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I am now studying undergraduate level real analysis, and the professor gives us an example of taking the indefinite integral of a real function by its Taylor series, and here is the example given by my professor:

Let $f(x) = e^x$, then by Taylor series, we have

\begin{equation} e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \cdots (1)\end{equation}

And then, take the indefinite integral of $(1)$, we get

$$\begin{aligned} \int e^x dx &= \int \sum_{n=0}^{\infty} \frac{x^n}{n!} dx \\ &= \sum_{n=0}^{\infty} \int \frac{x^n}{n!} dx \cdots (*) \\ &= \sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)!} + C \\ &= \sum_{n=-1}^{\infty} \frac{x^{n+1}}{(n+1)!} + (C -1) \\ &= \sum_{m=0}^{\infty} \frac{x^{m}}{m!} + D \\ &= e^x + D \end{aligned}$$

where $D$ is a constant.

Now my question is, for the step marked with $(*)$, why can we switch the order of $\int$ and $\sum$, and in general under what circumstances can we switch them? Can someone tell me which theorems and conclusions that I need to know to study this more deeply? Thanks!

ZYX
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    https://math.stackexchange.com/q/83721 – Sine of the Time Nov 17 '24 at 19:20
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    Theorems for switching integral and derivative are usually reserved for definite integrals, and rightfully so. If you consider $\int f(x),\mathrm dx$ as the set of antiderivatives of $f$, then infinitely summing these doesn't really make sense. You can even see this in your solution where you arbitrarily decide that the constant should somehow be outside of the sum even though you are infinitely summing over indefinite integrals, each with its own constant. What you should be thinking about instead is functions like $F(x)=\int_a^x f(t),\mathrm dt$, which by the FTC give you antiderivatives – Lorago Nov 17 '24 at 19:24
  • Uniform convergence; cf. baby Rudin p. 152 – J. W. Tanner Nov 17 '24 at 19:27
  • Since indefinite integrals are involved, the relevant concept is actually switching the order of derivative and summation. I regret not having time to expand on the topic in a proper answer, but the explanation that you seek is essentially the theorem on differentiating a power series. The topic of switching definite integrals with summation that the others are suggesting is mostly irrelevant, unless you are willing to go through the fundamental theorem of calculus, probably in both directions. – Adayah Nov 17 '24 at 20:06
  • And now even a user with a golden badge in calculus single-handedly closes the question, but doesn't notice that the answers he links to are all about definite integrals? And also all situated in some abstract measure-theoretic setting, most likely beyond the scope of knowledge of the OP? Sigh... – Adayah Nov 17 '24 at 20:33

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