Arctan definite integral-infinite limits

[iocal]

Hi guys. I was hoping to get some help on the following problem. The question asks me to prove that:

∫1/π *1/(1+x^2) dx sums to 1 over (-∞,∞)

In statistics, this is the Cauchy distribution density function and what we are asked is basically confirm that it fulfills the kolmogorov's law for distributions, i.e. that the probabilities over the whole set sum to 1. The main question is how to evaluate the arctan in infinite limits of integration.

Thank you in advance.

 

[pka]

∫1/π *1/(1+x^2) dx sums to 1 over (-∞,∞)
In statistics, this is the Cauchy distribution density function and what we are asked is basically confirm that it fulfills the kolmogorov's law for distributions, i.e. that the probabilities over the whole set sum to 1. The main question is how to evaluate the arctan in infinite limits of integration.

Because \(\displaystyle f(x)=\dfrac{1}{1+x^2}\) is an even function you know that \(\displaystyle \displaystyle\int_{ - \infty }^\infty {f(x)dx} = 2\int_0^\infty {f(x)dx} \).

You also know that \(\displaystyle \arctan(0)=0 \) and \(\displaystyle \displaystyle{\lim _{x \to \infty }}\arctan \left( x \right) = \frac{\pi }{2} \).

 

[iocal]

Because \(\displaystyle f(x)=\dfrac{1}{1+x^2}\) is an even function you know that \(\displaystyle \displaystyle\int_{ - \infty }^\infty {f(x)dx} = 2\int_0^\infty {f(x)dx} \).

You also know that \(\displaystyle \arctan(0)=0 \) and \(\displaystyle \displaystyle{\lim _{x \to \infty }}\arctan \left( x \right) = \frac{\pi }{2} \).

Clear and concise. Thanks!