Cauchy RV X has pdf f(x)=$c/1+x^2, x>0$ and $0, x\le0$ Cauchy RV X has pdf f(x)=$c/1+x^2, x>0$ and $0$ otherwise
a) Find c
$\int_0^\infty c/1+x^2$=arctan($\infty$)-arctan(0)=$\pi/2$
b) Compute E(1/$\sqrt{1+X^2}$)
$\int_0^\infty (1/\sqrt{1+(\pi/2)/(1+x^2)})*(\pi/2)(1/1+x^2)$
c) Find the pdf of $\sqrt{|X|}$
$P(Y\le y)=P(\sqrt{|X|} \le y)=P(|X|\le y^2)=\int_0^{y^2}(\frac{\pi}2)\frac1{1+x^2}$
Can someone help me check my work? Pretty sure this is all wrong except a cause the numbers are super funky
 A: a)
$$\int_0^{\infty}\frac c{1+x^2}\mathrm dx=c\frac{\pi}{2}$$
But the integral must equal $1$ since you integrate a continuous probability distribution on its support. So $c=\dfrac{2}{\pi}$.
b)
$$E(g(X))=\frac2\pi\int_0^{\infty}\frac{g(x)}{1+x^2}\mathrm dx$$
So
$$E\left(\frac{1}{\sqrt{1+X^2}}\right)=\frac2\pi\int_0^{\infty}\frac{1}{(1+x^2)^{3/2}}\mathrm dx$$
And note that a primitive of $\dfrac{1}{(1+x^2)^{3/2}}$ is $\dfrac{x}{\sqrt{1+x^2}}$.
c)
First note that $X\ge0$ so $\sqrt{|X|}=\sqrt{X}$. Then the usual LOTUS. Let $Y=\sqrt{X}$, for any measurable function $g$,
$$E(g(Y))=E(g(\sqrt{X}))=\int_0^{\infty} g(\sqrt{x}) f_X(x)\mathrm dx$$
Now with a change of variable $x=y^2$
$$E(g(Y))=\int_0^{\infty} g(y) f_X(y^2)\cdot 2y\mathrm dy$$
So $f_Y(y)=2yf_X(y^2)$.
A: Guide:
For part $b$, you should compute 
$$\frac{\pi}2\int_0^\infty \frac1{\sqrt{1+x^2}}\frac{1}{1+x^2}\, dx=\frac{\pi}2\int_0^\infty \frac{1}{(1+x^2)^\frac32}\, dx$$
For part $c$, seems fine if $y> 0$. Just differentiate with respect to $y$.
