When we can change $\int$ and $\sum$ for indefinite integral?

Question

I know, for example, that if the series $\displaystyle\sum_{n=1}^{\infty}f_n(x)$ consisting of integrable functions on a closed interval $[a, b] \subset \mathbb{R}$ converges uniformly on that closed interval, then its sum is also integrable on $[a, b]$ and $$\int_a^b \left(\sum_{n=1}^{\infty}f(x)\right)\, dx=\sum_{n=1}^{\infty} \int_a^b f_n(x)\, dx.$$

Or, Beppo-Levi lemma (we are working on $(X, \mathfrak{M}, \mu))$: Let $f_n:X \to [0,+\infty]$ be sequence of measurable functions on $X$. Then, for every measurable set $A \subset X,$ $$\int_A \sum_{n=1}^{\infty} f_n \, d\mu =\sum_{n=1}^{\infty} \int_A f_n d\mu.$$

And similar proposition.

But, we are working here with definite integrals. My question is: What we can say if we are working with indefinite integrals. Is this always true, or we have some restrictions?

Answer 1

A well known calculus theorem says: If $F_n:\>[a,b]\to{\mathbb R}$ is a sequence of differentiable functions that converges at the point $\xi\in[a,b]$, and if the sequence of derivatives $f_n:=F_n'$ converges uniformly on $[a,b]$ to some function $f:\>[a,b]\to{\mathbb R}$ then the $F_n$ converge in fact uniformly on $[a,b]$ to some function $F$, and one has $F'=f$.

Given this theorem and the sequence $(f_n)_{n\geq1}$ we can fix a point $\xi\in \ ]a,b[\ $ and choose for all $n\geq1$ the particular primitive of $f_n$ that vanishes at $\xi$. The $F_n$ so determined satisfy the assumptions of the above theorem, and the desired conclusion follows.

Answer 2

As the name suggests, there is not a unique indefinite integral.

For simplicity I will asume that $f_n$ is continuous and that $\sum f_n$ converges uniformly on $[a,b]$ to a continuous function $f$. Let $F_n$ and $F$ be indefinite integrals of $f_n$ and $f$ respectively. If I understand your question, you want to know wether $F=\sum F_n$. The answer in general is no. We have $$ F_n(x)=c_n+\int_a^xf_n(t)\,dt,\qquad F(x)=c+\int_a^xf(t)\,dt. $$ for some constants $c_n$ and $c$. It is true that under these conditions $$ \sum_n\int_a^xf_n(t)\,dt=\int_a^x\Bigl(\sum_nf_n(t)\Bigr)\,dt=\int_a^xf(t)\,dt. $$ But since the constants $c_n$ can be chosen arbitrarily, there is no reason to expect that $\sum_nF_n$ converges, or that if it converges, the sum is $F$.

The result will be true under the additional condition $\sum_nF_n(x_0)=F(x_0)$ for some $x_0\in[a,b]$.