Integrating functions with absolute values

[Mogarrr]

To find E |X| of a cauchy random variable, I need to integrate
\int_{-\infty}^{\infty}\frac1{\pi}\frac{|x|}{1+x^2}dx.

From the definition of absolute value, we have
\int_{-\infty}^0\frac1{\pi}\frac{-x}{1+x^2}dx + \int_0^{\infty}\frac1{\pi}\frac{x}{1+x^2}dx (I think).

But, the very next step in the textbook, which I'm trying to follow along, is
\frac2{\pi}\int_0^{\infty}\frac{x}{1+x^2}dx.

Unless I made some mistake, the reasoning here is that
\int_{-\infty}^0\frac1{\pi}\frac{-x}{1+x^2}dx = \int_0^{\infty}\frac1{\pi}\frac{x}{1+x^2}dx.

If this is true, why is this?

 

[Saitama]

Unless I made some mistake, the reasoning here is that
\int_{-\infty}^0\frac1{\pi}\frac{-x}{1+x^2}dx = \int_0^{\infty}\frac1{\pi}\frac{x}{1+x^2}dx.

If this is true, why is this?

What happens if you do the substitution $-x=t$?

 

[Pond Dragon]

To find E |X| of a cauchy random variable, I need to integrate
\int_{-\infty}^{\infty}\frac1{\pi}\frac{|x|}{1+x^2}dx.

From the definition of absolute value, we have
\int_{-\infty}^0\frac1{\pi}\frac{-x}{1+x^2}dx + \int_0^{\infty}\frac1{\pi}\frac{x}{1+x^2}dx (I think).

But, the very next step in the textbook, which I'm trying to follow along, is
\frac2{\pi}\int_0^{\infty}\frac{x}{1+x^2}dx.

Unless I made some mistake, the reasoning here is that
\int_{-\infty}^0\frac1{\pi}\frac{-x}{1+x^2}dx = \int_0^{\infty}\frac1{\pi}\frac{x}{1+x^2}dx.

If this is true, why is this?

Same idea behind $$\int_{a}^{b}\omega=-\int_{b}^{a}\omega.$$

 

[Mogarrr]

Thanks for your help guys. Understanding accomplished.

Let t=-x, then -dx=dt, \frac{-x}{1+x^2}=\frac{t}{1+t^2}, and the bottom limit of integration changes from -∞ to ∞. Thus

\int_{-\infty}^0\frac1{\pi}\frac{-x}{1+x^2}dx=(-1)\cdot \int_{\infty}^0\frac1{\pi}\frac{t}{1+t^2}dt = \int_0^{\infty}\frac1{\pi}\frac{t}{1+t^2}dt.

Since t is a dummy variable, the integrand can be rewritten with x=t.