I'm trying to solve this :
$$\frac{d}{dy}\int_{y}^{\infty} \int_{f(y)}^{\infty} (g(y)+h(x_2))f(x_2)dx_2 f(x_1)dx_1$$
In this case, can I change the order of Integration ? I will get this :
$$\frac{d}{dy}\int_{f(y)}^{\infty} \int_{y}^{\infty} (g(y)+h(x_2))f(x_1)dx_1 f(x_2)dx_2$$
p.s. $f(x)$ is the probability density function (i.e., $\int_0^\infty f(x_1)dx_1=1$ and $\int_0^\infty f(x_2)dx_2=1$).
I know if the limits are constants then the order can be changed like that, but what happens when the limits are infinite?
I searched online but couldn't find much helpful information.
Thank you!!!
Intuitive thinking makes me non-rigorously deduce that we can not change the Order , without knowing more about the Integration limits & the functions involved.
Let me try to Elaborate that :
When we have $I_1=\int_{l_{11}}^{l_{12}} \int_{l_{21}}^{l_{22}} f(x,y) dx dy$ with Constant limits , we have various well-known Criteria about changing the Order to make Integral $I_2$ where we know $I_1 \equiv I_2$.
Consider $I_3=\int_{l_{11}}^{\infty} \int_{l_{21}}^{\infty} f(x,y) dx dy$
Assume we change the Order to get Integral $I_4$ where we wish $I_3 \equiv I_4$
My thinking is that this is not Possible , because there is no 1 unique $\infty$ , unlike the Case of Constant limits.
$$I_5=\lim_{A \rightarrow \infty} {\int_{l_{11}}^{A} \int_{l_{21}}^{A} f(x,y) dx dy}$$
$$I_6=\lim_{A \rightarrow \infty} {\int_{l_{11}}^{A} \int_{l_{21}}^{A^2} f(x,y) dx dy}$$
$$I_7=\lim_{A \rightarrow \infty} {\int_{l_{11}}^{A} \int_{l_{21}}^{\sqrt{A}} f(x,y) dx dy}$$
$$I_8=\lim_{A \rightarrow \infty} {\int_{l_{11}}^{A} \int_{l_{21}}^{2A} f(x,y) dx dy}$$
Each of these Integrals will have Different value , even though all the Integration limits are $\infty$ , Hence when we change the Order , we can not know which $\infty$ is used where.
We are losing that information when we use Integration limit $\infty$ , hence we can not change the Order , without considering the Exact nature of the Integration limits & the functions.