# Calculus 2: Series Integration

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.

• You really should start every line from the second with an equals sign. – Lord Shark the Unknown Dec 15 '19 at 6:35
• Your third and fourth lines are identical. – Lord Shark the Unknown Dec 15 '19 at 6:40
• 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 $C=K_0-K_1+K_2+\cdots\pm K_n$. – Lord Shark the Unknown Dec 15 '19 at 6:42
• I presume that $(-1)^n$ is a typo for $(-1)^i$ throughout? – Lord Shark the Unknown Dec 15 '19 at 6:47
• You can get proper parentheses (and other paired delimiters) that adjust to the size of their content by preceding them with \left and \right. – joriki Dec 15 '19 at 8:31

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 $$C$$.