When can we swap the order of integration vs differentiation?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
6 replies · 20K views
pellman
Messages
683
Reaction score
6
What conditions does the real scalar function f(x,y) (on the particular range of integration) have to satisfy in order to put

[tex]\frac{d}{dx}\int{f(x,y)dy}=\int{\frac{\partial}{\partial x}f(x,y)dy}[/tex]

?
 
Physics news on Phys.org
pellman said:
What conditions does the real scalar function f(x,y) (on the particular range of integration) have to satisfy in order to put

[tex]\frac{d}{dx}\int{f(x,y)dy}=\int{\frac{\partial}{\partial x}f(x,y)dy}[/tex]

?

That relation holds when the limits of integration are not a function of x or y. If they are, you have to apply the Leibnitz rule. When you do apply the Leibnitz rule to a function of one variable, you end up with one term out of a possible 3 that is exactly what you wrote above. Google leibnitz rule... the other two possible terms (each corresponding to limits of integration) involve taking the derivative of the limits of integration with respect to the variable of integration.

For your problem, there would be more than 3 terms because the integral is a function of 2 variables... to deal with this you would just put brackets around the inner integral and then apply the Leibnitz rule twice, which will surely end up giving you more than 3 terms and second order derivatives.
 
oops sorry your integral is just a function of one variable.

If the limits of integration are not functions of x, the two operations commute.
 
I think there are additional cases where the interchange is possible.

I remember Max Rosenlicht's Analysis book has a whole section on

the interchange of limit operations, where he covers precisely your case.

Unfortunately, I don't have the book with me at this point. If you can't find

the book, let me know, I will try to find it myself. I think baby Rudin's book

also included a section on this topic.
 
7thSon said:
If the limits of integration are not functions of x, the two operations commute.

That makes sense. I was just afraid that it wasn't that simple. thanks.
 
I wonder, tho--I have not yet looked at the chapters I made ref. to--
if that condition is sufficient, or if it is also necessary. I will look it
up soon, hopefully.