I do not understand how$\int \sum_{n=0}^\infty (-1)^nx^{2n}dx = \sum_{n=0}^\infty (\frac{(-1)^nx^{2n+1}}{2n+1})+C$. Here are my steps:

$$\int \sum_{n=0}^\infty (-1)^nx^{2n}dx$$ $$=\sum_{n=0}^\infty (-1)^n\int x^{2n}dx$$ $$=\sum_{n=0}^\infty (-1)^n(\frac{x^{2n+1}}{2n+1}+K)$$

$$=\sum_{n=0}^\infty (\frac{(-1)^nx^{2n+1}}{2n+1}+C)$$

The constant $C$ cannot be separated from the series. It is a part of the sequence that makes up the series.

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    $\begingroup$ You really should start every line from the second with an equals sign. $\endgroup$ – Lord Shark the Unknown Dec 15 '19 at 6:35
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    $\begingroup$ Your third and fourth lines are identical. $\endgroup$ – Lord Shark the Unknown Dec 15 '19 at 6:40
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    $\begingroup$ When you do the integrations, you get an arbitrary constant for each, so it's perhaps best not to denote all by $K$. Maybe use $K_0,K_1,\ldots,K_n$. Then the final arbitrary constant will be $C=K_0-K_1+K_2+\cdots\pm K_n$. $\endgroup$ – Lord Shark the Unknown Dec 15 '19 at 6:42
  • $\begingroup$ I presume that $(-1)^n$ is a typo for $(-1)^i$ throughout? $\endgroup$ – Lord Shark the Unknown Dec 15 '19 at 6:47
  • $\begingroup$ You can get proper parentheses (and other paired delimiters) that adjust to the size of their content by preceding them with \left and \right. $\endgroup$ – joriki Dec 15 '19 at 8:31

There are three ways to fix this.

Perhaps the easiest and most direct one is to use a definite integral from $0$ to $y$ instead of the indefinite integral, then relabel $y$ to $x$.

Alternatively, if you want to stick with the indefinite integral: As pointed out in a comment, each of the integrations has its own integration constant $K_n$. Let $a_N=\int\sum_{n=0}^N(-1)^nx^{2n}$ be the partial sums of the original series, $b_N=\sum_{n=0}^N(-1)^n\frac{(-1)^nx^{2n+1}}{2n+1}$ the partial sums of the integrated series and $c_N=\sum_{n=0}^NK_n$ the partial sums of the integration constants. Then $a_N=b_N+c_N$, so $c_N=a_N-b_N$. Both $a_N$ and $b_N$ converge for $x\in(-1,1)$; hence $c_N$ converges. Denote its limit by $C$.

Or, think about what the indefinite integral actually means. It represents a class of functions that are antiderivatives of the function being integrated and differ by an additive constant. When you swap summation and indefinite integration, you're using the fact that summing any sequence of representatives of the infinitely many classes of antiderivatives on the right-hand side will yield an antiderivative of the integrand on the left-hand side, as long as you choose them so that the sum converges at all. So if you properly formulate a theorem that allows you to swap summation and indefinite integration, it would already be part of that theorem that a single additive constant appears on either side of the equation.


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