Question

How to use series to find definite integral?

Answer 1

Given

$\displaystyle\int_a^bf(x)\,\mathrm dx$

you can evaluate the integral by first expanding $f(x)$ as a series, say

$\displaystyle\int_a^b\sum_{n\ge0}c_nx^n\,\mathrm dx$

then interchange the order of integration/summation (provided Fubini's theorem holds; it usually will, so no need to worry greatly about this aspect) to write

$\displaystyle\sum_{n\ge0}c_n\int_a^bx^n\,\mathrm dx$

Then evaluating the integral yields

$\displaystyle\sum_{n\ge0}c_n\frac{x^{n+1}}{n+1}\bigg|_{x=a}^{x=b}$
$\displaystyle\sum_{n\ge0}c_n\frac{b^{n+1}-a^{n+1}}{n+1}$

Given an appropriate sequence $c_n$, you would then be able to evaluate the integral exactly, or at the very least find a partial sum that approximates the value of the integral to within a specified degree of accuracy.

Here's an example that demonstrates the procedure. Suppose we want to evaluate the definite integral

$\displaystyle\int_0^1\sin\pi x\,\mathrm dx$

Recall that

$\sin x=\displaystyle\sum_{n\ge0}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$

so that we can write the definite integral as

$\displaystyle\int_0^1\sin\pi x\,\mathrm dx=\int_0^1\sum_{n\ge0}\frac{(-1)^n(\pi x)^{2n+1}}{(2n+1)!}\,\mathrm dx=\sum_{n\ge0}\frac{(-1)^n\pi^{2n+1}}{(2n+1)!}\int_0^1x^{2n+1}\,\mathrm dx$

Integrating yields

$\displaystyle\int_0^1x^{2n+1}\,\mathrm dx=\frac{x^{2n+2}}{2n+2}\bigg|_{x=0}^{x=1}=\frac1{2n+2}$

and so we're left with

$\displaystyle\sum_{n\ge0}\frac{(-1)^n\pi^{2n+1}}{(2n+2)(2n+1)!}=\sum_{n\ge0}\frac{(-1)^n\pi^{2n+1}}{(2n+2)!}$

The trick now is to evaluate the sum. Well, recall that

$\cos x=\displaystyle\sum_{n\ge0}\frac{(-1)^nx^{2n}}{(2n)!}$

Our sum closely resembles this power series. In our sum, we have odd powers of $\pi$ in the numerator, but even factorials in the denominator. We can adjust for this by simply multiplying by $\dfrac\pi\pi$:

$\displaystyle\sum_{n\ge0}\frac{(-1)^n\pi^{2n+1}}{(2n+2)!}=\frac1\pi\sum_{n\ge0}\frac{(-1)^n\pi^{2n+2}}{(2n+2)!}$

Now, our denominators take the form $2!,4!,6!,\ldots$, while the cosine series proceeds with $0!,2!,4!,\ldots$ - in other words, our sum skips the first term of the cosine series. We can adjust for this as well, by adding and subtracting the same term of $1$. In terms of our summand, we can get $-1$ by plugging in $n=-1$, so we can write

$\displaystyle\frac1\pi\left(-1+1+\sum_{n\ge0}\frac{(-1)^n\pi^{2n+2}}{(2n+2)!}\right)=\frac1\pi\left(1+\sum_{n\ge-1}\frac{-1)^n\pi^{2n+2}}{(2n+2)!}\right)$

Then shifting the index by 1 so that it starts at $n=0$ gives

$\displaystyle\frac1\pi\left(1+\sum_{n\ge0}\frac{(-1)^{n+1}\pi^{2n}}{(2n)!}\right)$

and now our sum exactly resembles to the negated cosine series evaluated at $x=\pi$.

$\displaystyle\frac1\pi\left(1+\underbrace{\sum_{n\ge0}\frac{(-1)^{n+1}\pi^{2n}}{(2n)!}}_{-\cos\pi}\right)=\frac1\pi(1-\cos\pi)=\frac1\pi(1-(-1))=\frac2\pi$

We can verify that this result is correct:

$\displaystyle\int_0^1\sin\pi x\,\mathrm dx=\frac1\pi\int_0^1\sin\pi x\,\mathrm d(\pi x)=-\frac1\pi\cos \pi x\bigg|_{x=0}^{x=1}=-\frac1\pi(\cos\pi-\cos0)=\frac2\pi$