Question Regarding $\int_0^1\int_0^∞ e^{-xy}-xye^{-xy}dxdy$

Question

Specifically for $\int_0^{\infty}e^{-xy}-xye^{-xy}dx$, I see that this inner integral evaluates to $xe^{-xy}$ from x = zero to x = infinity, which after plugging in, comes out to be zero (according to my textbook). However, I stuck on how to evaluate $xe^{-xy}$ for x = infinity. I tried to solve it by rephrasing it to $$\lim \limits_{x \to {\infty}}\frac{x}{e^{xy}}$$ But I realized there is no l'hopital's rule for multiple variables.

Any hints or answers would be greatly appreciated. Thanks.

Answer 1

$$I=\int_{0}^{1} \int_{0}^{\infty} (1-xy)^{-xy}dx~dy$$ Let $xy=t$,then $$I=\int_{0}^{1}\frac{dy}{y}\int_{0}^{\infty} (1-t)e^{-t} dt=0$$

Answer 2

When you are integrating w.r.t $x$ you are treating $y$ as a constant. So L'Hopital's Rule is applicable and $\lim \frac x {e^{xy}}=\lim \frac 1{ ye^{xy}} =0$ for any $y>0$.

Answer 3

You can write $xe^{-xy}=x(e^y)^{-x}$. For any $y \in (0,1), e^y$ is a constant $k$ greater than $1$, and we know $\frac x{k^x} \to 0$. As changing the value of the integrand at one point does not change the integral, we can ignore what happens at $y=0$.