Given that $X$ and $Y$ are independent random variables and $Y>0$ and $E(X)=0$. We wish to compute $E(\frac{X}{Y})$.
The intended method
Since $X$ and $Y$ are independent, it is obvious that $X$ and $\frac{1}{Y}$ are independent also. Thus $$E(\frac{X}{Y})=E(X\frac{1}{Y})=E(X)E(\frac{1}{Y})=0E(\frac{1}{Y})=0$$
However we could also supply the following method:
Other Method
Let $Z=\frac{X}{Y}$ then $$E(Z)=E(\frac{X}{Y})=\int_{-\infty}^{\infty}\frac{x}{y}f_Z(\frac{x}{y}) dz=\int_{-\infty}^{\infty}\frac{x}{y}\int_0^{\infty} y f_Y(y)f_X(zy)dydz $$ Now we have that $$x=zy\Rightarrow dx=ydz$$ So in the integral we obtain $$\int_{-\infty}^{\infty}\frac{x}{y}\int_0^{\infty} f_Y(y)f_X(x)dydx = \int_{-\infty}^{\infty}xf_X(x)dx\int_0^{\infty}\frac{1}{y}f_Y(y)dy=E(X)E(\frac{1}{Y})=0$$
I am slightly unsure if this other method is entirely valid -- is this use of $dx=ydz$ entirely valid? I would have thought we need to use product rule since $y$ too is a function of $z$?
Is this method still correct?