.$$ Thus $$1=c\int_{-\infty}^{\infty}\frac{dx}{1+x^2}.$$ Step 2: Evaluate the improper integral. The antiderivative below has finite limits at both $-\infty$ and $\infty$, so both one-sided improper integrals converge separately. Their sum may therefore be computed using symmetric limits. $$\begin{align*} \int_{-\infty}^{\infty}\frac{dx}{1+x^2} &=\lim_{R\to\infty}[\arctan x]_{-R}^{R}\\ &=\frac\pi2-\left(-\frac\pi2\right)=\pi. \end{align*}$$ Step 3: Solve for $c$. $$1=c\pi\qquad\Longrightarrow\qquad c=\frac1\pi.$$ Because $c>0$, $f(x)\ge0$ for every real $x$. $$\boxed{c=1/\pi}$$