longrob
Oct8-11, 07:18 AM
Hi all
I am looking for a simple way to show that the mean of the Cauchy distribution us undefined. This is because this integral diverges:
\underset{-\infty}{\overset{\infty}{\int}}\frac{x}{x^{2}+a^{2 }}dx
Now, I know one proof which replaces the limits of integration with -x1 and x2. After carrying our the definite integration we are left with \frac{1}{2}\ln\left(\frac{a^{2}+x_{2}^{2}}{a^{2}+x _{1}^{2}}\right) and finally (by Taylor Series expansion) 2\ln x_{2}-2\ln x_{1}+smaller terms . Then allowing x1 and x2 to approach infinity shows that the intergral diverges.
My question is now: is it sufficient, on any level, just to look at the antiderivative \ln(a^{2}+x^{2}), state that it is an increasing function of x, and simply conclude from it that the integral diverges ?
Thanks
LR
I am looking for a simple way to show that the mean of the Cauchy distribution us undefined. This is because this integral diverges:
\underset{-\infty}{\overset{\infty}{\int}}\frac{x}{x^{2}+a^{2 }}dx
Now, I know one proof which replaces the limits of integration with -x1 and x2. After carrying our the definite integration we are left with \frac{1}{2}\ln\left(\frac{a^{2}+x_{2}^{2}}{a^{2}+x _{1}^{2}}\right) and finally (by Taylor Series expansion) 2\ln x_{2}-2\ln x_{1}+smaller terms . Then allowing x1 and x2 to approach infinity shows that the intergral diverges.
My question is now: is it sufficient, on any level, just to look at the antiderivative \ln(a^{2}+x^{2}), state that it is an increasing function of x, and simply conclude from it that the integral diverges ?
Thanks
LR