I misjudged this one at first

follow up video:
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[@asparkdeity8717]

a^n - c^n = (a-c)(a^(n-1) + a^(n-2)c + ... + c^(n-1)) for positive integer n. In particular, assuming b a positive integer, (1 - x^b)/(1-x) = (1 + x + ... + x^(b-1)), and it is just integrating a simple polynomial. This gives: (1/b) * [x + x^2 / 2 + ... + x^b /b ] (1,0) = (1/b) * (1 + 1/2 + ... + 1/b) = H(b) / b, where H(b) is the harmonic function (for natural b, truncated harmonic series at b). Your result for this integral also holds, and so we can equate the two: Σ(n = 1 to ∞) 1/(n(n+b)) = H(b)/b ==> H(b) = Σ(n = 1 to ∞) b/(n(n+b)) = Σ(n = 1 to ∞) [ 1/n - 1/(n+b) ]. Thus, H(b) = Σ(n = 1 to ∞) [ 1/n - 1/(n+b) ] (*) The LHS of (*) is only defined for positive integers b, but the RHS of (*) is defined for all real b except the negative integers. Hence, (*) is the real extension of the harmonic function to the real numbers, which is astonishing!

[@owlsmath]

Great stuff! So many interesting variations and ways for this one. I'm working on a quick follow up video to add some things I left out of the video.

[@MikeMagTech]

That worked out well. Nice job!

[@owlsmath]

@@MikeMagTech thanks Mike! Have a good day 😃

[@owlsmath]

Here's a follow up video with some more info on the solution and the methods: kzbin.info/www/bejne/eafPiGSpaNtjfNU

[@edcoad4930]

Slightly different; did the division to get 1+x+x^2...+x^b-1 which is a sum from 0 -> b-1 of x^n. Flipped summation and integration and got sum from n=0 to b-1 of 1/(n+1) multiplied by 1/b. The solution above telescopes down to the same if you apply partial fractions and avoids the need to sum to infinity.

[@owlsmath]

Nice. Makes sense. Requires the assumption that b is an integer right?

[@edcoad4930]

@@owlsmath true!

[@siddharthchabra9022]

@@owlsmath You solution yields a series result which is divergent, but we can prove than for b = Z+ the integral has a finite value.. something is missing

[@owlsmath]

hello @@siddharthchabra9022 why do you say it's divergent? It's similar to 1/n^2 which converges

[@owlsmath]

@@edcoad4930 i'm doing a quick follow up video on the integer scenarios

[@alexkaralekas4060]

Or just use the finite geometric series its the exact form and |x|

[@owlsmath]

Hi Alex. I think I see what you mean and then you have only one thing to integrate. But I think b needs to be an integer for it to work?

[@BoringExtrovert]

I feel like this could be solved by a much simpler factorization

[@owlsmath]

Interesting what do you have in mind? I don't think b needs to be an integer although I didn't think much about that part.

[@TGRRohit]

We could use GP formula ​@@owlsmath

[@Hussain-px3fc]

5:50 can you get a closed form of this sum using digamma function?

[@owlsmath]

yep great idea. Looks like it would be (digamma(b+1) + euler-mascheroni)/b. awesome :)

[@owlsmath]

also i have another way in the descripion in terms of harmonic numbers solution is H_b/b. The problem was given to me with the series solution so it kind of biased what I did in the video.

[@skylardeslypere9909]

I don't know if splitting the integral like that is allowed? Neither of them converge (as with the sums)

[@owlsmath]

Hi. At what time in the integral?

[@skylardeslypere9909]

@@owlsmath A first time at 2:13. The sum of x^n equals 1/(1-x) but the integral of 1/(1-x) from 0 to 1 is divergent. Same with the other integral. You get ∞ - ∞. At 3:48 we also see this by the fact that we have an infinite sum of 1/(n+1) and 1/(n+b+1) which both diverge to infinity (see: "harmonic series diverges"). If you have a function f+g that you integrate from, say, a to b, then you can only say that ∫ f+g dx = ∫ fdx + ∫ gdx (from a to b) when the right hand integrals are finite.

[@owlsmath]

@@skylardeslypere9909 Yep that's exactly it. I shouldn't break it up there. I think I say at the beginning the individual integrals are divergent but then I can't break it up that way.