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.
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.
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 ❤❤❤
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
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.
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
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.
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).
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.
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.
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
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))
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
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+)
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.
@@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.
@@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.
@@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.
@@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).
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
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
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 ?
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.
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.
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(??)
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...
UA-cam 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.
Im glad to be one of the 400
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.
@@vixguy same
It depends on what level of notification they've selected.
@@Marcel-yu2fwIt's 301. Numberphile did a video on it 12 years ago.
I graduate high school and finished my 4th year in college and my boy still posting bad ass Integral videos , what a dedication
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)
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.
Seems acceptable. We'd normally assume the thing of the differential is experiencing the bounds. So 0≤ln(x)≤1
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 ❤❤❤
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
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.
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
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
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.
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).
I think log is used in a lot of places outside of the US
@@Loyisnah, many places, if not all, use log for base 10 and ln for base e
Papa flammy just likes to use log
@@DatBoi_TheGudBIAS My teacher in highschool used ln while my professor at uni always used log
@@diobrando7642 weird... Might be common outside here. I've never used anything other than ln for base e, not anyone I know
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.
Nice followup Video! Really enjoyed it!
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.
cool video, I liked the homework problem as well
Wtf is wrong w/ me
It's summer holiday and instead of chilling I'm on my desk solving that goddam integral
Today is my birthday! 🥳🎉🎊🎉
No
Then this problem is your gift
solve the problem then 😃
Happy Birthday! 🎉
Happy birthday
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
Good Video! Hope all is well :)
Andrew’s Mom 💕
Great work Papa Flammable
Can you work out a video with the sqrt(dx), so called stochastic integral differential form
Doesn't "monic polynomial" mean that the leading coefficient is 1, not that the constant coefficient is 0?
DIFFERENTIAL FORMS MENTIONED 🗣🔥🔥🔥
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))
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
Yes! I went more in depth in my comment
@@blueslime5855 ok
what about the integrals' bounds
Yeah, he should've changed them
Anything published by Springer Verlag = Difficult & Expensive.
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?
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+)
integral of 0 to 1 of x log(dx)
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.
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!?
@@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.
@@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.
@@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.
@@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).
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
RS has nothing to do with substitution!
@@PapaFlammy69 Riemann stasis?
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
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😅
He fell off, that's why, it's expected
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 ?
How you upload video 54 years ago?
Ooh, looks like someone's trying to integrate a partition function back into a action
You using log(x) instead of ln(x) is killing me :(
so... unneat
I usually see log(x) used as log base 10
Challenge:
∫[0,1]lnxd(xlnx) = ∫[0,1](lnx)^2dx + ∫[0,1]lnxdx
= x(lnx)|[0,1] - ∫[0,1]lnxdx (use integral 1)
= 1 - (lim x->0- lnx/(1/√x)^2 (L'Hôpital rule)
= 1 - 2(lim x->0- √x)^2
= 1
I would suggest using ln for natural log, throughout the video I kept thinking that you were using the common log.
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.
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.
Cool!
0'th
i-th
you-th
0i-th
I6=int[x=0,1](ln(x))d(xlnx)
I6=int[0,1](ln(x)(1+lnx))dx
t=-ln(x)
dt=-dx/x
-e^-t•dt=dx
I6=int[0,♾️]((t^2-t)e^-t)dt
I6=2!-1!=2-1=1
u=x
v=ln(x)
du=dx
I=(xln(x))[0,1]-int[0,1](ln(x))dx
I=(xln(x)-xln(x)+x)[0,1]
I=1
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
ln x = log(x)/log(e) for any base, so it's irrelevant
Downgraded a lot from the golden period
He wasn't that good even back then, we only stayed for the occasional integral and annoying voice (cracks)
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(??)
at the end of the day notations are arbitrary use what you prefer
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...
log_b(x) = log(x)/log(b), so the base of log is irrelevant