The integral reminded me of the famous equation connecting the zeta function to the gamma function. If you take the integral from 0 to inf of x^t / (1+e^x) dx, and you differentiate wrt t and evaluate at t=0, you get the desired integral. Can’t find a way to work it out from that though because you’d have to evaluate zeta and gamma at -1 🤔
The integral you've described sorts out to \Gamma(t+1)\eta(t+1) so you have to evaluate the derivative at t=0. The rest will follow as it did in the video.
I'd want to say on a seperate comment, that this might be a candidate for my favorite maths 505 vids, in addition to the fact that the integral could be computed by defining I(s) = int (x^s/1+e^X)m and in light of a previous maths 505 video, this evaluates to - (Dirichelet eta(s+1)*gamma(s+1)) deriving by s on both sides and evauating at s = 0 should yeild the same result.
Yes indeed Although that video was the one with e^x-1 in the denominator that evaluated to the product of the gamma and zeta functions, the technique you're describing is right on the money.
At 2:30 you say there are no issues with boundedness or convergence. Maybe that's true as x -> infinity, but the log term diverges as x -> 0 and the exponential tends to 1 there.
@@TheEternalVortex42 Yes, Fubini's theorem requires that the integral converges and is finite, not the integrand as such. It's a subtlety that I think doesn't get enough attention.
@davidblauyoutube yes you're right the case for x->0 deserved mention. A look at the graph of the integrand shows that the area converges rapidly as x->0 so the convergence of the integral justifies the switch up.
By splitting up the domain of integration [0,∞) into [0,1]∪[1,∞) at the first place and turn the integrand into the series in question, collecting successive terms in that series; it could be deduce that all the terms in the series are either always positive or negative, then by Beppo Levi Lemma, we can switch the order of integration and summation on each integral. Finally, by combining two series and the integrals within, the switch up at 2:30 is justified.
Im trying to find a way to make the original integral look like the negative integral from 1 to 2 of lnx/x dx, as it also evaluates to (ln2) ^2/2 But even though they seem likely, I can't seem to be able to link them
Audio fine. I've found proofs of the integral form of the EM constant, but i don't get the "intuition/motivation" that led to it. Any suggested references, anone? And of course, another elegant solution as usual.
Is there any reason this looks like the antiderivative of x at x = ln2? Maybe something to do with the lnx/(1+e^x) which in the limit is similar to lnx/sinhx? Maybe in the limit it is the antiderivative of the antiderivative of the dirchlet integral sinx/x -> 1? And the x to ln2 transformation relates to a substitution involving e^x -> x?
can I suggest an integral? integral 1 to inf log(log(x))/(1+x^2) and iin general this family of integral of log(log(log...(x)))-(1+x^n) I was never able to figure out how to integrate those ahah
Not my favorite piece on the channel. It might feel satisfying for the author of the video to operate the calculations (he couldn’t have made that any clearer…), but the sheer volume of outside references make it quite cumbersome for the interested non-professional viewer who has to familiarize himself with the concepts. It also destroys intuition and makes the entire argument, albeit impressive, rather formal and dull. I always thought the target group of this channel were interested outsiders- the other videos on that channel i saw catered far better to that group.
I wanted to introduce a bit of analytic number theory. But just an introductory bit to encourage the viewer to search and study relevant material. It's a really interesting subject so its worth the lookup.
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This is really on rather advanced level , but contains important mathematical tools.
Whooosh. I tried, but eventually I couldn't keep up. I learned a lot attempting to follow along though.
The integral reminded me of the famous equation connecting the zeta function to the gamma function. If you take the integral from 0 to inf of x^t / (1+e^x) dx, and you differentiate wrt t and evaluate at t=0, you get the desired integral. Can’t find a way to work it out from that though because you’d have to evaluate zeta and gamma at -1 🤔
The integral you've described sorts out to \Gamma(t+1)\eta(t+1) so you have to evaluate the derivative at t=0. The rest will follow as it did in the video.
@@maths_505I did this question using Feynman technique. However the answer came out to be -1.5 ln2.
@@sohomsen2922May I ask where did you put the parameter?
I'd want to say on a seperate comment, that this might be a candidate for my favorite maths 505 vids, in addition to the fact that the integral could be computed by defining I(s) = int (x^s/1+e^X)m and in light of a previous maths 505 video, this evaluates to - (Dirichelet eta(s+1)*gamma(s+1)) deriving by s on both sides and evauating at s = 0 should yeild the same result.
Yes indeed
Although that video was the one with e^x-1 in the denominator that evaluated to the product of the gamma and zeta functions, the technique you're describing is right on the money.
@@maths_505 oh yes; but since we have a + here the sum would have an additional (-1)^s term which makes the sum congruent to eta.
At 2:30 you say there are no issues with boundedness or convergence. Maybe that's true as x -> infinity, but the log term diverges as x -> 0 and the exponential tends to 1 there.
Integral of log x converges on (0, a) though so it's ok
@@TheEternalVortex42 Yes, Fubini's theorem requires that the integral converges and is finite, not the integrand as such. It's a subtlety that I think doesn't get enough attention.
@davidblauyoutube yes you're right the case for x->0 deserved mention. A look at the graph of the integrand shows that the area converges rapidly as x->0 so the convergence of the integral justifies the switch up.
There is still the issue of the alternating sum of such integrals needing to converge absolutely. I could be missing something not too hard.
By splitting up the domain of integration [0,∞) into [0,1]∪[1,∞) at the first place and turn the integrand into the series in question, collecting successive terms in that series; it could be deduce that all the terms in the series are either always positive or negative, then by Beppo Levi Lemma, we can switch the order of integration and summation on each integral. Finally, by combining two series and the integrals within, the switch up at 2:30 is justified.
Bro taught me math more than my school,
btw whats the software or application you use?
Im trying to find a way to make the original integral look like the negative integral from 1 to 2 of lnx/x dx, as it also evaluates to (ln2) ^2/2
But even though they seem likely, I can't seem to be able to link them
Yes,michael penn solved it like this,with nice tricks this integral become to - integral from 1 to 2 of lnx/x.
Awesome Solution. Thank you.
Do you watch Michael penn?
Not much. Great content though. I've watched alot of flammable maths though.
Old flammy was GOATED
I'm also a huge fan of dr peyam
@@maths_505Not OP, but I'm sure the question was due to him solving the same equation today completely differently.
Audio fine. I've found proofs of the integral form of the EM constant, but i don't get the "intuition/motivation" that led to it. Any suggested references, anone? And of course, another elegant solution as usual.
Not as of now but that's a nice question to search upon.
Is there any reason this looks like the antiderivative of x at x = ln2? Maybe something to do with the lnx/(1+e^x) which in the limit is similar to lnx/sinhx? Maybe in the limit it is the antiderivative of the antiderivative of the dirchlet integral sinx/x -> 1? And the x to ln2 transformation relates to a substitution involving e^x -> x?
can I suggest an integral?
integral 1 to inf log(log(x))/(1+x^2) and iin general this family of integral of log(log(log...(x)))-(1+x^n) I was never able to figure out how to integrate those ahah
The first one is the Vardi integral that I've solved here on the channel. I'll investigate the general case too.
用费曼积分发也可以吧，let I(t)=x^t/1+e^x，we can get the answer at I’(0)😅
I = ln(1/2)*ln(sqrt(2))
Great!! Thx so much!! Could you recommend us a book on analytical number theory covering the second part of this vid? Thank you again!
Well I recently started teaching myself the subject and I'm using Apostol's book
Thanks for this great video.Very good example of tricky integral. What is the exact title of Apostol book that you use?
undefined solution.
👌👍
To be fair, you left the real work in the dark.
There is a much easier and clearer way to compute the derivative of Eta at 1.
am i the only one thats not getting any audio?
Yup it's only you bro😂
It is a reasonablly common issue, it is always solved by reloading the video
@@maths_505 android doing android things
Professor Penn stole your integral :D
Its fine😂
How can you blame him when the integral so damn good😂
Lost me after 3~4mins...
Not my favorite piece on the channel. It might feel satisfying for the author of the video to operate the calculations (he couldn’t have made that any clearer…), but the sheer volume of outside references make it quite cumbersome for the interested non-professional viewer who has to familiarize himself with the concepts. It also destroys intuition and makes the entire argument, albeit impressive, rather formal and dull. I always thought the target group of this channel were interested outsiders- the other videos on that channel i saw catered far better to that group.
I wanted to introduce a bit of analytic number theory. But just an introductory bit to encourage the viewer to search and study relevant material. It's a really interesting subject so its worth the lookup.