A change in the order of integration

[cristianbahena]

Homework Statement

I don't understand the change order of integration in the following sequence for calculate the integral. Can you help me?

Relevant Equations

A change order of integration

 

[WWGD]

There is a reference to #3 , seems like a footnote, n the third line. Can you show us a screenshot of the footnote or see if it is something else?

 

[MathematicalPhysicist]

The change of variables is elementary:
from the first integral after the first equality you get:
$$U_0\le E \le \bar{U}$$
$$U_0 \le U \le E$$
so if you want to change the order of integration, i.e. first on $dE$ you have: $U \le E \le \bar{U}$, since $U \ge U_0$ and $U_0 \le U \le \bar{U}$, since $E\le \bar{U}$.

 

[Delta2]

Our double integral is $$\int\int \frac{F(U)}{\sqrt{\bar U -E}\sqrt{E-U}}dEdU$$ and if we assume that the unknown function $F(U)$ is such that the double integral exists, then we can integrate first with respect to E and then with respect to U and because $F(U)$ does not depend on E the above double integral is equal to :
$$\int (F(U)\int \frac{1}{\sqrt{\bar U -E}\sqrt{E-U}}dE)dU$$
it turns out that $\int \frac{1}{\sqrt{\bar U -E}\sqrt{E-U}}dE$ has a value that does not depend on U so we can write the above expression as $$\int F(U) dU\times \int \frac{1}{\sqrt{\bar U -E}\sqrt{E-U}}dE$$

 

[archaic]

The change of variables is elementary:

Do you realize how demeaning this is when said to someone struggling with something? It adds nothing and causes harm.