Respuesta :
Given
[tex]\displaystyle\int_a^bf(x)\,\mathrm dx[/tex]
you can evaluate the integral by first expanding [tex]f(x)[/tex] as a series, say
[tex]\displaystyle\int_a^b\sum_{n\ge0}c_nx^n\,\mathrm dx[/tex]
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
[tex]\displaystyle\sum_{n\ge0}c_n\int_a^bx^n\,\mathrm dx[/tex]
Then evaluating the integral yields
[tex]\displaystyle\sum_{n\ge0}c_n\frac{x^{n+1}}{n+1}\bigg|_{x=a}^{x=b}[/tex]
[tex]\displaystyle\sum_{n\ge0}c_n\frac{b^{n+1}-a^{n+1}}{n+1}[/tex]
Given an appropriate sequence [tex]c_n[/tex], 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
[tex]\displaystyle\int_0^1\sin\pi x\,\mathrm dx[/tex]
Recall that
[tex]\sin x=\displaystyle\sum_{n\ge0}\frac{(-1)^nx^{2n+1}}{(2n+1)!}[/tex]
so that we can write the definite integral as
[tex]\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[/tex]
Integrating yields
[tex]\displaystyle\int_0^1x^{2n+1}\,\mathrm dx=\frac{x^{2n+2}}{2n+2}\bigg|_{x=0}^{x=1}=\frac1{2n+2}[/tex]
and so we're left with
[tex]\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)!}[/tex]
The trick now is to evaluate the sum. Well, recall that
[tex]\cos x=\displaystyle\sum_{n\ge0}\frac{(-1)^nx^{2n}}{(2n)!}[/tex]
Our sum closely resembles this power series. In our sum, we have odd powers of [tex]\pi[/tex] in the numerator, but even factorials in the denominator. We can adjust for this by simply multiplying by [tex]\dfrac\pi\pi[/tex]:
[tex]\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)!}[/tex]
Now, our denominators take the form [tex]2!,4!,6!,\ldots[/tex], while the cosine series proceeds with [tex]0!,2!,4!,\ldots[/tex] - 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 [tex]1[/tex]. In terms of our summand, we can get [tex]-1[/tex] by plugging in [tex]n=-1[/tex], so we can write
[tex]\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)[/tex]
Then shifting the index by 1 so that it starts at [tex]n=0[/tex] gives
[tex]\displaystyle\frac1\pi\left(1+\sum_{n\ge0}\frac{(-1)^{n+1}\pi^{2n}}{(2n)!}\right)[/tex]
and now our sum exactly resembles to the negated cosine series evaluated at [tex]x=\pi[/tex].
[tex]\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[/tex]
We can verify that this result is correct:
[tex]\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[/tex]
[tex]\displaystyle\int_a^bf(x)\,\mathrm dx[/tex]
you can evaluate the integral by first expanding [tex]f(x)[/tex] as a series, say
[tex]\displaystyle\int_a^b\sum_{n\ge0}c_nx^n\,\mathrm dx[/tex]
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
[tex]\displaystyle\sum_{n\ge0}c_n\int_a^bx^n\,\mathrm dx[/tex]
Then evaluating the integral yields
[tex]\displaystyle\sum_{n\ge0}c_n\frac{x^{n+1}}{n+1}\bigg|_{x=a}^{x=b}[/tex]
[tex]\displaystyle\sum_{n\ge0}c_n\frac{b^{n+1}-a^{n+1}}{n+1}[/tex]
Given an appropriate sequence [tex]c_n[/tex], 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
[tex]\displaystyle\int_0^1\sin\pi x\,\mathrm dx[/tex]
Recall that
[tex]\sin x=\displaystyle\sum_{n\ge0}\frac{(-1)^nx^{2n+1}}{(2n+1)!}[/tex]
so that we can write the definite integral as
[tex]\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[/tex]
Integrating yields
[tex]\displaystyle\int_0^1x^{2n+1}\,\mathrm dx=\frac{x^{2n+2}}{2n+2}\bigg|_{x=0}^{x=1}=\frac1{2n+2}[/tex]
and so we're left with
[tex]\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)!}[/tex]
The trick now is to evaluate the sum. Well, recall that
[tex]\cos x=\displaystyle\sum_{n\ge0}\frac{(-1)^nx^{2n}}{(2n)!}[/tex]
Our sum closely resembles this power series. In our sum, we have odd powers of [tex]\pi[/tex] in the numerator, but even factorials in the denominator. We can adjust for this by simply multiplying by [tex]\dfrac\pi\pi[/tex]:
[tex]\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)!}[/tex]
Now, our denominators take the form [tex]2!,4!,6!,\ldots[/tex], while the cosine series proceeds with [tex]0!,2!,4!,\ldots[/tex] - 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 [tex]1[/tex]. In terms of our summand, we can get [tex]-1[/tex] by plugging in [tex]n=-1[/tex], so we can write
[tex]\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)[/tex]
Then shifting the index by 1 so that it starts at [tex]n=0[/tex] gives
[tex]\displaystyle\frac1\pi\left(1+\sum_{n\ge0}\frac{(-1)^{n+1}\pi^{2n}}{(2n)!}\right)[/tex]
and now our sum exactly resembles to the negated cosine series evaluated at [tex]x=\pi[/tex].
[tex]\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[/tex]
We can verify that this result is correct:
[tex]\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[/tex]
