The real magic happened when we took the upper bound to be same limiting variable as the one associated with the exponential function. Needs some elbow grease to cover this up properly but all is well as the exponential converges uniformly and we have dominated convergence.
It helps if your function is continuous, yeah. Incidentally, this same logic is what I used the first time I wrote a proof of the Fundamental Theorem from scratch - taking the limit arguments to be the same works if you have a continuous (in this case, absolutely convergent) function, which is also the condition for using the fundamental theorem (or indeed any integral)
That does indeed work, exp(-x) >= (1-x/k)^k on [0, k]. You also need to also be careful about the stray infinities being added/subtracted by the 1/x integrals and explicitly define your goal to be the limit of the difference and NOT the difference of the limits.
6:50 I have my doubts on the justification from changing upper bound to n as well just because it is also approaching infinity. It would be like replacing a double limit lim n->inf lim m->inf with just one limit feels incirrect
(1-x/n)^n converges uniformly to e^-x for any finite interval [0,m]. A bit more tricks show that [(1-x/n)^n-1]/x does aswell. Then the integral converges uniformly (this is a basic theorem) from 0 to m for every m>0. Likewise, it's not hard to show by comparison that the integral from 0 to infinity converges, to a value I(n). Therefore, there exists an m_0 such that for all m>m_0 the integral from 0 to m_0 is less than epsilon/2 from I(n). We are interested in limit n to infinity of I(n), call it J. There exists an n_0 such that I(n) is less than epsilon/2 from J for n>n_0. Now, for every n>n_0 and n>m_0, we have the integral from 0 to n is withing distance epsilon of J. So, the integral from 0 to n converges to the wanted result. Sorry for the formatting, just wanted to write something. I have no idea if I used uniform convergence, I think its needed but I can't see where, since I don't really deal with a function of x. This is just a quick response.
If he were to show all his work, the upper limit of the integral would be a limit m -> infinity. The tricky bit is showing that the pair of limits converges and that, therefore, setting m=n produces the same value as the value of the limit-pair.
At time of composing this there are 4 posts here already. The original and three replies. Mine is fourth reply making it the fifth post. My thoughts were to go forth but I really went fourth? 🙂 Or fifth that went fourth. Countable things are like that no? Anyway - I agree with all of previous posters. Many magical and mysterious things happen about that slight of hand when countable infinities go to uncountable infinities. Probability and stats, and distribution functions and density functions seem to depend on it. What a wonderful area that must have exercised great minds in the past and still does here. As an aside I did wonder what would happen using Taylor Series (more correctly McLaurin Series no?) and product of infinite sums.
A small note on the name Mascheroni, I looked it up because I also got it wrong. He's Italian, so the sche is pronounced like ske. It happens to look like the German sch trigraph but that a coincidence. The ch is a little quirk of Italian spelling - they have a rule where ce and ci sounds like English che/chi, so to avoid it you put an h, so keeps its /k/ sound. btw, Spanish disagrees, in Spanish like in English, ch is always /tʃ/
Love that bit at about 7:12. Many deep almost magical things seem to happen when crossing from realm of countable infinities to uncountable infinities. It must be worth a doctorate studying that single thing alone but I suppose it has been done already. In that case the doctorate could be simplification and cataloguing these things for educators to fill pedagogical study materials? Anyway it does seem an important area - well presented by the one - the only - The one and the only Michael Penn! Take it away Mike ...
Not sure if this is another possible way to solve this, but you could get a pretty neat "hint" at the solution if you take the derivative of the inside. This gives d/dx [f(x)] = ∫e^-x(1/x-lnx)dx And it's right there, only one step in (after simplifying) Initially, i had very little justification for it, but I guessed the answer would include γ (euler mascheroni constant) somehow, given the nature of how e^-x --> 0 and lnx--> ∞ I figured that their product would be some finite value. Otherwise, this wouldn't make a great TH-cam video. My guess was only by dumb luck but I'm still proud, and I wonder if theres another route to the solution that would include taking the derivative step i mentioned above, and then moving things, substituting in γ, and then getting the rest to vanish.
just consider \int_0^\infty x^t \exp(-x) dx which equals \Gamma(t+1). Your integral is simply the derivative of this w.r.t. 't' evaluated at t=1 namely \Gamma'(1)=-\gamma
If you already know the derivative of the Gamma function, then yes, you can use that. But this video obviously was intended to show a way to derive the result without knowing the derivative of the Gamma function. (Proving that result would be quite difficult, too.)
I knew the goal, so I found the replacement of ∞ with n to be genius, but it still felt like cheating for a second, even though I knew it would work out.
Nice! also we can use integral(e^(-x)x^t-1 dx which is gamma function and by taking derivative whith respect to t and t=1 we get polygamma function of 1 which is minus eulers mascheroni constant
If you already know the derivative of the Gamma function, then yes, you can use that. But this video obviously was intended to show a way to derive the result without knowing the derivative of the Gamma function. (Proving that result would be quite difficult, too.)
5:35 the integral you write here does not converge if you leave the upper limit as infinity. We have (1-x/n)^n being a polynomial of degree n, which explodes in the limit of x to infinity (unlike e^-x), so obviously the integral does not converge. You should replace the infinite upperbound with n first, and *then* rewrite the integrand. Formally you would still need to justify why you can rewrite the integrand, but at the very least the equality you assert would be *true* and not false...
"Today we are looking at an integral that looks fairly simple but has a really nice trick in its evaluation." I think that sentence is missing something in the middle... :)
More than year ago he recorded video about it th-cam.com/video/UEqI9GKYozU/w-d-xo.html This integral appears when you try to calculate Laplace transform of ln(t)
If you don't add the constant 1 you are facing to a converging problem when trying to perform integration by parts on interval [0,1] for exp(-x)ln x becaucse limit of -exp(-x)ln x when x is approaching 0 is +infinity. if you add the constant 1, limit (1-exp((-x))ln x when x is approaching 0, is 0. Anti-derivative is defined up to a constant on an interval.
If you already know the derivative of the Gamma function, then yes, you can use that. But this video obviously was intended to show a way to derive the result without knowing the derivative of the Gamma function. (Proving that result would be quite difficult, too.)
I don't like the explanation too much. This is a trick which is hard to use, since it is not exactly clear when to use it, or even what to do (choose the constants so that one of the terms after the integration by parts is zero?). Also, the fact that Euler's constant appears in the end is nice, but without any intuition about what happens here it doesn't give too much. I would appreciate in videos like this either explaining more about the "trick" (e.g. when and how to use) or alternatively a broader context about the integral and its connection to the Euler constant.
Wrong! Contour integration sadly proven quite useless again. After learning the trick, I tried to apply it to several real-life physics problems. Nah, nah, nah.
The real magic happened when we took the upper bound to be same limiting variable as the one associated with the exponential function. Needs some elbow grease to cover this up properly but all is well as the exponential converges uniformly and we have dominated convergence.
It helps if your function is continuous, yeah. Incidentally, this same logic is what I used the first time I wrote a proof of the Fundamental Theorem from scratch - taking the limit arguments to be the same works if you have a continuous (in this case, absolutely convergent) function, which is also the condition for using the fundamental theorem (or indeed any integral)
How does dominated convergence work in this case? let f_n(x) = [(1-x/n)^n]/n for x < n, and f_n(x) = 0 otherwise?
That does indeed work, exp(-x) >= (1-x/k)^k on [0, k]. You also need to also be careful about the stray infinities being added/subtracted by the 1/x integrals and explicitly define your goal to be the limit of the difference and NOT the difference of the limits.
WOW! I absolutely did not expect this constant to pop up here!? Incredible problem.
Also, 4:26 proof by "when all of the dust settles"
It looks like it's a fun one. It's great to learn from you, Michael.
6:50 I have my doubts on the justification from changing upper bound to n as well just because it is also approaching infinity. It would be like replacing a double limit lim n->inf lim m->inf with just one limit feels incirrect
Yup I had sirens going off in my head during that step too
(1-x/n)^n converges uniformly to e^-x for any finite interval [0,m]. A bit more tricks show that [(1-x/n)^n-1]/x does aswell. Then the integral converges uniformly (this is a basic theorem) from 0 to m for every m>0.
Likewise, it's not hard to show by comparison that the integral from 0 to infinity converges, to a value I(n). Therefore, there exists an m_0 such that for all m>m_0 the integral from 0 to m_0 is less than epsilon/2 from I(n).
We are interested in limit n to infinity of I(n), call it J. There exists an n_0 such that I(n) is less than epsilon/2 from J for n>n_0. Now, for every n>n_0 and n>m_0, we have the integral from 0 to n is withing distance epsilon of J.
So, the integral from 0 to n converges to the wanted result.
Sorry for the formatting, just wanted to write something. I have no idea if I used uniform convergence, I think its needed but I can't see where, since I don't really deal with a function of x. This is just a quick response.
If he were to show all his work, the upper limit of the integral would be a limit m -> infinity. The tricky bit is showing that the pair of limits converges and that, therefore, setting m=n produces the same value as the value of the limit-pair.
At time of composing this there are 4 posts here already. The original and three replies. Mine is fourth reply making it the fifth post. My thoughts were to go forth but I really went fourth? 🙂 Or fifth that went fourth. Countable things are like that no?
Anyway - I agree with all of previous posters. Many magical and mysterious things happen about that slight of hand when countable infinities go to uncountable infinities. Probability and stats, and distribution functions and density functions seem to depend on it.
What a wonderful area that must have exercised great minds in the past and still does here.
As an aside I did wonder what would happen using Taylor Series (more correctly McLaurin Series no?) and product of infinite sums.
@@Noam_.Menashe[(1-m/n)^n]/m doesn't converge when m approaches infinity.
A small note on the name Mascheroni, I looked it up because I also got it wrong.
He's Italian, so the sche is pronounced like ske.
It happens to look like the German sch trigraph but that a coincidence.
The ch is a little quirk of Italian spelling - they have a rule where ce and ci sounds like English che/chi, so to avoid it you put an h, so keeps its /k/ sound.
btw, Spanish disagrees, in Spanish like in English, ch is always /tʃ/
Great! Would just add that it coincides with the Laplace Transform of ln(x) evaluated at S=1🍻
Or the derivative of the Laplace transform of 1 eval. at 1
The result is astonishing, but the proof lacks rigour in many passages.
Wow I was trying to understand that integral when you came to cluch and save me 💪
The maximum of the integrand occurs when x = the reciprocal of the omega constant (or, 1/W(1) where W is the Lambert WW function).
Love that bit at about 7:12. Many deep almost magical things seem to happen when crossing from realm of countable infinities to uncountable infinities.
It must be worth a doctorate studying that single thing alone but I suppose it has been done already.
In that case the doctorate could be simplification and cataloguing these things for educators to fill pedagogical study materials?
Anyway it does seem an important area - well presented by the one - the only - The one and the only Michael Penn!
Take it away Mike ...
You haven't justified the swapping of the limit and the sign of integral.
Not sure if this is another possible way to solve this, but you could get a pretty neat "hint" at the solution if you take the derivative of the inside. This gives
d/dx [f(x)] = ∫e^-x(1/x-lnx)dx
And it's right there, only one step in (after simplifying)
Initially, i had very little justification for it, but I guessed the answer would include γ (euler mascheroni constant) somehow, given the nature of how e^-x --> 0 and lnx--> ∞ I figured that their product would be some finite value. Otherwise, this wouldn't make a great TH-cam video.
My guess was only by dumb luck but I'm still proud, and I wonder if theres another route to the solution that would include taking the derivative step i mentioned above, and then moving things, substituting in γ, and then getting the rest to vanish.
just consider \int_0^\infty x^t \exp(-x) dx which equals \Gamma(t+1). Your integral is simply the derivative of this w.r.t. 't' evaluated at t=1 namely \Gamma'(1)=-\gamma
If you already know the derivative of the Gamma function, then yes, you can use that. But this video obviously was intended to show a way to derive the result without knowing the derivative of the Gamma function. (Proving that result would be quite difficult, too.)
@@bjornfeuerbacher5514 but then you need to know that infinite series in the end which means you most likely know the \Gamma derivative anyways
@@amirb715 I've known the infinite series in the end for many years before I learned what the derivative of the Gamma function is.
I knew the goal, so I found the replacement of ∞ with n to be genius, but it still felt like cheating for a second, even though I knew it would work out.
Nice! also we can use
integral(e^(-x)x^t-1 dx which is gamma function and by taking derivative whith respect to t and t=1 we get polygamma function of 1 which is minus eulers mascheroni constant
If you already know the derivative of the Gamma function, then yes, you can use that. But this video obviously was intended to show a way to derive the result without knowing the derivative of the Gamma function. (Proving that result would be quite difficult, too.)
5:35 the integral you write here does not converge if you leave the upper limit as infinity. We have (1-x/n)^n being a polynomial of degree n, which explodes in the limit of x to infinity (unlike e^-x), so obviously the integral does not converge. You should replace the infinite upperbound with n first, and *then* rewrite the integrand. Formally you would still need to justify why you can rewrite the integrand, but at the very least the equality you assert would be *true* and not false...
For what reason can the limit be taken outside the integral?
I submitted this as a suggestion problem years ago, but it must have gotten lost. This is pretty similar to how I evaluated it.
my first thought was to use a taylor expansion of e^-x and then group terms.
Do you teach math for a living?
(youtube counts, but I'm trying to ask if you work for some university or school, too)
Since youtube is more lucrative it is a matter of time quitting the job in university.
See Trefor Bazett. Or others. There are many.
Which is the value of the Laplace transform of ln(x) when s = 1.
So many steps where you don’t justify their equality, one step is alright but doing it for an entire 9 min vid is not
He assumes that you know the theorems and stuff. You can prove most of them. What did you not understand for example?
"Today we are looking at an integral that looks fairly simple but has a really nice trick in its evaluation."
I think that sentence is missing something in the middle... :)
More than year ago he recorded video about it
th-cam.com/video/UEqI9GKYozU/w-d-xo.html
This integral appears when you try to calculate Laplace transform of ln(t)
I=L{ln(x)}|[s=1]
L{ln(x)}=-(ř+ln(s))/s
I=-ř, where ř denotes the Euler-Masceroni constant
alternatively, I(a)=int[0,♾️](x^(a-1)•e^-x)dx
I'(a)=int[0,♾️](x^(a-1)•e^-x•ln(x))dx
I=I'(1)
I(a)=Ř(a)
I'(a)=Ř'(a)
I=Ř'(1)=-ř
Ans:-γ
I didn't understand why he said that the integration of the function e(-x) is 1 - e^(-x).
In 02:35
If you don't add the constant 1 you are facing to a converging problem when trying to perform integration by parts on interval [0,1] for exp(-x)ln x becaucse limit of -exp(-x)ln x when x is approaching 0 is +infinity. if you add the constant 1, limit (1-exp((-x))ln x when x is approaching 0, is 0. Anti-derivative is defined up to a constant on an interval.
just use gamma and digamma's definition
If you already know the derivative of the Gamma function, then yes, you can use that. But this video obviously was intended to show a way to derive the result without knowing the derivative of the Gamma function. (Proving that result would be quite difficult, too.)
I don't like the explanation too much.
This is a trick which is hard to use, since it is not exactly clear when to use it, or even what to do (choose the constants so that one of the terms after the integration by parts is zero?). Also, the fact that Euler's constant appears in the end is nice, but without any intuition about what happens here it doesn't give too much. I would appreciate in videos like this either explaining more about the "trick" (e.g. when and how to use) or alternatively a broader context about the integral and its connection to the Euler constant.
Γ’(1)
"Fun fact!"
That was a fun way for me to test my brain, seeing if I could follow along through all the steps.
Here comes the negative Euler Mascheroni constant
No its called the Oily Macaroni constant
@@paridhaxholli As papa flammy calls it
i smell contour integration
Wrong! Contour integration sadly proven quite useless again. After learning the trick, I tried to apply it to several real-life physics problems. Nah, nah, nah.
Sorry, I don't buy it !!!
The result is probably true, but we need more justifications ...
You only did the calculus part !!!
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