I graduate high school and finished my 4th year in college and my boy still posting bad ass Integral videos , what a dedication
@SockChecked4 ай бұрын
I started watching you about 5 years ago and stopped because I quit on my applied maths major and switched to neuroscience. I still have the shirt and a signed note from you on a PETA card as a memory from the past when I won that giveaway. Glad to see you doing so well bro! Keep up the wonderful videos ❤❤❤
@TortoogaThePenguin4 ай бұрын
KZbin acting like they totally pushed this to your 356k subs. And only 400 of them decided to watch after a whole hour of the video coming out.
@vixguy4 ай бұрын
Im glad to be one of the 400
@Marcel-yu2fw4 ай бұрын
Well the issue might also be a delay in the view counter. I think there was some weird reason why it stops at around 300-400 views even though way more people have seen it.
@Cr4zy-50004 ай бұрын
@@vixguy same
@tomkerruish29824 ай бұрын
It depends on what level of notification they've selected.
@tomkerruish29824 ай бұрын
@@Marcel-yu2fwIt's 301. Numberphile did a video on it 12 years ago.
@Duckallister4 ай бұрын
int_0^1 lnx d(lnx) can directly be integrated to 1/2 (lnx)^2 |_0^1 without sub since they are the same, and while it diverges, it does so toward -infty because the 0 is on the lower bound
@Wutheheooooo4 ай бұрын
But doing so is a good practice, you are suppose to integrate with respect to a variable, not everything possible. I didn't watch the video btw, so not sure if I was taking to the point.
@matyasandras90384 ай бұрын
i interpret d(lnx) with bounds 0 and 1 as integrating y=lnx=0 to y=lnx=1 which in x are the bounds 1 to e. this gives different results from the video
@ricardoparada53754 ай бұрын
It almost slipped my mind that differentiability is on an open interval so we didn’t need to worry about the bounds for dlog(x)
@ytpanda3984 ай бұрын
Nice :) felt topical for me since I'm learning about the measure for the inner product space for Hermite Polynomials atm, would be cool to see more of this stuff - maybe a bit funkier
@sussybawka99994 ай бұрын
It's really fascinating. A lot of us flammy bois went and studied math, but we're still coming back for the nostalgia. I won't lie, I do miss the more "out there" videos, but those are obviously a lot more effort to produce.
@janeknowakowski57324 ай бұрын
My take on the challange problem. Int by parts: Int(udv)=uv-int(vdu) u=logx du=1/x dx v=xlogx I=x(logx)^2-int(logxdx)=x(logx)^2- x(logx-1) Pluging in the bounds of integration (and using L'Hopital twice) we get: (0-0+1)-(0-0+0)=1 🎉🎉 Please let me know if i made any mistakes.
@waqarpunisher89924 ай бұрын
Wtf is wrong w/ me It's summer holiday and instead of chilling I'm on my desk solving that goddam integral
@mr.inhuman79324 ай бұрын
Nice followup Video! Really enjoyed it!
@Ve-trik14 ай бұрын
How you upload video 54 years ago?
@threepointone4154 ай бұрын
One might consider d(xlnx) = (1+lnx) dx, and so (Here ∫ means an integral from 0 to 1) ∫ lnx d(xlnx) = ∫ lnx(1+lnx) dx = ∫ lnx dx + ∫ (lnx)^2 dx which is (Here you might invite the Gamma Function) -1 + 2 = 1
@PixelVoyager7774 ай бұрын
Mr. Flammy, In the fourth problem, integral(0 to 1) ln x d(ln x), the limits mean ln x = 0 to ln x = 1, so when you rewrite d(ln x) as 1/x dx, the limits must change to x = 1 to x = e (Since, ln(x) = 0 => x = 1 and ln(x) = 1 => x = e) Now, the new integral is (1 to e) ln(x)/x dx, here, we use the substitution ln(x) = t, so dx/x = dt. Changing limits again on substitution as 0 to 1 (Since ln(1) = 0 and ln(e) = 1), we finally end up with integral (0 to 1) t dt, which is simply t^2/2 from 0 to 1, which on simplification is just 1/2. Please take a look at this solution and point out any mistakes.
@xinpingdonohoe39784 ай бұрын
Seems acceptable. We'd normally assume the thing of the differential is experiencing the bounds. So 0≤ln(x)≤1
@MOTIVAO4 ай бұрын
Great work Papa Flammable
@Wielorybkek4 ай бұрын
cool video, I liked the homework problem as well
@natelewis33814 ай бұрын
Hey flammy boy, couldn’t we also solve these types of nonlinear spaced integrals by restructuring the kernel to be in terms of the differential? I.e: x = exp(log(x))
@brandonklein14 ай бұрын
Subtle comment but when your differential is a function, you should clarify what variable is in the integral bounds! Integral from 0 to 1 log(x) d(log(x)) = 1/2 if log(x) is the variable of integration
@blueslime58554 ай бұрын
Yes! I went more in depth in my comment
@brandonklein14 ай бұрын
@@blueslime5855 ok
@void22584 ай бұрын
I had students who got very confused and sent me emails about this video because you used "log" for logarithm base e instead of "ln" ("log" without a base typically represents logarithm base 10 for which 1/x is not the derivative).
@Loyis4 ай бұрын
I think log is used in a lot of places outside of the US
@DatBoi_TheGudBIAS4 ай бұрын
@@Loyisnah, many places, if not all, use log for base 10 and ln for base e Papa flammy just likes to use log
@diobrando76424 ай бұрын
@@DatBoi_TheGudBIAS My teacher in highschool used ln while my professor at uni always used log
@DatBoi_TheGudBIAS4 ай бұрын
@@diobrando7642 weird... Might be common outside here. I've never used anything other than ln for base e, not anyone I know
@bsmith62764 ай бұрын
From what I know, ln is typically used in introductory level calculus classes to distinguish it from base 10 log, but once you get to real analysis then log means natural log since that is the only log you commonly use at that point.
@the_eternal_paradox4 ай бұрын
DIFFERENTIAL FORMS MENTIONED 🗣🔥🔥🔥
@maureendotson46344 ай бұрын
Good Video! Hope all is well :) Andrew’s Mom 💕
@rick41354 ай бұрын
Can you work out a video with the sqrt(dx), so called stochastic integral differential form
@thatguyrandy18592 ай бұрын
So for ∫x dlog(x) on [0,1], is it ok to rewrite x as exp(log(x)) and do normal integration with respect to log(x)? I think it works out the same in calculation, but I don't think that's mathematically correct.
@A.A3824 ай бұрын
In the 3rd integral couldn't we write x as e^logx and the integral would be just e^logx? Then e^log1 - (lim x--->0 e^logx)=1-e^(-inf)=1? Or is it wrong because one when we don t have dx we can't make x= e^logx since its dlogx and two since limx-->0 logx does not exist(unless x>0 and x tends to 0+)
@DatBoi_TheGudBIAS4 ай бұрын
Idk if it works, but can we be phisicidts and just multiply the 3rd integral by dx/dx, yielding the integral from 0 to 1 of x d(lnx)/dx dx?
@tomkerruish29824 ай бұрын
I think the way to attack the challenge problem is using the identity W(x ln x) = ln x, where W is the Lambert W function. Edit: consulting Wikipedia, it looks like the answer is W(1) - 2 + 1/W(1), where W(1) has the approximate value 0.567 and is known as the omega constant.
@82rah4 ай бұрын
Papa flammy just spent 20 minutes showing how to deal with integrals like the challenge problem. Why on earth bring the Lambert function into this!?
@tomkerruish29824 ай бұрын
@@82rahIn every other case, it was easy to express the integrand as a function of the variable over which we were integrating. For example, x d(ln x) becomes exp(u) du, ln x d(ln x) becomes u du, and x ln x d(ln x) becomes u exp(u) du. The only way I know how to express ln x as a function of x ln x is through the use of the Lambert W function. Honestly, I think he was pulling a fast on us. If you have a simpler answer, I would sincerely like to see it.
@zvezdanzvezdov15304 ай бұрын
@@tomkerruish2982 While you can do that for some of these, you can also just do it exactly how Papa Flammy did it in the video. Namely by using the fact that for f = f(x), you have df = f'(x)dx. In this case, it's f(x) = xln(x) which gets you df = d(xln(x)) = (ln(x) + 1)dx. Then the integral of ln(x) d(xln(x)) from 0 to 1 becomes the integral of ln(x)(ln(x) + 1)dx from 0 to 1, which is more or less straightforward and yields 1. This way you can tackle the most general case of the integral of f(x) d(g(x)), which is equal to the integral of f(x) g'(x)dx, as Papa Flammy showed in the previous video.
@tomkerruish29824 ай бұрын
@@zvezdanzvezdov1530 What are the new bounds on the integral? The lower bound is x ln x = 0, so let's say x = 1, but what is the solution for x ln x = 1? We can exponentiate each side, getting x^x = e. I don't know the solution for this, but I bet it involves the Lambert W function.
@zvezdanzvezdov15304 ай бұрын
@@tomkerruish2982 The bounds are supposed to refer to x, not the function. In other words: integral of f(x) d(g(x)) from x = a to b integral of f(x)g'(x) dx from x = a to b Look up "Riemann-Stieltjes integral" on wikipedia. In the first section "Formal definition" it's made immediately clear, that the bounds are w.r.t. x, not g(x).
@jaikumar8484 ай бұрын
Hello sir ! If you find a laplace transform of time domain function F(t) and if poles of that transform F(s) lies in right half plane...do does it means that for at least one value of t in time domain our function F(t) is touching infinity ?
@Hobbychemiefreak4 ай бұрын
Doesn't "monic polynomial" mean that the leading coefficient is 1, not that the constant coefficient is 0?
@douglasstrother65844 ай бұрын
Anything published by Springer Verlag = Difficult & Expensive.
@mrjoshmtz974 ай бұрын
Today is my birthday! 🥳🎉🎊🎉
@iLinked4 ай бұрын
No
@anantasantra-bu5qo4 ай бұрын
Then this problem is your gift
@Issacnewton_4 ай бұрын
solve the problem then 😃
@yashprajapati88574 ай бұрын
Happy Birthday! 🎉
@schizoframia48744 ай бұрын
Happy birthday
@o_s-244 ай бұрын
what about the integrals' bounds
@blueslime58554 ай бұрын
Yeah, he should've changed them
@alexwarner38032 ай бұрын
@@blueslime5855yeah, no he shouldn't have. Misinformation. Bounds stay the same.
@blueslime58552 ай бұрын
I looked it up back then and it was just a weird noration, so what he did in the video is correct, but if you interpret it as a normal integral, it's wrong
@alexwarner38032 ай бұрын
@@blueslime5855 yeah. It's the Riemann-Stieltjes integral. Not like a u-sub.
@alexwarner38032 ай бұрын
@@blueslime5855 Lebesgue is better anywayz
@ParadoxProblems4 ай бұрын
Ooh, looks like someone's trying to integrate a partition function back into a action
I don't know what happened to your views But maybe because some audience don't have school now Because i started watching you when i had math exam😅
@stefanalecu95324 ай бұрын
He fell off, that's why, it's expected
@gnbxyznv4 ай бұрын
Hi youtube :) I recently stumbled over the following integral \int_ ho^\infty \dfrac{s \, e^{-\frac{s^2}{\kappa}}}{\sqrt{\text{cosh}(s) - \text{cosh}( ho)}} \: \text{d} s where $\kappa$ and $ ho$ are positive constants. Unfortunately I am not very deep in integration theory so I can't really evaluate how hard this integral really is. But at first glance it doesn't seem too crazy does it? If anyone seeks a challenge and wants to help solving this integral or telling me why it is not possible to solve I would greatly appreciate it.
@royalefighter01594 ай бұрын
I've just quickly put it into the Wolfram Alpha Add On for ChatGPT and it doesn't seem like there is any straightforward analytic solution in any closed form, though the AI can't really guarantee it. I've also just had a try at solving it for a bit and I also don't seem to get anywhere.
why do you keep using log(x) instead of ln(x)? it is so confusing. log(x) will be always log_10(x) for me
@stefanalecu95324 ай бұрын
ln x = log(x)/log(e) for any base, so it's irrelevant
@blueslime58554 ай бұрын
I believe you made a mistake, in the 3rd problem, the bounds for the integral are 0 and 1 for u=ln x, not x, so for x they will become 1 and e (x=e^u=e^ln x) so you'll get e-1 this way because the integrand became 1 with respect to x like what was shown in the video. Another way you could do it is keeping the integral bounds as they are. d(ln x)=du, x becomes e^u, therefore you have the integral of e^u from 0 to 1 which also yields e-1 For the 4th integral, you said that ln x is a dummy variable here which is correct, so you can replace it with t, but again, you shouldn't be changing the bounds I believe, so you get the integral of tdt from 0 to 1, which is ½ For the same reasons, the 5th integral should give 1. I may be wrong, I'd appreciate it if you explained, however, I'm fairly confident in what I wrote simply by imagining doing a normal substitution like we usually do and call it a new variable
@PapaFlammy694 ай бұрын
RS has nothing to do with substitution!
@blueslime58554 ай бұрын
@@PapaFlammy69 Riemann stasis?
@blueslime58554 ай бұрын
I just looked it up and it seems that the bounds are for x, that's not really clear here, but yeah, it solves the issue. Otherwise, it seems like substitution and that the bounds are for ln x, not x
@davidalexandrov2924 ай бұрын
Downgraded a lot from the golden period
@stefanalecu95324 ай бұрын
He wasn't that good even back then, we only stayed for the occasional integral and annoying voice (cracks)
@television-channel4 ай бұрын
writing “ln x” for natural log takes less time then “log x” for natural log, and oh you know what every normal person uses ln(x) for logarithmus naturales(??)
@hornkneeeee4 ай бұрын
at the end of the day notations are arbitrary use what you prefer
@hackergaming68694 ай бұрын
Why write log(x) and not just ln(x)? With ln everyone will know what you mean without listening to you but with log many might think its log10 because many calculators use it this way...
@stefanalecu95324 ай бұрын
log_b(x) = log(x)/log(b), so the base of log is irrelevant