Hahaha Step 1: Let's make this easier to understand by eliminating the integral altogether. Step 2: Let's make this easier to understand by reintroducing an integral.

We can easily generalize this by putting a>1 instead of 2. We'll get the result of ln(a/(a-1)). The fact that a>1 has been used many times, from the inequality pushing the floor to n+1, to the convergence of the series

10:40

For the sum at the end, I reindexed the sum and wrote 1/2 as -1*-1/2. Then, I brought another -1 Inside the sum by multiplying by -1 outside the sum. This gave me -(sum from n=1->inf of (-1)^(n+1)*(-1/2)^n / n. This is the series expansion of -ln(1+x) at x = 1/2. This gives -ln(1-1/2) = -ln(1/2) = ln(2).

This evaluation invokes so many different concepts I for sure would not have been able to solve it myself in an appreciable amount of time.

Very interesting! I just used the initial substitution u = 1 - log_2(x) and got the same result.

That was great. Thank you, professor!

What I ended up doing is a substitution, 1-log_2 x = y, to simplify the floor function to make the series easier to work with.

wow, after watching so many your video, what a surprise I can do it on my first trial with the exact approach!

A sketch of the graph of the integrand at the start might have more clearly motivated the intro of -ve powers of 2. Ingenious soln as always

Very cool video and the best part of it is that I managed to do the integral myself :)

¡Impresionante!

Amazing!

I came up with something that might be a cool integral. It's the integral from 1 to infinity of dx/[1+2+3+4+...+floor(x)]

lb(x) can be used for log of x in base 2.

Hi micheal how are you ,bravoo it's a beatiful solution

This was beautiful

I just used the substitution x=2^-u, from which it follows dx=-ln(2)2^-u du as well as the integral bounds infty and 0, so after switching the bounds, you get I=ln(2) int_0^infty 2^-u 1/floor(1+u) du. I think this is a lot easier to split into a sum, because you can just split it into the integrals from n to n+1: I=sum_{n=0}^infty ln(2) int_n^{n+1} 2^-u 1/floor(1+u) du Now n < u < n+1, so floor(1+u)=1+n. That gives you: I=sum_{n=0}^infty ln(2)/(1+n) int_n^{n+1} 2^-u du And the integral int_n^{n+1} 2^-u du is easily calculated as int_n^{n+1} 2^-u du = -1/ln(2) [2^-u]_n^{n+1} = -1/ln(2) (2^-(n+1) - 2^-n) = 1/ln(2) 2^-(n+1) so in total you get I=sum_{n=0}^infty ln(2)/(1+n) 1/ln(2) 2^-(n+1) =sum_{n=0}^infty 1/(n+1) (1/2)^(n+1) And this is the exact same result as what was achieved at 7:30. Personally I found that this approach felt a lot less tricky than choosing the correct integral bounds from the start and showing that the floor function behaves correctly for those. Basically, simplify the contents of the floor() function as far as possible, and that way it'll give you the summation more directly.

What happened to the logró base 2 during the video?

Rewriting that 1/2 as an "X" is a tremendously clever move .. I wonder if some of the kids didn't get that. -- also dominated convergence, that uh .. seems like something most of them would simply have never heard of. At least, traditionally a lot of the best students at finding integrals are not math majors, right?

I set y=2/x to make it int_2^{\infty}. No need for limits.

I tried it by newton lebnitz rule bt still difficult to sove for me plz need its solution to

Can we generalize this!!?

Hard ❌ Interesting ✅

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