I am working through a DE text book and am on Fourier series / half-range series. I am stuck on understanding how to integrate a series. I have the following:
$$y^{(4)}(x) = \frac{2w_o}{EI\pi} \sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin\left(\frac{n\pi x}{L}\right),$$
where $E$, $I$, $w_0$ and $L$ are all constants. The boundary conditions make all the constants of integration equal to zero.
My final answer is (which agrees with the text book):
$$ y(x) = \frac{2w_oL^4}{EI\pi^5} \sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^5}\sin\left(\frac{n\pi x}{L}\right).$$
I don't understand why I don't have to re-index the series, or how to generally take an integral of a series. I just integrated the inside of the series, factored out constants, and put the final answer back in a series summation. I would appreciate if anyone can help clarify what I actually should be doing. I guess my question is how do I do the following:
$$ \int\Big(\sum\limits_{n=1}^{\infty}f(x)\Big) dx.$$
Thank you in advance!
EDIT:
Take care to note that there are several theorems (Fubini, Dominated Convergence, etc.) that tell you when an infinte sum can be exchanged like this. I was not aware of the importance of these theorems at the original time of writing.
Original:
The integral operator is linear. $\int a(x)+b(x)\,dx\equiv\int a(x)\,dx+\int b(x)\,dx$. So the integral of a sum is the sum of integrals... and exactly the same goes for the derivative of a sum being the sum of derivatives!
$$\int (\sum_{n=1}^\infty f(x,n))\,dx=\sum_{n=1}^\infty\int f(x,n)\,dx$$
For example, the expansion for the natural log goes as follows:
$$\ln(1+x)=\int_0^x\frac{1}{1+t}\,dt=\int_0^x(1-t+t^2-t^3+\dots)\,dt=\int_0^x\left(\sum_{n=0}^\infty (-1)^nt^n\right)\,dt\\=\text{(the key Fubini-style re-ordering)}\sum_{n=0}^\infty\int_0^x(-1)^nt^n\,dt=\sum_{n=0}^\infty\left((-1)^n\frac{x^{n+1}}{n+1}\right)=\sum_{k=1}^\infty(-1)^{k-1}\frac{x^k}{k}\text{ (unnecessary but sensible re-indexing)}$$
The integral of a summation series is just the summation of each term, and you can integrate term-by-term or integrate the general rule, if such a thing is possible to do. That was a demonstration of integrating a power-series, with re-indexing included but as a last, arbitrary step! I didn't have to re-index it, I just wanted to. Integrating a product series involving $\prod$ is more difficult, but $\sum$ series are straightforward by the linearity of integration.
A more explicit showing of the re-ordering:
$$\int_0^x\left(\sum_{n=0}^\infty (-1)^nt^n\right)\,dt=\int_0^x(1-t+t^2-t^3+\dots)\,dt=\\\int_0^x1\,dt+\int_0^x-t\,dt+\int_0^xt^2\,dt+\dots=\sum_{n=0}^\infty\int_0^x(-1)^nt^n\,dt$$