Another way: Split the original integral into one from -1 to 0 and one from 0 to 1. In the first one, change x to -x. Then combine the integrals again; the same simplification will happen.
I'm pretty sure this method and the method splitting into odd/even functions are exactly equivalent (at least for all "nice" functions). The 2 and the 1/2 will always cancel.
Yes! It also generalises: if you have an integral from 1/A to A, you can split it into two integrals; one from 1/A to 1 and one from 1 to A. Then do a substitution of x -> 1/u in either one, and recombine. An example of this is the integral from 0 to infinity of ln(x)/(x^2+1) dx (though you need to worry a little about convergence when you limit A to infinity)
Yes, that's much quicker. It requires you to spot that that's going to work, though. The method he used is the natural thing to do if you are just following your nose and don't already know what the answer is going to be.
@@thomasdalton1508 It's a fairly standard trick when you have something of the form 1/(y + 1) + 1/(1/y + 1), which is effectively what the terms in the video are, with y = e^(1/x)
That construction of f0+f1 seems to fall into that 'just stupid enough to work' category... It's like, duh, of course you can multiply by 1 and add zero, but the effect is unexpected.
This techinuqe is critically important in analytic number theory and in combinatorics. Also, I dont believe you :p, there's just no way that's true. Maybe if you're just out of calc 2 or something
@@MrMctastics I read an entire two volume book on complicated integrals and this technique didn't show up. I'll take your word for this because I've yet to take analytic number theory though.
I teach a signals and systems course, and instruct my students they should always look at the function first without just jumping into the math, in the hopes of simplifying their approach to the solution. When it comes to the Fourier series and transforms, this can make a very crazy looking function so much easier. Students get amazed how something so simple can save them hours of work. I'm very glad you cover this approach here. Will show my students this video!
Oh neato, the even-odd decomp; Just learned this! This is onna those tricks that i wish was taught like the kings property, Weierstrauss sub, and complex subs.
This works for any bounds of integration, as long as they're symmetric around 0. So, one can write: sin(x) = integral between -x and +x of cos(t)/(e¹ᐟᵗ+1) dt.
@@oom_boudewijns6920 I dont think he does. If you are referring to f0 and f1, that is just notation for a function. Could be replaced with g and h. The Idea is just to use notation similar to f to denote hes splitting f into two similar functions.
students here in india learn this in 12th grade, in fact we learn even more of these integral "tricks" to solve questions as quickly as possible, they are used very oftenly in competitive exams aswell.
No, but at least heuristically, the integrand has only a jump discontinuity at 0, so it can be integrated over the interval: As x→0−, the denominator approaches 2, so the integrand approaches ½, while as x→0+, the denominator approaches +∞, so the integrand approaches 0+.
A slightly easier simplification involves removing negative exponents from the right term: Then you would have 1/(e^(1/x)+1)+e^(1/x)/(1+e^(1/x)), which more obviously simplifies to 1.
In the sense of a principal value, it is, but to be more careful, you do need to show that the integrand is sufficiently well-behaved near singularities.
It's a long story but I do appreciate and take on-board this feedback. trust me when I say it's not on purpose or what I'm aiming for, just how it shook out this round. -Stephanie MP Editor
Huh, I'd learned you could explicitly decompose a function into even+odd from a linear algebra exercise, but I didn't realize you could use it to help solve y-axis-symmetric integrals since you can both ignore the odd part and only worry about half of the even part.
why would anyone say this integration trick is "too powerful", this technique only applies to a relatively small number of definite integrals, and most aren't going to work out this nicely
I have a couple of Paul Nahin's books. One is about i and the other is about Euler's Fabulous Formula. If I recall, he is a retired professor of electrical engineering from the University of New Hampshire
Interesting how any even function e(x) can be integrated with 1/(1+exp(1/x)) over a symmetric interval to yield half the integral of e(x) over that interval. Actually this can be generalized even more: replacing 1/(1+exp(1/x)) with 1/(1+exp(o(x)) has the same effect, where o(x) is any odd function
What are you talking about? No Mathematician hates tricks. Mathematicians want students to know when and why the trick works. The trick is fine. However, it only works for symmetric integrals when f(x)+f(-x) is easier to integrate than f(x) which is rarely the case except for functions that were odd in the first place or were easy enough to integrate without symmetry.
seeing the f0 function: (f(x)+f(-x))/2 obviously reminds me of the definition of cosine with complex exponentials: (e^(ix)+e^(-ix))/2 but also reminds me of Binet's Formula for the fibonacci numbers: (phi^n+(-phi)^(-n))/sqrt(5) which motivates me to try and see if binet's formula can be written in a form involving a complex argument for cosine (real argument for a hyperbolic cosine?). that would be a really neat little identity; the fibonacci numbers in terms of complex trigonometry
You can write the Fibonacci numbers as follows: (e^(n*ln(phi)) + e^(-n*ln(phi))/sqrt(5) = 2cos(i*n*ln(phi))/sqrt(5) =2cosh(n*ln(phi))/sqrt(5) Thus we get from the definition of the Fibonacci numbers: cosh(n*ln(phi)) = cosh((n-1)*ln(phi)) + cosh((n-2)*ln(phi)) Hope it helps!
Limit doesn't exist, but both left and right-sided limits exist and are finite, so you can split this integral into two convergent parts, which makes it fine.
You have not shown that the integral EXISTS. E^(1/x) is UNDEFINED at x=0. You simply ignored that issue. I am not saying that your solution is incorrect, but that you never demonstrated that it existed.
When x goes to 0 1/x is inf from the right and -inf from the left and e^1/x goes to inf from right and 0 from left so cosx/(e^1/x+1) goes to 1 from left and 0 from right very rapidly. So it is easy to see the integral exists. There are also tests to see if the integral converges,and michael did it before.i do not think he should show this evry time.the goal of this videos is to solve nice tricky integrals this is not math lecture.
hyperreal numbers (or other infinitesimals) are IMO extremely interesting but I don't think they'll help one bit in computing antiderivatives of elementary functions...
@@LucasDimoveo ooh nice, it's actually good to not put that much pressure on a student, here we have to study this and much more in order to clear entrance of college to get admission
Jon Rogawski Calculus gives you all the basic tools with plenty of practice problems. A little light on multivariable, but that's ok. I would then recommend a book like schaums outline of calculus to keep you fresh and give you more practice.
Does anyone want to try this as a general method in similar integrals? Substitute, x+1/x = u x-1/x = v Then, use Arithmetic Mean - Harmonic Mean inequality on x and 1/x. Then, calculate u^2 + v^2 in terms of powers of x and replace the powers of x in the inequality with powers of u and v. You will get something like u^2 >= 4-v^2. Then you calculate the integrals only in terms of u, followed by calculating only in terms of v and compare the results. You should get integrals that are greater than or less than the original integrals. Sometimes, it's a good idea to discuss what they mean physically. (Take care of the 2nπ periodicity).
