# Interchange the order of integration and summation

I read some questions, where people usually discuss the interchange of definite integral and summation; for examples, here, here and here. But I would like to ask about the interchange of indefinite integral with summation, the reason is very practical. I am not sure if I read all the similar questions in SE, so if someone finds the answer somewhere else, please show me the link, and we could close this question.

The statement of question: I have a function $$F(z)$$, which is represented by an indefinite integral $$F(z)=\int d z\; f(z),$$ and the integration can not be carried out analytically. I would like to study the zeros and singularities of $$F(z)$$ based on knowing the zeros and singularities of $$f(z)$$. For example, if I know that $$z=0$$ is a singular point of $$f(z)$$, and $$f(z)$$ can just be expanded by Laurent series, say $$f(z)=\sum_{n=-\infty}^{\infty} a_n z^n.$$ My question is in what conditions the following equality holds $$\int d z\;\sum_{n=-\infty}^{\infty} a_n z^n = \sum_{n=-\infty}^{\infty} \int d z\; a_n z^n+c,$$ where $$c$$ is an arbitrary constant.

Generally the Laurent series is not necessary, the expansion could be in any other forms, say asymptotic expansion or fractional expansion.

• I am afraid that the textbook you might be using does the change of order in indefinite case. – Kumar Nov 13 '20 at 9:27
• the problem is local so generally, you need local uniform convergence; however, in general, the relation between the zeros of $F$ and $F'$ is not easy to discern (eg $F'=e^{z^2}$ has no zeroes, while $F$ has quite a lot of zeroes), while in the analytic case it is true that if $F'$ has a singularity, $F$ must have a singularity too but again the situation is tricky as isolated singularities may not remain so $1/z, \log z$ are typical examples – Conrad Nov 13 '20 at 13:20
• "the expansion could be in any other forms, say asymptotic expansion or fractional expansion" So this is not just an indefinite integral, it's an indefinite question. – user436658 Nov 13 '20 at 21:02
• @Conrad, from intuition, $F$ and $F'$ may have singularities simultaneously, and either the type and location may change. But is there any theorems support it? – user142288 Nov 14 '20 at 0:05
• Well if $F$ is non-singular near a point, $F'$ surely is so $F$ definitely has singularities where $F'$ does; the other way is trickier in the sense that if $F$ has a (non-removable) isolated singularity, $F'$ does too by inspection, but if $F$ has a branch point, $F'$ may have just an isolated singularity – Conrad Nov 14 '20 at 2:07

You need compact convergence of the series, that is, convergence on compact sets. But we also need some more preliminaries:

First, I'm not going to talk about the indefinite integral, but just about antiderivatives, simply because we will need definite integrals down the line, and I think it will be less confusing if we reserve the integral sign for the definite ones. Second, the expression $$\sum\int a_nz^n\mathrm dz$$ is a bit icky, since the integrals all come with an arbitrary additive constant attached, which can make the sum diverge. We will have to specify which antiderivative we are choosing specifically. Third, we will only be considering connected domains, for simplicity's sake. You will soon see why. Fourth, interchanging sums and integrals or sums and antiderivatives is really about interchanging limits and integrals/antiderivatives, since infinite sums are just sequences/their limits written down in a particular way. So the real question is: If $$f_n\to f$$, does $$F_n\to F$$, where $$F_n,F$$ are suitably chosen antiderivatives of $$f_n$$ and $$f$$? And in what manner do they converge?

Now with these remarks out of the way, we can say the following (which is about sequences of functions, but series are sequences, so you can apply it to series exactly the same way):

Let $$D\subseteq\mathbb C$$ be a connected domain. Let $$f_n:D\to\mathbb C$$ be holomorphic for all $$n\in\mathbb N$$, and let $$f_n$$ converge to $$f:D\to\mathbb C$$ uniformly on compact subsets of $$D$$. Also, let all $$f_n$$ have an antiderivative on $$D$$.

Then the function $$F:D\to \mathbb C,~z\mapsto\int_{z_0}^z f(w)\mathrm dw,$$ where the integral goes along any arbitrary path from a fixed $$z_0\in D$$ to $$z$$, is well defined and an antiderivative of $$f$$. We also have $$F_n\to F$$ uniformly on compact sets, where $$F_n$$ is an analogously defined antiderivative of $$f_n$$.

Proof:

First note that because $$f_n\to f$$ uniformly on compact subsets, $$f$$ is holomorphic. Also, since $$D$$ is connected and open, it is also path connected, so
a path from $$z_0$$ to $$z$$ is guaranteed to exist. And since the functions $$f_n$$ admit an antiderivative, the integral $$F_n(z)=\int_{z_0}^z f_n(w)\mathrm dw$$ does not depend on the path and is thus well-defined. Due to uniform convergence of the integrand to $$f$$ on the arbitrarily chosen path (it's compact), the integral converges to $$F(z)=\int_{z_0}^z f(w)\mathrm dw,$$ which thus also doesn't depend on the path and is then well-defined. If we can show that the convergence is uniform on compact subsets, then $$F$$ is holomorphic and $$F_n'\to F'$$ uniformly on compact subsets, too. But since $$F_n'=f_n$$ and $$f_n\to f$$, we will then have $$F'=f$$, so $$F$$ is an antiderivative of $$f$$. So we show this uniform convergence on compact subsets:

Note that any compact subset of $$D$$ can be covered by a finite number of compact discs. And if a function converges uniformly on a finite number of sets, then it also converges uniformly on their union. So it is sufficient to show uniform convergence on compact discs. Let $$\overline{U_r}(z_\ast)\subset D$$ be such a disc with radius $$r$$ centered at $$z_\ast$$. On this disc we have

\begin{align} \vert F(z)-F_n(z)\vert&=\left\vert\int_{z_0}^z f(w)-f_n(w)\mathrm dw\right\vert\\ &=\left\vert\int_{z_0}^{z_\ast} f(w)-f_n(w)\mathrm dw+\int_{z_\ast}^z f(w)-f_n(w)\mathrm dw\right\vert\\ &=\left\vert F(z_\ast)-F_n(z_\ast)+\int_{z_\ast}^z f(w)-f_n(w)\mathrm dw\right\vert\\ &\leq\vert F(z_\ast)-F_n(z_\ast)\vert + \left\vert\int_{z_\ast}^z f(w)-f_n(w)\mathrm dw\right\vert\\ &\leq \vert F(z_\ast)-F_n(z_\ast)\vert + r\sup_{w\in\overline{U_r(z_\ast)}}\vert f(w)-f_n(w)\vert. \end{align}

The last estimation gives a bound that doesn't depend on $$z$$ and goes to $$0$$ (the first term because $$F_n\to F$$ pointwise, the second because $$f_n\to f$$ uniformly on the compact disc). So $$F_n\to F$$ uniformly on the disc, and by the argument above, on any other compact set as well.

• First, thanks; second I really read your reply several time to find the answer of my question; third, let me make sure that we understand each other right by giving a simple example, say I have a sequence $\ln^n(x)$，I would like to know the singularities of the function $F(x)=\int \sum \ln^n(x)$. So based on your statement I can not do anything, because $\ln^n(x)$ is not holomorphic at $x=0$, am I right? – user142288 Nov 13 '20 at 10:32
• $0$ is not important for this, since $0$ is not in the domain. You're only concerned with the domain of the functions, which is $D=\mathbb C\backslash(-\infty,0]$. On $D$, the functions are all holomorphic. If the series converges uniformly on compact subsets of the domain, then the series itself is also holomorphic and you can interchange the sum and the integral. – Vercassivelaunos Nov 13 '20 at 10:53
• Thanks, I got it. But you did not solve my problem, I wanna know if $z=0$ is a singular point of $F(z)$ from $f_n(z)$, you just simply kick it out from the beginning by definition. Whatsoever thanks. – user142288 Nov 13 '20 at 10:59
• Well, if $f_n$ converges to $f$ uniformly on compact subsets, then the singularities of $f_n$ are also singularities of $f$, since the domains of $f$ and $f_n$ are exactly the same, and singularities are only characterized by the domain of the function (isolated boundary points of the domain). Same goes for the antiderivatives: if $F_n$ are antiderivatives of $f_n$ on the entire domain, then the singularities of $F_n$ are exactly the singularities of $f_n$, and then the singularities of $F$ are exactly those of $F_n$, which are those of $f_n$, which are those of $f$. – Vercassivelaunos Nov 13 '20 at 11:29
• Could you show me how does a $f_n$ with singularities can converge to a $f$ uniformly at the singular point included in a compact subset? For example, for $f_n=\ln^n z$ with a compact domain (you can pick any one) . – user142288 Nov 13 '20 at 12:03