Can we change order of integration and differentiation here. Assume that $P(x,y)$ have continuous first order partial derivatives. Let $\int P(x,y) dx$, denote the antiderivative with respect to x.
Is it then true that $\frac{\partial}{\partial y}\int P(x,y)dx=\int \frac{\partial P(x,y)}{\partial y}dx$?
I am struggling with how I can show this. Does it help if we restrict ourselves to a rectangle?
Update:
In the comments I got a tips of a Wikipedia article: http://en.wikipedia.org/wiki/Leibniz_integral_rule. I tried to modify the proof to my situation, can someone check if it is correct?
Proof:
Let $\{y_n\}$ be a sequence that goes to zero, but is never actually zero.
We then have:
$\frac{\partial }{\partial y}\int P(x,y) dx=\frac{\partial}{\partial y} [\int_a^xP(t,y)dt +C]=lim _{n \rightarrow \infty} \frac{\int_a^xP(t,y+y_n)dt-\int_a^xP(t,y)dt}{y_n}=lim_{n \rightarrow \infty}\frac{\int_a^x[P(t,y+y_n)-P(t,y)]dt}{y_n}$.
Because of the same reason here: http://en.wikipedia.org/wiki/Leibniz_integral_rule#Proof_of_basic_form that the term inside the integral is bounded, we get that our intgral is bounded.(P and it's partial derivatives are continuous, we can restrict ourselves to a closed local place, so we get compactness). Since all the terms are Riemann integrable, they are alse Lebesgue integrable, and hence we can use the bounded convergence theorem to get(they didn't say anything about riemann vs: Lebesgue in the article, but I assume we have to do that?):
$=\int_a^xlim_{n \rightarrow \infty}\frac{P(t,y+y_n)-P(t,y)}{y_n}dt=\int_a^x\frac{\partial  P(t,y)}{\partial y}dt=\int\frac{\partial P(x,y)}{\partial y}dx +C$
So in conclusion we have that:
$\frac{\partial }{\partial y}\int P(x,y) dx=\int\frac{\partial P(x,y)}{\partial y}dx +C$, but when working with antiderivatives, we can overlook the constant, so we get:
$\int\frac{\partial P(x,y)}{\partial y}dx $
Is this proof correct?
 A: I think your proof has the right idea, but I would advise against using indefinite integrals in a formal proof - you would be better off looking at a specific antiderivative. Namely, assume $P$ is defined on $[a,b]\times[c,d]$ and let
$$Q(x,y):=\int_a^xP(t,y)\,dt,$$
so $\frac{\partial Q}{\partial x}(x,y)=P(x,y)$. Your question becomes about whether we have
$$\frac{\partial Q}{\partial y}(x,y)=\int_a^x\frac{\partial P}{\partial y}(t,y)\,dt.$$
This is, of course, a question about interchanging limits, and as in your proof, the bounded convergence theorem is the way to go. We have
$$\frac{Q(x,y+h)-Q(x,y)}{h}=\int_a^xR_h(t,y)\,dt$$
where $R_h(t,y)=\frac{P(t,y+h)-P(t,y)}h$. Clearly $R_h(t,y)\to\frac{\partial P}{\partial y}(t,y)$ as $h\to0$. Moreover, by the intermediate value theorem, $R_h(t,y)=\frac{\partial P}{\partial y}(t,c)$ for some $c$ between $y$ and $y+h$. Since $P$ is $C^1$, $\frac{\partial P}{\partial y}$ is bounded. This is enough to apply the bounded convergence theorem and conclude.
In terms of Riemann vs Lebesgue integral, I would say as a general rule you should always assume you're working with the Lebesgue integral (once you know it) except for very specific circumstances, such as discussing the convergence of an improper integral which is not absolutely convergent. It is simple to show that if $f$ is Riemann integrable (on a bounded interval) then it is also Lebesgue integrable, and we have many more tools available with Lebesgue integration, so we may as well use them.
