I did this as a pure Feynman problem and it was an entertaining ride!

Excellent technique

The integral doesn't match the result for t=1. Perhaps we could constrain t>1 for the result to apply.

christmas integral!!

nice

I was a tad concerned at the e to the e part when I started working on this, so I was happy to see how you resolved it. Well played!

the music is bad

Even more magnificent on a second look

Upon substituting the parameter a by square root of i, how do you justify taking the principle root instead if the other one? The same question applies to the case of taking the principle fourth root of i in another video where your integrand was presented with x^8. You picked the principal fourth root of i instead of one of the other three fourth roots.

Outstanding technique

Magnificent technique

Fantastic

Excellent

Can you also find the sum of 1/4^3 + 1/7^3 + 1/10^3 + 1/13^3+ ....?

I just think about this integral ,could we just use laplace transform ,for L[tf(t)]=-F'(s) ,plus L(sinx)=1/(1+s^2) ? Seem it is easier to get the Sum representation .

Super Cool ~~~!

Awesome! I

Excellent

another method you could use is to rewrite sin(n) as Im[ e^{in} ] and then invoke the series expansion -ln(1-x) = (n ≥ 1) Σ x^n / n then after expanding through Euler's formula you get -Im[ ln( 1 + cos(1) + isin(1) ) ] the imaginary part of the natural logarithm of a complex number = principle argument (angle) of the complex number so if z = 1 + cos(1) + isin(1) , arg(z) = arctan( sin(1) / (1 + cos(1) ) this is actually equal to 1/2; tan(x/2) = sin(x) / (1 + cos(x)) if you expand through the half angle identities so answer is -1/2

Fantastic

Multiplying by x, taking the derivative twice of a sum -- new (to me) methods

5*e from the last result minus 5*e from the original sum equal to 0? sum [n from 0 to inf] ((n^3-5)/n!) = 0 Heh, it looks weird 🧐

Excellent 👌

I=int x sin(nx) dx is a simple integration by parts problem. Set u=x, dv=sin(nx) dx. It follows that du=dx, v=-1/n cos(nx). I=int udv = uv - int vdu = -x/n cos(nx) + int 1/n cos(nx) dx = 1/n² sin(nx) - 1/n x cos(nx) + C Evaluate on (-π, π): I = -1/n π cos(nπ) + 1/n (-π) cos(nπ) = -2π/n cos(nπ) = 2π/n (-1)^(n+1)

When α≠0, sin(αx)cos(βx) is an odd function, so integrating over (-π, π) will yield 0. This proof doesn't need to assume that α is a positive integer (which you weren't sure about), and it does work for almost all real values of α, but α=0 is an exception.

I like the way you solve integrals w and how everything is so clear Big support 🖤🖤

I am addicted to these videos! Thanks. Did I find a typo? On the right side of the board, the last line starts ... . f(t) = 2te^(-t^2) Integrand du = ... . But on the line before. the t went into the integrand. And, I think it's still under the integral - in the du. So I think the factor t in front of the integral is extra. It seem's to disappear in the next step.

Wow that was intense! Impressive amount of work for your viewers! I really wasn't trying to make you paranoid about your algebra. I would have been happy with " f'' was an ansatz, and once we find it, I checked it with some crazy algebra". For a while with all the Feynman's trick hype the last couple years, I've been trying to wrap my head around two things: _ 1. The "quick and dirty" mindset you mentioned in your last comment. I am not a physicist so I believe them when they say "we just use this to find the right idea, then check it with empirical data" (although I have no idea if any physicist actually uses this trick for any physics problem). But this seems to me to be math channel, technically, and you yourself in admitted in an video that the technique didn't quite work because some DE was undefined somewhere, and had to verify it with a complex integral. But then doesn't that mean you just needed a different technique? I don't mind quick and dirty if the Leibniz rule applies and you just skip the details in the video. But in that earlier video, I remember checking that dominated convergence couldn't apply, so then you have to do something else. I'm not picking on you, this is a compliment, because you're the only channel that seems receptive. (Some channels do arrive at answers for integrals that have no solution!) What is the point if you don't know it's right? (This is not a rhetorical question. _ 2. Otoh, it almost works? I've checked a lot of channels and probably at least 30% of the problems I see either don't meet the criteria for the Leibniz rule (usually due to infinite intervals) or they fail to include the endpoint of their parameter's interval (usually requires Fatou''s lemma). But I've only seen a couple problems where the answer isn't verifiable some other way. I can buy that meeting dominated convergence is just the most convenient requirement and not the only one, but you'd think more would fail if that's the best we have. It's just sorta weird. You might not have any insight there, I'm just wondering. _ Btw, where do you find all these problems? I've looked for practice before and never found anything more than a single SE post with like 12 problems. Did you find that textbook Feynman liked? Sorry for the late reply after you did all that work right away. I was trying to articulate all this.

Sweet!

Quite a fine piece of work

Excellent 👌

For f(a) expressition ,why do we use the one form of Beta function ,and f(a)=Beta (1/2,1/2 + a/2) ,a lot easier ,right ?

Unique integral

Could you please state a rule that allows the result of the nth derivative of f(t) evaluated at t=1 to apply by substituting of the discrete variable n by the continuous variable a/2?

I think this integral shoule be standard Beta function with 2times derivative to t .

great~

perfact thinking ~!

Good stuff

Great proof! Perhaps mathematical rigor may require to prove lim f(c1, t+h) as h->0 is precisely f(b(t),t).

Excellent 👌

Or f(t)= ∫_0^1 (ln⁡ x) ln(1-tx)dx Then we calculate f'(t), which with a series expansion leads us to: f'(t)= - ∑_(n=1)^∞ t^(n-1) ∫_0^1(x^n)ln⁡ x dx and after an IBP we get the same result

Excellent

Hi 🙂Wondering whether it is possible to solve it in an alternative way with f(t) = arctan(tx)/(x+1), but it seems a real mess once developed...

Great work!

Fantastic

Or, after an IBP: I = 2 ∫_0^∞ x dx/(e^x-1) = 2 ∫_0^∞ x e^(-x) dx/[1-e^(-x)] f(t)= 2 ∫_0^∞ x^t e^(-x) dx/[1-e^(-x)] f(1) = I Through the series expansion of "1/[1-e^(-x)]" and the sub x(n+1) = u, we get: f(t)=2 ∑_(n=0)^∞ t!/(n+1)^(t+1) I = 2 ∑_(n=0)^∞ 1/(n+1)^2 = π^2/3

f(t) and f'(t) are undefined for all values of t. How can we be sure of the sum identity for f''?

Minute 3:20 The summation expression indexed by n for f(t) diverges for all values of t. The solution becomes problematic from this point.