exchange of order derivatives integrals

sanctus · 06-28-2007

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):

$\partial_t \int^b_a f(t,x)=\int^b_a\partial_t f(t,x)$

Thanks to anyone who remembers it. I think it has to be something like
$f \in C^1$

Erasmus00 · 06-28-2007

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 · 06-28-2007

Thanks Erasmus very much, actually b=t, but to simplify and not confound people I didn't write it. the real thing is:
$\partial_t \int^t_{t_{in}} dt' n^i\partial_if(t',{\bf x}-{\bf n}(t-t'))$

After hours of trying and with the help of my prof i came to the result:
$n^i\partial_if(t,{\bf x})-n^i\partial_i \int^t_{t_{in}} dt' n^i\partial_if(t',{\bf x}-{\bf n}(t-t'))$

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)...

exchange of order derivatives integrals

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