Exchange of order derivatives integrals

[sanctus]

I don't remember what is the condition on f to have the following (and didn't find in my books, eve if I remeber having done it a few years ago):

 

[math] \partial_t \int^b_a f(t,x)=\int^b_a\partial_t f(t,x)[/math]

 

Thanks to anyone who remembers it. I think it has to be something like

[math] f \in C^1[/math]

[Erasmus00]

On top of f being in C1, the integral needs to be convergent. Also, a and b need to be independent of t (or else you end up with some new terms).

-Will

[sanctus]

Thanks Erasmus very much, actually b=t, but to simplify and not confound people I didn't write it. the real thing is:

[math]\partial_t \int^t_{t_{in}} dt' n^i\partial_if(t',{\bf x}-{\bf n}(t-t'))[/math]

 

After hours of trying and with the help of my prof i came to the result:

[math]n^i\partial_if(t,{\bf x})-n^i\partial_i \int^t_{t_{in}} dt' n^i\partial_if(t',{\bf x}-{\bf n}(t-t'))[/math]

 

First term being the additional term you said, second a term resulting of the exchange,without remembering the justification you gave now.

 

I mainly got lost with all dependencies of t and which ones are explicit (and hence derived by a partial derivative)...