Error in double integration $\int_0^ \infty dy \int_0^ \infty e^{-2x-y} dx $ Hello I need some help on this one...
$$
\int_0^ \infty dy \int_0^ \infty  e^{-2x-y} dx 
$$
After using substitution $t=-2x-y$, $dt = -2$ for solving the inner one I get:
$$
\int_0^ \infty \frac{1}{2}dy 
$$
Which is infinity? (and that's nonsense as this is part of another question in this case)
Wolfram Alpha says it should be $1/2$. Where did I go wrong?
 A: Your integral is separable:
$$\int_0^{\infty} dy \, e^{-y} \, \int_0^{\infty} dx \, e^{-2 x} = \frac12 $$
A: In general: if you can isolate your variables in double integration, isolate them. In particular,
$$\int\limits_{0}^{\infty}\int\limits_{0}^{\infty}e^{-2x-y}\text{ d}x\text{ d}y = \left(\int\limits_{0}^{\infty}e^{-y}\text{ d}y\right)\left(\int\limits_{0}^{\infty}e^{-2x}\text{ d}x\right)\text{.}$$
Your change of variables does not work in this case, mainly because change of variables in double integration is different from change in variables in the univariate case. See this link.
A: You don't need a substitution all you have is 
$$
\int dy \int dx f(x,y) 
$$
Where $f(x,y) = h(x)g(y)$
Thus you can split the integral as
$$
\left(\int h(x) dx\right)\left(\int g(y)dy\right)
$$
A: As other answers suggest, you can split this into a product of two integrals.
However, your method will work as well; letting $t=-2x-y$ as you did will give
$\displaystyle\int_0^{\infty}\left[-\frac{1}{2}e^{-2x-y}\right]_{x=0}^{x=\infty}dy=\int_0^{\infty}-\frac{1}{2}\left(0-e^{-y}\right)dy=\frac{1}{2}\int_0^{\infty}e^{-y}dy=\frac{1}{2}\left[-e^{-y}\right]_0^{\infty}=\frac{1}{2}$
