Many places I have seen when solving integrals you change a lot of it into sums.
Finding $\int_{0}^{\pi/2} \frac{\tan x}{1+m^2\tan^2{x}} \mathrm{d}x$
Is just an example.
So in general, how do you solve integrals (CLOSED FORM) by using series?
Thanks!
(fOr example integrate $\displaystyle \int_{0}^{4\pi} \sin(x) dx$ using SERIES)?
Thanks!
The general method is to find a power series for the integrand, and then use uniform convergence of the polynomials $(x-c)^n$ to swap the integral sign and the sum. So: if we have a definite integral
$$\int_{a}^{b} f(x) \, \mathrm d x$$
and $f(x)$ admits a Taylor series around $c \in [a,b]$ that converges (uniformly) on $[a,b]$, then we have that $f(x) = \sum_{n=0}^{\infty} d_n(x-c)^n$ for some sequence $d_n$ and thus
$$\int_a^b f(x) \, \mathrm d x = \int_a^b \sum_{n=0}^{\infty} d_nx^n \, \mathrm d x= \sum_{n=0}^{\infty} \int_a^b d_nx^n \mathrm d x = \sum_{n=0}^{\infty} \frac{d_n}{n+1}(b^{n+1} - a^{n+1})$$
One applies the series method when that final infinite sum is easy to evaluate. Now, I have glossed over the idea of uniform convergence somewhat in this answer. A series of functions $\sum_n f_n$ converges uniformly to $f$ if for any $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $x \in [a,b]$ and for all $n > N$ we have $|\sum_{i=0}^n f_i(x) - f(x)| < \epsilon$. Contrast this with the notion of pointwise convergence.
In order to swap the infintie sum and the integral (in general, $\sum_n \int f_n$ to $\int \sum_n f_n$) as I have done, we require that the sequence of functions $f_n$ be uniformly convergent. Thankfully, a powerful theorem states that whenever a power series converges, it converges uniformly. Thus, this method is justified.
Update: Here is an example. Consider the series for $\sin(x)$ around $0$: $$\sin(x) = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$
So we have that $$\int_{0}^{2\pi} \sin(x) \,\mathrm d x= \sum_{n=0}^{\infty}(-1)^n \int_0^{2\pi} \frac{x^{2n+1}}{(2n+1)!} \, \mathrm d x = \sum_{n=0}^{\infty} (-1)^n\frac{(2\pi)^{2n+2}}{(2n+2)!} = -\left(\sum_{n=1}^{\infty} (-1)^{n}\frac{x^{2n}}{(2n)!}\right)$$
Recalling that the Taylor series for $\cos(x)$ is $\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}$, we have
$$\int_{0}^{2\pi} \sin(x) = -\left(\sum_{n=0}^\infty \frac{(-1)^n(2\pi)^n}{(2n!)} - 1\right) = - (\cos(2\pi) - 1) = 0$$
Now this may seem a little unsatisfying, as it seems like we just used that $\int \sin = -\cos$ in a really roundabout way. And that is essentially true, for this trivial example. However, when the series method is applied to much more complicated functions, then the results are non-trivial.
