Time derivative of definite integral

  • Thread starter mmwave
  • Start date
211
2
Hi,

Physics books gloss over math. Sometimes it bothers me.

Given a separable function of time and position f = h(x)g(t) then

d / dt of [inte] h(x)g(t)dx = [inte] h(x) dg(t)/dt dx

Where the derivative in the second integral is a partial deriv.

Why the chain rule does not apply is glossed over by the statement that the definite integral of f(x,t) with respect to x results in a new function that is a function of t only. I believe that, but since x can be a function of t I can't see why the chain rule does not apply. Can someone please explain this to me?

To be more specific the book says:

d/dt of [inte] x * psi(x,t)dx = [inte] x * dpsi(x,t)/dt dx
where again the second integral has partial deriv. w.r.t. time t. The '*' above just means times here.
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,845
17
x cannot be a function of t because x is a dummy variable of integration.

(However, you do pick up some extra terms that bear resemblance to the chain rule if the limits of integration depend on t)
 
211
2
Originally posted by Hurkyl
x cannot be a function of t because x is a dummy variable of integration.

(However, you do pick up some extra terms that bear resemblance to the chain rule if the limits of integration depend on t)
Thanks Hurkyl,

The limits are definite ( plus/minus infinity and psi(x,t) goes to zero as x goes to infinity). If x can't be a function of t, then the chain rule says x*dpsi/dt + dx/dt * psi and the second term is zero. Surely the book would have said so, so I must have messed something up in the presentation.

Can you confirm that if f(x,t) is separable into h(x)g(t) then it makes no sense for x to depend on t? It would not be separable in fact if x were a function of t?

Sorry to be dense but if everything I've written is correct the author has made a mountain out of a molehill.
 

HallsofIvy

Science Advisor
Homework Helper
41,731
884
If your integral is F(t)= [int](from a to b) f(x,t)dx then
dF/dt= [int](from a to b) df/dt dx where "df/dt" is the partial derivative. (In your "separable" example, you don't need partial derivative.)

More generally "LaGrange's Formula" is:
If F(t)= int(from a(t) to b(t)) f(x,t)dx the
dF/dt= int(from a(t) to b(t))df/dtdx+(db/dt)f(t,b(t))-(da/dt)f(t,a(t))

Notice that, here, the limits of integration are also functions of t.
 

Related Threads for: Time derivative of definite integral

  • Last Post
Replies
23
Views
4K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
6
Views
2K
Replies
6
Views
2K
Replies
9
Views
2K
Replies
5
Views
5K
Replies
11
Views
823

Hot Threads

Top