Calculus: Integrating Functions Using Power Series

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Well, I thought I would show you something that I genuinely find just fantastic. You know, when we study all different techniques of integration, we see that in fact most functions cannot be integrated. There's no way of actually untangling the integral of some really complicated functions. You get functions that aren't so complicated that sometimes can be very, very difficult, if at all possible, to integrate with the human arm. I wanted to show you that, in fact, there's a way of getting around that, which is really intriguing. It's using the calculus of power series.
I want to immediately jump to an important integral that is used an awful lot in life. That's the integral from 0 to x of . Now, notice that if there were a t out in front, this would be an easy integral. It would just a u to u substitution. But, sadly, that t is missing. So, how can you evaluate this integral? Well, it's not clear what technique you would use to actually crack that. However, using power series and the fact that we're allowed to integrate power series along the interval of convergence, allows us to actually crack this. So, using the integration fact that we've seen, and remembering that I actually know what the infinite series is for . How do I do that? Well, remember that e^x = . So, therefore, ^is just going to be what I get if I plug in x^2 wherever I see x. So, I see So, that's great. Because what that means is, all I have to do now to integrate this, I just have to integrate this. Integrating that is actually much, much easier. So, this will equal - what's the integral of term wise? Well, the integral of term wise is just the sum, n going from 0 to infinity. What do I do? I add 1. So, I see x^2n+1 and then I divide by it. So, I have the n!(2n + 1). Without any effort at all, I just integrated this extremely difficult function into this. What's the answer? And the surprise is - the trick is that it's a hard integral to do. Now we see why, because the answer is actually an infinite series. But, there it is. If you want to evaluate, estimate it with the Taylor polynomial, only go up to a finite number, you can absolutely do that without a problem. You can go up to, instead of infinity, go up to 1,000 or 10 or whatever. You can estimate roughly what this function looks like, what this is, from any particular x. So, this is absolutely amazing that you can now use power series to integrate functions that otherwise would have been hard to integrate.
I thought, just for fun, I would show you one last example. I think this is just great. You can finally integrate almost anything you want, almost. How about integrating cos(t^4)dt? That seems nearly impossible. How would you possibly crack that? If there t^3 out here, a u to u substitution would be very prudent, but there is no such thing. So, unfortunately, this seems fatal until I remember that there's an easy formula for cos(x). Cos(x) just equals that. So, if cos(x) is that, I can produce what cos(x^4) is. Cos(x^4) would be just what I get when I replace x by x^4. So, this is the sum, . So, computing this integral is nothing more than integrating each of these terms separately. If I integrate each of these terms separately, I see summation, n going from 0 to infinity... and then integrating each of these terms, these are just constant, so I keep them out in front. So, I have (-1)^n x^8n+1 and then I just divide by that. So, that's (2n)!(8n + 1).
Look how easy it is to actually do integrals now. It's so simple. There's no such thing as a free lunch. So, where is the expense? The expense is that the answer now is an infinite series. The answer now, in fact, is a power series. But, that's OK. That's absolutely fine. We are actually comfortable with power series. We understand power series now. So, it's absolutely OK. It also provides some insight into why these integrals are generally difficult. The answer is because there's no simple way of expressing the answer. Right? The answer or the integral requires us to express that as an infinite series, as an infinite power series. So, finally, we're able to look at functions whose integrals first seem to be completely beyond our understanding or scope. Now we see that, in fact, using the power of power series, we can actually represent those answers, represent those integrals and evaluate them in terms of power series. This, in some sense, is one of the real summits, one of the real important aspects. It really gives rise to the power of power series. We can crack difficult calculus questions and express the answers in terms of this, now, natural way, in terms of power series.
Well, congratulations. I hope you're beginning to appreciate the power series as much as I do. I'll see you soon.