A viewer suggested integral.
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- Опубликовано: 5 сен 2024
- We evaluate a nice integral suggested by a viewer.
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14:09
who are you???
Where can I suggest integrals to Michael?😀
@@i.am.jihoonk I’m just a memer
@@ld1ego_733 Post in the comments, but Michael can miss it so you may have to repost. I know some people who kept posting the same suggestion for several days and eventually got the video.
@Good Place to Stop is right. You can also email me -- i won't list my email here, but it is easy enough to find online. The one bad thing with this channel getting bigger is that I tend to miss lots of nice suggestions, especially when I am busy with other things. I promise it is never due to a lack of my interest or gratitude though. Thanks eveyone!!
U have the coolest integral thumbnails on YT!
You did a video on this topic, I think
The Bose integral
Thanks, I'll never tell my secret!! j/k: I hand place all of the characters using canva and use a color pallet generator to make sure the colors work together.
@@angelmendez-rivera351 Ofcouse
I said he did a video about this "topic" as a whole
@@MichaelPennMath you only use canva for the thumbnails?
It is important to note that decomposing the infinite series into 2 parts trick only works due to absolute convergence.
In 12:05 the summation of the odd squares should go from 0 to infinity.
I too deal with this Trauma inflicted by Michael's mistake
Or use 2n-1
@skulliam4 That would be more elegant and, since he usually excludes zero from the naturals, a bit more consistent.
@@MAULIKPATELnamste jajajjsjsjjajajjs
It is important correct the mistake in the next step by making another small mistake "in the opposite direction" :)
That integral is related to a "special" version of the bose integral.
We have for the bose integral:
∫₀∞ xⁿ/(eˣ-1) dx = Γ(n+1) ζ(n+1)
We have an similar result if we switch the minus with a plus:
∫₀∞ xⁿ/(eˣ +1) = Γ(n+1) η(n+1)
So for n = 1 we get the result in the video :)
ya that is what i was thinking too and i wonder if he will pull out polylog suddenly some how
To add, that "special" version is sometimes called the Fermi Integral or Fermi-Dirac integral, and it is related to the behavior of fermions in the same way that the Bose integral is related to bosons. And writing this comment is the first time I realized that bosons are named after Bose and fermions are named after Fermi.
At 12:05, I think you wanted (2n-1) in the first series. At n=1, the first term is 3 in the parentheses. You are leaving out 1.
You are correct, but in the next step he rewrites that sum as (sum of all n's) - (sum of all even n's) so it gets to the right answer
I was able to solve this using contour integration. I struggled with the integral form and the contour at first, but once I had them right the solution was straightforward. As a plus, it had advantage of not needing the Basel relation.
Rewrite the function to be integrated as x*exp(-x)/1+exp(-x), then apply infinite geometric series to the denominator and integrate. Much simpler I think!
That's also how I did it. I agree: simpler.
changing the denominator to e^x -1 yields a very interesting integral whose solution is that of the Basel Problem!
i have unsuccessfully been trying to solve it using the Leibniz Rule. would be cool if you could do a video on it
I think the same method used in this video ought to work.
If found the general solution to this one by accident while working on another class of integrals :)
The integral of x^n/(1+e^(bx) ) from 0 to infinity equals n!/b^(n+1) *((2^n-1)/2^n ) *ζ(n+1), where ζ(n) is the Riemann function. Pretty cool.
Another cool exercise you could do is finding the value of the sum of the form:
F(n)/n! from n=0 to n=infty
Where F(n) represents the n'th Fibonacci number. The result is quite pretty (compared to what I expected at least).
What do mathematicians do when they meet on a conference?
They integrate.
At 4:44, since the integrals are indefinite, don't you have to include a "+ C" then show that C=0 to set those equal like that? (It's an easy step by checking with u=0.)
Very nice video! Might have been nice to mention we can split the alternating sum into the positive and negative parts, because the sum absolutely converges (because it is the Basel sum). Keep it up Michael!
To split the summation in 12:24 it is important to notice that the original series converges absolutely
Another approach..expand the 1/(1+e^x) as a geometric series and use the gamma integral formula to simplify further..you ultimately evaluate the same sum anyways...
We can't expand 1/(1+e^x) as a geometric series because e^x>1 for x>0.
However we have x/(1+e^x) = xe^(-x)/(1+e^(-x)) and now we can expand 1/(1+e^(-x)) as a geometric series to get x/(1+e^x) = sum_{n=0}^{\infty} (-1)^n x e^{-(n+1)x} dx, which we can integrate termwise because of uniform convergence.
A quick integration by parts gives int_0^{\infty} x e^{-(n+1)x} dx = 1/(n+1)^2, so we get the same result at the end.
Thank you Michael Penn!
You can get the indefinite integral in a reasonably straightforward way using the polylogarithm function. The integral evaluates to x^2/2 - x ln(1 + exp(x)) - Li_2(-exp(x)). Using the power series definition of Li and the MacLaurin expansion of ln(1 +x) you can show that d/dx Li_2(-exp(x)) is -ln(1 + exp(x)) and then it's essentially integration by parts.
The summand for the sum of the squares of the reciprocals of the odd numbers should be 1/(2*n-1) in order to make the first term in the expansion of the summation 1.
love these type of integrals, great work
Good job as always, prof. Penn, I love the integrals which involves series. I sent you 2 months ago (IIRC) some problems coming from the colombian math olympiad and I'd love to see one of these problems solved on your channel. Keep up the good job!
Could u plz post them in the comment section, so that we can try :)
9:35-9:48 -- one of the problems with doing problems like this in Calculus 2 is that you have to hand wave past measure theoretic considerations. I used to say that non-mathematicians do it this way and hope that mathematicians can verify the witchcraft.
I suggest a somewhat different way:
1/(Expx +1) = Exp[-x]/(1+Exp[-x])=Exp[-x]-Exp[-2x]+Exp[-3x] ...... Then calculate the Integral of x*Exp[-n x] from 0 to inf., which by Integration by parts gives 1/n^2 .The series you obtain has the value pi^2/12.
Integral and series cannot be exchanged in general. But in this case can be done because of uniform convergence of the geometric series.
At 1:36 use just the known:
integrate (f'(x)/f(x))=log(abs(f(x))+C with f(x)=e^x+1.
He does the same thing, except he doesn’t plug it into that formula but derives it.
It's not magical. He shows where the property comes from
think there is a mistake in the odd sum, it should start from n=0.Anyway it doesn't make a change in the result.
Yeah you are right :) I was just about to comment on the same thing.
Or make it 2n-1 at 12:04
Actually, it would have changed the result, but he makes another mistake later that cancels out this mistake.
12:03 it has to be (2n-1)² because the 1 is excluded. But at the end it's still correct
I just spotted the same error. He makes another error later that cancels this one out.
Done a comparison test with ∫xexp(-x).dx, since 0≤x/(1+exp(x))≤x/exp(x).
This integral ∫xexp(-x).dx between x=0 to x=∞ which can be easily shown to =1
Therefore our problem integral converges to a value ≤1.
I know doesn't solve it but at least gives you something to check you answer with!
I love your lessons
5:41
girlfriend: we let U = x 😕🥲
More geometric puzzle problems please
At appeoximately 12:15, when splitting the sum into n even and odd, is there not a small mistake? The first term, when n=1, will not be included when summing from 1 with the expression (2n+1), which will start at 3. Shouldn’t the denominator be (2n-1) instead?
Exactly i realised too ( we re right ????)
n must be evaluated from 0 to i finity in the first part to get (1)
Well, Things clear up at 12:48 in the video, while splitting up the term.
@@MAULIKPATELnamste Doens't change the fact that you are still missing a +1 constant term, by neglecting the (2n+1) = 1 (when n=0) term, which evaluates to 1. So basically, the final answer should be 1 + pi^2 / 12, unless I am missing something.
@@Hvidbergen The correct answer is pi^2/12. What you're missing is that he went on to make another mistake that canceled out this mistake. It was on the very next step/line.
Excellent video! A suggestion for something different, find the nth derivative of 1/(1+x^2).
Michael Pls. For tool one just multiply and divide by e^(-x) and it becomes a trivial integral.
Thank you sir you are the inspiration behind maths and science channel
Just what I needed for a homework... thx
The first part integration 1 over e to the x +1 . I think it can be done in an another process by multiplying e to the neg x in both the numerator and denomination
Beautiful integral. Hash marks to denote end of proof reminds me of Ancient Egyptian where that represents the sound "ai". Excellent!
you earned my subscription!
interesting!! i did a different approach. and that by multiplying first by exp(-x) and interpreting the quotient 1/1+exp(-x) as a geometric series and then proceeded by IBP...
Sir I am requesting you to solve a bunch of permutations, combinations and probability problems and also cover some Applied word problems in calculus.
What a beautiful integral
11:55 to 13:05 a literal example of two wrongs making a right: the sum for "2n+1" should've been from 0 to infinity. :-B
Hi,
Great! I think I proofed the reverse of this : that the alternate sum of the n squared equals that integral.
For fun:
2 "so let's may be go ahead and",
3 "ok, great".
I got an integral for you: integrate e^(-x-a/x)/(b+c/x) from 0 to +inf where a, b, and c are constants.
I would like to suggest this difficult problem:
Show that the sum from n=1 to n=infinity of 1/(n^(1+|sin(n)|)) diverges.
I don't see how it diverges, |sin(n)| averages 0.6 and the zeta function converges for all n>1.
@@paul_w try it yourself with software
@@stenzenneznets I did and looking at the partial sums it seems convergent to around 3.3 . But obviously this is no proof, it could diverge very slowly. But again I don't see how it would diverge (cf previous post).
I can prove it diverges, but I world like to see how Michael manages to do it
An interesting one... wasn't able to solve it on my own... learnt something new.. thank you Michael
His way calculating integral Int(1/(e^x+1),x)
reminds me how i calculated integral Int(x/sqrt(e^x+(x+2)^2),x)
why not just add and subtract xe^x on the numerator of the original integral, and then it becomes a simple integration by parts problem
Why not write constant of integration when finding the antiderivative? Or maybe why it has to be 0?
At 12:18, wouldn’t he also have to add 1, because the case when n=1 is not considered in his splitting of the sum? Edit: I’ve seen another comment regarding this, anothe solution would be to sum the odd terms from 0 to infinity instead. Either way, it’s a minor error but would result in an answer of pi^2/12 + 1. Could someone clear up if I am mistaken or he is?
At 12:48 onwards, the summations clear up automatically.
@@MAULIKPATELnamste yeah, thanks noticed that. I guess his error was more of a mis-use of the equals sign
@@tomatrix7525 He just made a small mistake writing the sum. No need to get complicated.
The correct answer is pi^2/12. What you're missing is that he went on to make another mistake that canceled out this mistake. It was on the very next step/line.
I think at 12:00 that sum will go from n=0 to infinity
Excellent. Suggestion: converge of sum(1/n^(2+cos(n))).
Hello Michael, I would like to suggest a problem from a 1962 Moscow Mathematical Olympiad: Given a_0, a_1, ... , a_n. It is known that
a_0=a_n=0;a_k-1-2a_k+a_k+10 for all k = 1, 2, ... , k-1.
Prove that all the numbers are nonnegative
.
Your videos on math are amazing!
It is not clear what "a_k-1-2a_k+a_k+10 for all k = 1, 2, ... , k-1" means.
@@ricardocavalcanti3343 It means a sub k, 2a sub k and so on.
In case what a_o or 2a_k means, it means a sub 0 and 2a sub k.
@@keshavb3128 But what is the condition on a_k-1-2a_k+a_k+10?
@@ricardocavalcanti3343 that a_0=a_n=0;a_k-1-2a_k+a_k+10 is only true when k=1,2,..., k-1 and are nonnegative.
Hey sorry if you only welcome integral suggestions but my suggestion is on number theory. prove that n^5 and n always have the same last digit for n being a positive integer. i've tried for the first 10 positive integers and it seems to hold, but i've failed to generalize.
Thank u mr helped me a lot🙏🏽🙏🏽🙏🏽🙏🏽🙏🏽🙏🏽🙏
I did it in 3 minites in my head using the gama and eta functions...
Proving Math is amazing. QED.
Hard integrals are always great
Integrate 1/x^5 and 1/(x^5+1)
Not to be one of "those people" (ok I am one of those people), but you are integrating xe^{-x} / (1 + e^{-x}), which expands to a geometric series sum 1 to infinity of (-1)^{n+1} x e^{-nx}. You can scale the integral of a given term by making the substitution u = nx, getting the sum from 1 to infinity of (-1)^{n+1}/(n^2) times the integral from 0 to infinity of ue^{-u}. The integral is 1 so you get just the sum from 1 to infinity of (-1)^{n+1}/(n^2) which is pi^2/12 as you do at the end there.
❤❤❤can you please make a video solving Int_0^inf x^3/(e^x-1) dx ??? it is used to derive Stefan-Boltzmann law from plancks law. I struggled with this. not able to crack the nut. The solution is (pi^4)/15. found by wolfram alpha, what is not realy satisfying. Guys like Planck and all these physicists of the early 1900 century are so talented, just like you Prof. Penn..... Best wishes from Germany
Hi ! Here's a quite quick problem :
Find all integers n (if they exist) s.t. (2^{4n+2}+1)/65 is a prime number.
Good luck !
Quick to state, not to solve! :-)
Here is my solution:
(1) For F(n) = (2^{4n+2}+1)/65 to be an integer (not necessarily prime), n must be of the form n=3k+1, where k is a nonnegative integer. Indeed, 2^{4(3k+1)+2}+1 = 2^{12k+6}+1 = 64^{2k+1}+1 = (-1)^{2k+1}+1 (mod 65) = 0 (mod 65).
(2) Next, we note that 2^{12k+6}+1 = (2^{4k+2}+1)(2^{8k+4}-2^{4k+2}+1) = f(k)g(k). These factors have the following divisibility properties:
(2.1) For any nonnegative k, f(k) is a multiple of 5: f(k) = 2^{4k+2}+1 = 4x16^{k}+1 = (-1)x1^{k}+1 (mod 5) = 0 (mod 5). Moreover, if k=3m+1, where m is a nonnegative integer, then f(k) is also a multiple of 65: f(3m+1) = 2^{4(3m+1)+2}+1 = 0 (mod 65) by (1).
(2.2) If k=3m or k=3m+2, where m is a nonnegative integer, then g(k) is a multiple of 13. Indeed, g(k) = 2^{8k+4}-2^{4k+2}+1 = 16^{2k+1}-4x16^{k}+1 = 3^{2k+1}-4x3^{k}+1 (mod 13) = (3^{k+1}-1)(3^{k}-1) (mod 13). Therefore,
(2.2.a) g(3m) = (3^{3m+1}-1)(3^{3m}-1) (mod 13) = (3x27^{m}-1)(27^{m}-1) (mod 13) = (3x1^{m}-1)(1^{m}-1) (mod 13) = 0 (mod 13);
(2.2.b) g(3m+2) = (3^{3m+3}-1)(3^{3m+2}-1) (mod 13) = (27^{m+1}-1)(9x27^{m}-1) (mod 13) = (1^{m+1}-1)(9x1^{m}-1) (mod 13) = 0 (mod 13).
Having established the preliminary results above, now we can show that there is no integer n such that F(n) = (2^{4n+2}+1)/65 is a prime number. For this, we have to examine the following cases:
(3.1) If n0, it follows from (2.1) and (2.2.a) that F(n) = F(9m+1) = f(3m)g(3m)/65 = 5Mx13N/65 = MN, where M and N are integers greater than 1; therefore, F(n) is not prime in this case;
(3.2.d) In a similar way, if k=3m+2 and m>=0, it follows from (2.1) and (2.2.b) that F(n) = F(9m+7) = f(3m+2)g(3m+2)/65 is not prime either;
(3.2.e) Finally, if k=3m+1 and m>0, it follows from (2.1) that F(n) = F(9m+4) = f(3m+1)g(3m+1)/65 = 65MxN/65 = MN, where M and N are integers greater than 1; therefore, also in this case F(n) is not prime.
These cases exaust all possibilities, thus proving that F(n) is not prime for any integer n.
@@ricardocavalcanti3343 Well done ! That was "the right way" to solve this exercise. (There were actually 2 intermediate questions that I deleted, otherwise it was too simple :-) ).
12:32 where is the number 1 in the odd set ??
has been pointed out by others, that it should start at n=0
3:02 Where do I buy the chalkboard with the self-rewriting checkbox?
why do you split series 1/(2n+1)^2 like that?
I thought you would differentiate the RHS of the first tool.
Mr Penn can you do a video about 1986 Q6 IMO please
Can you make a video about Glauert Integral?
What about using residues?
Good thank.......
nobody here?
I’m here 🙋
@@goodplacetostop2973 yooo lol you are always here
@@prithujsarkar2010 As always 👍
Not to be that guy, but you forgot your limits on the integral at 7:36. I'm going to keep watching and see if you catch it. Yep, you caught it.
You've lost me when you started to solve the second question. That's a bit too advanced for me.
Is it possible to do this with contour integration as well?
in theory yes, but the problem is that the function has infinitely many poles (i*pi*(2k+1) for all integers k) and moreover the integral starts at 0, so you would have to find an appropriate contour that you can calculate.
I did it
Good.
These are jee mains questions 😂😁🙂
Hello, can you explique me the zeta function , thanks
Some 9 ads in 14min video :( . Got a headache from all those interruptions.
стендап не твое парень завязывай
im early too