Video

floor(-2024pi)=-floor(2024pi)-1 I=int[-2024pi,ceil(-2024pi)](floor(x))dx (everything else cancels) width=ceil(-2024pi)+2024pi height=floor(-2024pi) I=floor(-2024pi)•(ceil(-2024pi)+2024pi) I=({2024pi}-1)•floor(-2024pi)

I=int[0,♾️](int[1,pi](1/(1+a^2•x^2))da)dx I=int[1,pi](int[0,♾️](1/(1+a^2•x^2))dx)da I=int[1,pi](pi/2a)da I=pi/2•ln(pi)

This is another of those problems where looking at it as two integrals you end up with the difference of two logarithmically divergent expressions. You can see that there's a problem if you put u = pi x in the first one, which transforms it into the second, giving zero. Since arctan(pi x) > arctan (x) for all x in [0, pi/2] that clearly can't be right. The problem is that the terms in the integrand both tend to pi/(2x) for large x, hence the divergence. So I tried writing y = arctan(a x) - arctan(x), so tan(y) = x(1-a)/(1 + a x^2) so the integrand becomes 1/x arctan(x(1-a)/(1 + a x^2) ). I couldn't do much with that, but it's clear that it's convergent, since the large x behaviour ~ 1/x arctan( (1-a)/(a x) ) ~ 1/x^2 * (1-a)/a. Then I gave up and used the Feynman thingy as the title suggested. ☺

Fastest way: (i think) use king's symmetry property to switch sin to cos then use identity 1+cos(x)=2cos²(x/2) so then we get 1/2 (0 to π/2) ∫ sec²(x/2) dx which easily integrates to tan(x/2) then applying bounds gives us tan(π/4) - tan(0) = 1

dude, i was just visualizing the thing and approximated it to be between 0.4 and 0.5 as a wild guess. what a lucky shot

This is actually an integral first evaluated by Euler about 275 years ago, if not more. All you have to do is a substitution x=2t and no other principles are necessary. For people who suggested any other methods (such as by parts), I have to disappoint you, this function does not have elementary antiderivative.

I learned another approach using a double integral. Use x<=y<=πx as the interval and tan(y)/x as the integrand.

I see a hyperbolic claim about the effectiveness of Feynman's trick, I click

It's also the same as the integral of sinx / (1 + cosx) And you can do that in your head to get ln( 1 + cos(π/3)) = ln( 3/2) Still with you almost daily, keep up the great work, you're nailing it! 😊👍

Just finished vector calculus and linear algebra. What in the world is the floor function

I enjoyed that. I've liked Weierstrasse substitution since I first learned about it.

Why not use the half angle formulas? 1-cos(x) = 2(sin(x/2))^2 sin(x) = 2sin(x/2)cos(x/2) The integrand becomes tan(x/2).

I kept getting myself confused with the bounds until I drew a number line and marked various points, setting floor(2024 pi) = N. Then it was easy enough to calculate the three contributions.

Here is the link to that previous video where I looked at the case with integer bounds and bounds going to infinity: czcams.com/video/D5kwCC_Pa9A/video.html

The sum can be represented in terms of the digamma function: S = -1/3Σ1/k - 1/(k+1/3) = -γ/3 - 1/3(-γ + Σ1/k - 1/(k+4/3-1)) = -1/3(ѱ(4/3) + γ) = -1/3(ѱ(1/3) + γ) - 1 By multiplication formula (z=1/3, k=3): ѱ(1/3) + ѱ(2/3) + ѱ(1) = 3(ѱ(1) - ln3) ѱ(1/3) + ѱ(2/3) = -2γ - 3ln(3) By reflection formula (z=1/3): ѱ(2/3) - ѱ(1/3) = πcot(π/3) = π/√3 So we have the linear relation: ѱ(1/3) + ѱ(2/3) = -2γ - 3ln(3) ѱ(2/3) - ѱ(1/3) = π/√3 ѱ(1/3) = -γ - 3ln(3)/2 - π/2√3 Hence S = ln(3)/2 + π/6√3 - 1 Alternatively you can use the Gauss digamma theorem

Integration by parts with u=arctan(x) , dv = 1/x^2 dx if we want to use numerical integration use substitution u = 1/x and equality arctan(x)+arctan(1/x) = pi/2 , for x > 0

俺やったら出されても解けんな…

Beautiful!

A nice integral Thank you so much

I did not see the limits in the thumbnail😢 and was solving the indefinite integral, I finally came to a point where I got the integral of (xcosx)^(-1), I am pretty sure this is unsolvable. If anyone knows how to solve please help😅

It equals approximately -0.14839396162690884587003100526504 .

😭😭WOW, THAT WAS BRILLIANT SIR❤❤

I've been working on an integral similar to this one but in the sum reverse the order of the terms and substitute a natural number k instead of 3. So far it doesn't seem to have a nice solution but I'll still try to genralise because I have nothing else to do with my life.

This is easy af bro

It's a great video ! The only thing I would like to see more is (even if it is simple and short) an explanation of why we can switch the sum and integral. It makes the video more satisfying for me, It kid of "legitimates" the result. thank you :)

Well done! 👌 (Maybe you would like to work a bit on your intonation as it sounds a little bit monotonous. But that may be just an interpretation by my own ears. ✌️)

That's fascinating ❤❤❤

Nice solution. I would do the following: Define three series, A,B,C defined as the sums of respectively 1/(3n-1), 1/(3n), 1/(3n+1), each from n=1 to n=N. Then A+B+C=H_(3N)-1 and B=H_N/3 where H denotes the harmonic numbers. Now the series A-C is nice and can be easily computed to be 1-π/(3√3) using the cotangent residues trick from complex analysis. Then we can find the series we want, C-B, as ((A+B+C)-3B-(A-C))/2. From the approximation H_N≈ln N+γ we find lim H_(3N)-H_N=ln 3. So in total we get (ln 3-1-(1-π/(3√3)))/2 which agrees with your answer.

Nice job!

Technically you should have left the -n at the last step instead of putting in -Infinity, then gotten to e^0+e^(-n), and then take the limit to get e^0=1. Convergence on limits if you take it sooner has very stingy rules, and I'm sure they don't apply automatically to bounds of integration

Ugh. I didnt care for that - and i can vouch that having arctan in the floor function makes it impossible! (The thumbnail is kinda hard to read so I started with your first handwritten version). One other comment is that ln(1/u) = - ln(u) so you didnt really need the y sub.

I gave this problem to my 10 year old cousin and he got it right. "1?!" Bro's gonna win the bee this year frfr

Imagine being a calc 1 student who just got into trig integrals and on the exam next day, there's a question that needs the long method to prove why the integral of 1/(x^2+1) is arctan(x) 💀

One could write the integral as I = 4 \int^(2pi)_(0) dx 1/[3 + cos 4x ]. Defining 4x=t, I = \int^(8pi )_(0) dt 1/(3 + cos t) = 4 \int^(2pi)_(0) dt 1/(3+cos t). If z=e^it, This can be converted to a contour integral over the unit circle C. I = 8/i \int_{C} dz/(z^2+6z+1). Using the residue theorem, I = 8/i (2 pi i) 1/(2√8) = 2 √2 pi.

The given integral, I = (d/dt)^n|t=a-1 [ \int^1_0 dx x^t], as d/dt x^t = x^t ln x. Now \int^1_0 dx x^t = 1/(t+1). So, I = [n! (-1)^n]/a^(n+1).

The constant value with ln(-1) is not completly trivial in the sense you are working with the complex logarithm and to make it properly defined you have to remove a half line from the domain, usually from 0 to minus infinity, so its not completly obvious (still its constant I know, but it should be mentioned that it is not completely trivial)

POV: Me trying to solve the 2 mark question in my exam

It’s easier to understand things intuitively if you know the underlying logic

you could also do complex factoring (1+i)(1-i) and do partial fractions. BPRP did it this way for fun. It's also a cool way of doing it!

Can we get Gamma function ?

Love it

I usually prefer the longer methods, and trying diffetent techniques, etc. It can be a great learning experience.

Hello how can I contact with you because I have some questions about learn mathematics

Nice approach.

getting used to the amount of variables

It is important to state that this is an improper integral at 0

1 integration by parts with du = 1/xdx , v = ln(1-x) or substitution u=1-x 2. Geometric series expansion 3. Change order of summation and integration 4. Integration by parts once again

Another integral from -a to a of even(x)/(1+b^odd(x)). very common