all. I'm trying to check in the following distribution is normalized, and am having a difficult time integrating it. If anyone can give me a clue on where to begin the integration, that be great.
The function in question is the Cauchy probability density function. I.e.,
$$ p(x) = \frac{a}{\pi (a^2 + x^2)}$$
Beginning my integration, I have:
$$\begin{align}\int_{-\infty}^{\infty} | p(x) |^2 dx & = \int_{-\infty}^{\infty} \bigg | \frac{a}{\pi (a^2 + x^2)} \bigg |^2 dx \\ & = \bigg ( \frac{a}{\pi} \bigg )^2 \int_{-\infty}^{\infty} \bigg | \frac{1}{a^2 + x^2} \bigg |^2 dx \end{align}$$
Unfortunately, this is as far I got. Would factoring out the $a^2$ term and using u-substitution for $u=x/a$ work for this?
Thanks.
You could write your integral using the reduction formula as
$${\displaystyle\int}\dfrac{1}{\left(x^2+a^2\right)^2}\,\mathrm{d}x =\dfrac{x}{2a^2\left(x^2+a^2\right)}+\class{steps-node}{\cssId{steps-node-4}{\dfrac{1}{2a^2}}}{\displaystyle\int}\dfrac{1}{x^2+a^2}\,\mathrm{d}x $$ Solve ${\displaystyle\int}\dfrac{1}{x^2+a^2}\,\mathrm{d}x$ by using the change of variable $u = \frac{x}{a}$, which will become $${\displaystyle\int}\dfrac{1}{x^2+a^2}\,\mathrm{d}x={\displaystyle\int}\dfrac{a}{a^2u^2+a^2}\,\mathrm{d}u=\class{steps-node}{\cssId{steps-node-5}{\dfrac{1}{a}}}{\displaystyle\int}\dfrac{1}{u^2+1}\,\mathrm{d}u=\dfrac{\arctan\left(u\right)}{a}=\dfrac{\arctan\left(\frac{x}{a}\right)}{a}$$ Therefore $${\displaystyle\int}\dfrac{1}{\left(x^2+a^2\right)^2}\,\mathrm{d}x=\dfrac{\arctan\left(\frac{x}{a}\right)}{2a^3}+\dfrac{x}{2a^2\left(x^2+a^2\right)}+C$$ Taking the integral limits, you get $${\displaystyle\int_{-\infty}^{\infty}}\dfrac{1}{\left(x^2+a^2\right)^2}\,\mathrm{d}x=\frac{\pi}{2a^3}$$