a nice integral and an important ignored constant

[@freecky1621]

It is still unknown that if Catalan's Constant is rational or not. There's many interesting expression of Catalan's Constant. One of my favorite is integral of 2*ln(2*cos(x)) from 0 to pi/4.

[@MathSolvingChannel]

I got another way to solve this problem😎: kzbin.info/www/bejne/l5-lkGqvpLh2mck

[@samueldevulder]

4:14 a sqrt() is missing. We should have sqrt(1+y²) as the length of the hypotenuse.

[@mabrouksalah4138]

Sin(×)=2y/(1+y.y) ^2

[@yassinezanned9837]

@@mabrouksalah4138 actually if you add the missing square root you'll get the form 2y/(1+y.y) which Pr.Penn found because it will be the product of two square roots in the bottom.

[@blackdeath39muffin45]

Catalan's constant is really useful when dealing with integrals related to polylogarithms, it's related to the Lerch transcendent and the Legendre Chi function as well

[@MathSolvingChannel]

2:57 Weierstrass substitution, I did another method to solve this problem, Link: kzbin.info/www/bejne/l5-lkGqvpLh2mck

[@goodplacetostop2973]

11:04 Good Place To Stop 11:06 Behind the scenes?

[@karolakkolo123]

@@kostasbr51 he probably has a computer script which makes a comment automatically whenever a new video is posted lmao

[@karolakkolo123]

@einstein9073 yeah but he's ALWAYS first and I doubt that he just sits there all the time waiting for a notification to pop-up. Or maybe he does...

[@MathSolvingChannel]

a bot?😂

[@ChaineYTXF]

superb. And thank you for writing arctan(x) and not tan^-1.

[@perrydimes6915]

Neat trick with the overlapping audio + video tracks. I don't think I noticed that before but it keeps up a neat rhythm.

[@Roger_Mansuy]

It seems a bit tricky to "hide" the uniform convergence argument to justify the series/integral inversion

[@JCCyC]

I started playing with the integral you obtained by variable substitution, but to any upper bound x rather than 1. The resulting function, for big numbers, can be approximated as the constant arctan tends to (pi/2) multiplied by the integral of 1/x, aka ln. Therefore, if you divide that function by ln x, the resulting function tends to pi/2 on infinity.

[@MathFromAlphaToOmega]

In general, I don't think there are nice formulas for sums like 1/a^2k-1/(n-a)^2k+(n+a)^2k-(2n-a)^2k+... where the sum is over integers congruent to +/-a mod n. Similarly, sums with all plus signs don't usually have closed forms when the powers are odd. This has to do with special values of L-functions, which I'm not an expert on, but maybe there are counterexamples to this rule.

[@theimmux3034]

How do we know we aren't just too stupid to find a closed form representation of that series? How do we know that it is a new constant of it's own? Like what if we never solved the Basel problem and just called it a new constant?

[@TheEternalVortex42]

We don't know. It could be that one day someone finds a different expression for it in terms of other constants.

[@manucitomx]

I very much enjoy these substitutions. Thank you, professor.

[@gregsarnecki7581]

I was very recently looking for a solution to the integral of x/sinx after reading a paper by Silagadze from 2018, "Sums of generalized harmonic series for kids from five to fifteen". In it he gives the expression 7*zeta(3)=integral of {(πx/sinx) - (x^2/sinx)}dx; alas from π to 0, not π/2 to 0. Still, your video came at the perfect time! Keep up the great videos!!

[@pierreabbat6157]

How does Catalan's constant compare to Occitan's constant, Valencian's constant, or Balear's constant?

[@chessematics]

2:29 WEIERSTRAẞ!! Another underrated method in calculus 2.

[@waldomcnaldo7306]

I took Prof. Bradley's discrete mathematics course while I was there. It's cool to see some of his work has been featured in a video!

[@holyshit922]

I would calculate it following way Integration by parts with u = x and dv = 1/sin(x)dx (Here 1/sin(x) is quite easy to integrate when we use double angle formula 1/sin(x) = 1/(2sin(x/2)cos(x/2)) = 1/(2sin(x/2)/cos(x/2)cos^2(x/2)) 1/sin(x) = 1/(2tan(x/2)cos^2(x/2))) Now I would use substitution u = sin(x)/(1+cos(x)) Then finally I would use power series (In fact it is geometric series with common ratio -u^2)

[@Fortitudeize]

If you do this integral by substituting the complex exponential identity for sin(x) you find a nice proof for the value of sum(n=0,inf) 1/(2n+1)^2 by comparing the real and imaginary parts of the integral. It also uses the Leibnitz-Madhava sum.

[@SabaSa6a]

It turns out that integral of atan(y)/y over 0 to x is a special function called Inverse tangent integral Ti_2(x). It is easy to derive its power series representation (by using that of inverse tangent function): Ti_2(x) = 1 - x^3/3^2 + x^5/5^2 - x^7/7^2 + ... and then Ti_2(1) = G

[@insouciantFox]

Leibniz rule is simpler here: Int_0^1(arctany/y)dy = int_0^1int_0^1(y²z²+1)^-1dydz = int_0^1int_0^1sum((-1)ⁿy²ⁿz²ⁿ)dydz Because f(y) and f(z) are ind. and equal, = sum (-1)ⁿ[int_0^1(y²ⁿzdy)]²= sum (-1)ⁿ/(2n+1)² = G ■

[@IbrAhMath]

Did I miss it or why was it okay to change the series and the integral?

[@Wabbelpaddel]

Continuous function in the sum and uniformly converging sum

[@DeanCalhoun]

y is bounded between 0 and 1, and z is bounded between 0 and y. This means z

[@Willtedwards]

Could you have also opted for an IBP to bring tan(x/2) into play?

[@adandap]

Can this be done more directly using the Laurent series for cosec(x)? Seems like you should be able to, though the series for cosec isn't all that pretty, and involves Bernoulli numbers.

[@bobzarnke1706]

SInce the initial substitution y=tan(x/2) wasn't obvious, I integrated by parts, knowing that ∫csc(x)dx = ln(tan(x/2)). This makes the integral equal to -∫[0,π/2] ln(tan(x/2)) dx, where the substituting y=tan(x/2) is more obvious and gives -∫[0,1] ln(y) dy/(1+y²) (which I later learned was a known integral for G). How to integrate this wasn't clear but, after watching the video and learning that the result was an infinite sum, I expanded this to -∫[0,1] (ln(y) - y²ln(y) + y⁴ln(y) - ...) dy, each term of which can be integrated using ∫yⁿ‾¹ln(y)dy = yⁿ(ln(y) - 1/n)/n; evaluating the result at the limits uses lim[x->0] xⁿ ln(x) = 0. (Alternatively, replacing ln(y) by ∫[y,1] dz/z makes it equivalent to the above after changing the order of integration,)

[@lukandrate9866]

-int[0,1] ln(y)•dy/(1+y²) can be again integrated by parts and then you just use the arctan series

[@tonywong8677]

I solved it by using complex integral. Z = exp ix, x = -i ln Z, sin x = 1/2i * ( exp ix -exp -ix ), Z from 1 to i, then equating the real part of the integral to get the same answer. I find out another interesting result when I equated the Imaginary part to zero. It shows that Summation of (-1)^n/(2n+1) from 0 to infinity is equal to pi/4. Which means pi can be written as an infinite series. Is there anything wrong with my calculation ?

[@pineapplegodguy]

I did a monte carlo integration of 1/2 x/sin(x) in the (0, pi/2) interval to find out if the result matched the constant... and it does not! import random from math import pi from math import sin somma = 0 for i in range(10000): x = random.uniform(0,pi/2) f = (1/2)*x/sin(x) somma += f integra_f = somma/10000 print(integra_f) the program returns something in the range of 0.5827 instead. any idea why?

[@pineapplegodguy]

ok if anyone is interested, it's simply a matter of multiplying the integral by the length of the interval (so, pi/2) the following code gives the correct answer import random from math import pi from math import sin somma = 0 for i in range(10000): x = random.uniform(0,pi/2) f = (1/2)*x/sin(x) somma += f integra_f = (pi/2)*somma/10000 print(integra_f)

[@michelebrun613]

At minute 4:28 you should add a square in the drawing

[@__hannibaal__]

In past i studied too much look like this integral with more parameter : result is fascinated

[@sushildevkota350]

This concept of converting integral into summation form and finally the integral is the best approach for tough calculus problem. So this question has been solved by me from this idea. I know this idea from olympiad and penn's technique.

[@erikross-rnnow5517]

Lovely video :)) Maybe I'm just slow, but I'd love to have you talk us through some of the later calculations just a tiny bit more since I get lost without pausing the video to ponder how you got to the next equation. This is not necesarily a critique of your style because the lack of deeper explanations cut through to the interesting connections between the infinite series and a somewhat arbitrary looking integral at first, anyways, just an idea from a lowly pleb.

[@thehokkanen]

Could you evaluate the integral of arctan(y)/y from 0 to 1?

[@maxwellsequation4887]

He literally just did

[@thehokkanen]

@@maxwellsequation4887 I mean evaluate the exact answer

[@davidblauyoutube]

@@thehokkanen The exact answer is Catalan's constant. It is unknown if this constant is rational.

[@coc235]

@@thehokkanen It doesn't have a closed form aother than just G, it's approximately equal to 0.916

[@richardheiville937]

No close form does exist for this, this is a constant like e, Pi.

[@Bodyknock]

One thing I'm curious about but can't seem to find the answer to is why they chose the letter G for Catalan's constant? Obviously there's overlap in constant labels (Gauss' constant is also G, Cahen's Constant is a C, etc), but often if it's a capital English letter it's the first letter of the name of the constant. There must have been a reason for the G, I wonder what it was?

[@ojasdeshpande7296]

26 letter of the alphabet seem too less to us to use repeated notations and also Greek. Symbols

[@brahimsebbata9036]

Fine intégrale good explication

[@d4slaimless]

Why, oh why we use the serious expansion...

[@ojasdeshpande7296]

Cebecoz y not

[@mathematicsmi]

I already published the solution of this question before 5 months.. good solution btw

[@The1RandomFool]

Complexifying sin and using geometric series to evaluate the integral leads to yet another solution to the Basel problem. It is present in the imaginary part, which must be 0.

[@cernejr]

Approx 0.916

[@redwanekhyaoui7232]

Hello Michael. You usually switch intergals and sums (interchanging) but I think we are not allowed to do this always. It is under some conditions. Could you please make a video about this situation with an easy clear example? 😊 Keep up the good content!

[@InstigationMex95]

If the sum does not contain a variable for which we are integrating you can take it outside of the integral like a constant :)

[@vaddiparthyyyogeswara6921]

Good observation. Even though it may be a correct one, explanation can not be skipped

[@miro.s]

It would be nice to show it every time when it is applied after It’s a good place to stop

[@failsmichael2542]

If it works, it works.

[@mohamedfarouk9654]

When you think it may be irrational, then it is.

[@Pablo360able]

Do you think abs(ln(i^i)/pi) is irrational? Don't use context clues.

[@giuseppemalaguti435]

Non ho capito perché l'integrale di artangy/y l'hai fatto così complicato.... Non potevi fare un unica serie di potenze di arctg diviso y senza fare un doppio integrale? Grande Michael

[@DavidSavinainen]

You could do, if you know the series representation of arctan(z) by heart. Not everyone does, but most people know the geometric series, so he uses that to derive the arctan(z) series.

[@spudhead169]

2nd November is Trying to be Tom Scott day.

[@laurentpaget5388]

x = 4.(3-2.sqrt(2)) = .686

[@Fun_maths]

Hi, love the math and all but I have s different question, how do you make your thumbnails?

[@MichaelPennMath]

I use coolors.co/ to make a color scheme and place everything in canva "by hand". For mathematical symbols, I copy and paste the unicode.

[@Fun_maths]

@@MichaelPennMath Thank you!

[@txikitofandango]

video duration is 11:11

[@inyobill]

Isn't there a discontinuity of the function at x = 0?

[@anonymous-xm4gx]

Sir can you please tell what is the utility of this integral representation?

[@adityaekbote8498]

Art is its own practicality

[@maxwellsequation4887]

@@adityaekbote8498 yes

[@nandeesh2ninad]

I am faster