Simple Trig Subs won't Help you here...

[@NPCooking69]

Hey my bois and grills, thanks for watching

[@Hevander75]

Would love some more integrals using complex analysis

[@cristofer2794]

He hecho una imagen i made a picture of the integral resolved, a board and a known character of the loquendo videos of gta sa el tio gillipollas (the uncle asshole). imgur.com/a/FCvBh0n

[@samirgeiger1042]

Hey you said you would link something about your master thesis, but i cant seem to find it. It would interest me a great deal if you could maybe do that :D Grüße aus Mannheim ;)

[@matron9936]

Yay! Like back in the old days again.

[@PackSciences]

I tried to do it by myself. I did: 1. Use Bioche's rules: (here, the Bioche's rules fail to get the sine as a valid change of variable, because it doesn't work well to predict Taylor expansions later in the problem). No conditions of the Bioche's rules is verified, therefore the change of variable is u = tan(x/2). cos(x) = (1-t²)/(1+t²), sin(x) = 2t/(1+t²), tan(x) = 2t/(1-t²), dx = 2dt / (1+t²), the bounds become respectively 0 and +infinity (it is not really a problem). One can note that log(cos(x)) = - 2 arctanh(t²) (I am not sure it's that helpful) The integral then becomes: int_0^infty of - 2 arctanh(t²) (log(2t) - log(1+t²)) 2 dt (1-t²)/ ((1-t²)*2t) = -2 int arctanh(t²) (log(2t) - log(1+t²))/(t) dt Then I realized arctanh(t²) isn't comfortable. -2 int arctanh(t²) (log(2t) - log(1+t²))/(t) dt = - int (log(1+t²) - log(1-t²))*(log(2t) - log(1+t²))/(t) dt Then I tried to distribute: - int (log(1+t²) - log(1-t²))*(log(2t) - log(1+t²))/(t) dt = int ((log(1+t²))² - log(1+t²)log(2t) + log(1-t²)log(2t) - log(1-t²)log(1+t²))/t dt = int ((log(1+t²))² + log((1-t²)/(1+t²))log(2t) - log(1-t²)log(1+t²))/t dt Then I say "ok reference integral time" so I hoped on WA and checked: int (log(1+t²))² dt = -Li3(t²+1) + Li2(t²+1) log(t²+1) + 1/2 log(-t²) log(t²+1) + C int log((1-t²)/(1+t²))log(2t)/t dt = A gigantic expression I am not going to copy here int log(1-t²)log(1+t²)/t dt = A gigantic expression I am not going to copy here either And then at that point I said, nevermind, let's go back to earlier expressions. -2 int arctanh(t²) (log(2t) - log(1+t²))/(t) dt I used the Taylor expansion of arctanh near zero. arctanh(t²) = sum_1^infty of t^(4k - 2)/(2k-1) We get: -2 int (log(2t) - log(1+t²)) sum t^(4k-3)/(2k-1) dt Interverting sum and integral works. I know that int x^n log(x) dx exists and has a closed formula for all n. I know that int x^n log(1+t²) dx does too so I said, ok good enough (there is still the problem of what to do with the sum over k, but I assumed it was going to simplify nicely). If anyone wants to pursue this solution, feel free, but it pure calculation.

[@khemirimoez8661]

Nerd alert nerd alert

[@brucedienst7553]

Me, In Calc 1/2 😳

[@איתןגרינזייד]

I am just impressed that you actually wrote all of this

[@debajyotisg]

This may seem trivial, but finding interesting integrals, and then solving them and presenting them in a captivating fashion takes time, effort and love for the art. Thank you for what you do! Love this series a lot.

[@Yellowgary]

As soon as I saw that log(1-t^2)log(t)/t I knew Riemann was popping up somewhere. Very nice apery boi

[@aleksandervadla4840]

I tried u-sub and integration by parts. Needless to say, I am in severe pain right now.

[@sansamman4619]

0:43 : "Intermediate difficulty"

[@Noam_.Menashe]

I substituted ln(sinx( and from there it was pretty simple.

[@mudkip_btw]

Aahh yes we bac with the integarahlls, very noice one!

[@mudkip_btw]

This one is really spicy :3

[@nnniv]

0:17 you gotta make a video with that accent

[@mdmohiuddin7089]

I went all the way to the end of the video just to make sure that this is not a 17 min hahahahahaha video or something :3

[@grantplunkett8410]

Here at work on a Saturday morning, this makes it better Thanks Flammy

[@reeeeeplease1178]

Well you can also just let u=ln(sinx), du=1/tanx dx Using this sub, the integrand becomes u*ln(1-e^2u) and from there you can replace the ln by its series expansion (just like in the video) and after changing the order of summation and integral (again, just like the video), you end up with -1/2 * [Series from 1 to inf] of 1/n * [Integral from -inf to 0] u*e^(2n u) du Using integration by parts, the integral evaluates to just -1/4n^2 Multiplying this by the remaining -1/2n, you also end up with the series 1/8n^3 You can replace the ln by its series representation since |e^2x| < 1 for -inf < x < 0

[@Noam_.Menashe]

Exactly what I did.

[@tatjanagobold2810]

Yeah finally integrahals to keep my brain fit during the holidays :D

[@PapaFlammy69]

:D

[@gustavgadehebsgaard5727]

I love your integheral videos. Thank you so much, papa

[@ricardoparada5375]

Man I was missing the old integral videos. This is very nice

[@Roarshark12]

Wow, this was one thoroughly-explained integration, thanks!!

[@chan3912]

Every day I wake up, hoping that it'll be a day without pain, but then I look at your videos, and my dreams are shattered.

[@timenbobnar3148]

omg boi i had a similar problem while studying for analizis exam ....1. year on faculty of Math and Pyzsics... keep up the good work tnx :)

[@angelmendez-rivera351]

You can avoid needing to interchange the summation and integral with a limit as ε -> 1 by instead using the substitution earlier. Let -η = log(t), so -dη = dt/t, and t^(2k) = exp(-2kη). The interval of integration (0, 1) becomes (♾, 0), but the factor -1 from expanding log(1 - t^2) into -Σ{k > 0, x^k/k} transforms the interval into (0, ♾). The integrand simplifies to η·Σ{k > 0, exp(-2kη)/k}. Now, interchanging the summation and the integral is trivial, because there is no singularity, and uniform convergence is more obvious.

[@DendrocnideMoroides]

just press Alt+236 to get this infinity sign: ∞

[@JuanLopez-rl7ry]

I did this a little easier but mostly the same method int_0^(π) ln(cos(x))*ln(sin(x))*cos(x)/sin(x) dx Step 1) Let's convert cos(x) = sqrt(1 - sin^2(x)) which is valid between 0 and pi, which we are happy int_0^(π/2) ln(sqrt(1 - sin^2(x)))*ln(sin(x))*cos(x)/sin(x) dx = 1/2 * int_0^(π/2) ln(1 - sin^2(x))*ln(sin(x))*cos(x)/sin(x) dx ; Step 2) Make u = sin(x); du = cos(x) dx; a = sin(0) = 0; b = sin(π/2) = 1; 1/2 * int_(0)^(1) ln(1 - u^2)*ln(u)/u du; Step 3) Make z = ln(u); dz = 1/u du; a = -∞; b = 0; u = exp(z); 1/2 * int_(-∞)^(0) z* ln(1 - exp(2z)) dz; Step 4) Let y = -z; dy = -dz; a = ∞; b = 0; 1/2 * int_(∞)^(0) y* ln(1 - exp(-2y)) dy = -1/2 * int_(0)^(∞) y*ln(1 - exp(-2y)) dy ; We know that ln(1 - q) = sum_(n = 1)^(∞) (-1)(q^n)/n; Step 5) Let q = exp(-2y) ln(1 - exp(-2y)) = sum_(n = 1)^(∞) (-1)(exp(-2y*n))/n; which is valid when (0 < y) which is excellent because that is bounds of the integral. Step 6) Plug in the summation -1/2 * int_(0)^(∞) y*sum_(n = 1)^(∞) (-1)(exp(-2y*n))/n dy; 1/2 * sum_(n = 1)^(∞) {(1/n)*[int_(0)^(infinity) (y)(exp(-2y*n)) dy]} Step 7) Integration by parts; u = y; du = dy; dv = exp(-2y*n) dy; v = exp(-2y*n)/(-2n); int (y)(exp(-2y*n))/n dy = -y*exp(-2y*n)/(2n) + int exp(-2y*n)/(2n) dy = (-y*exp(-2y*n)/(2n)) - (exp(-2y*n)/(4*(n^2))); int_(0)^(∞) (y)(exp(-2y*n))/n dy = [limit y->(∞) [(-y*exp(-2y*n)/(2n)) - (exp(-2y*n)/(4*(n^2)))]] - [limit y->(0) [(-y*exp(-2y*n)/(2n)) - (exp(-2y*n)/(4*(n^2)))]] int_(0)^(∞) (y)(exp(-2y*n))/n dy = 1/(4*(n^2)) when n > 0 which is always true in this case as the summations goes from (n => 1). Step 8) plug back into the summation 1/2 * sum_(n = 1)^(∞) [(1/n)*(1/(4*(n^2)))] = 1/8 * sum_(n = 1)^(∞) (1/n^3);

[@abdusabdud8218]

You are my best math teacher because I like this types of vdo

[@jakubpuchatek3029]

finally papa doin some integaralz

[@antonienewman9379]

Finally integrals ..

[@let1742]

damn this is so impressive when you do it

[@PapaFlammy69]

:)

[@HAL-oj4jb]

Solving integarals really is your superpower

[@linkeshifttaste1455]

Very nice!

[@lucascruz3977]

Is there any book where I can find similar problems?

[@tgx3529]

If I use the substitution log(sin x)= t, then the substitution t=-v, I get the integral from 0 till infinity from (1/2)* v* log(1-exp(-2v))dv, I can write log(1-exp(-2v)) as series -exp(-2v(n+1))/(n+1), functions fn(v) =v*exp(-2v(n+1))/(n+1) convergent uniformly, lim sup |fn(v)-f(v)|=lim |[1/(2e*(n+1)^2] |=0, (functions fn(v)-f(v) have supremum for v=1/2(n+1), there is f(v)=0), finally I will get the same result .

[@WhattheHectogon]

infniti buoy

[@harsh_t]

:)

[@samirgeiger1042]

Is it neccesarry to turn the log(t)t^(2k-1) part into the gamma function? wouldnt intgeration by parts be way easier?

[@famillemagnan1313]

Samir Geiger of course

[@natealbatros3848]

Hey papa here's maybe a nice little question you might like. Create a function that gives out the sum of its numbers (for example f(2541)=2+5+4+1=12) I did find a solution to it but want to see if you have a more elegant solution.

[@angelmendez-rivera351]

You mean a function that gives that sum of the digits in the standard decimal representation of a number?

[@natealbatros3848]

@@angelmendez-rivera351 yes I think

[@angelmendez-rivera351]

matan guedj What is your solution?

[@aadfg0]

Recursion makes this easy. f(0) = 0, and for n>0 we define f(n) = (n mod 10) + f(floor(n/10)) = n - 10floor(n/10) + f(floor(n/10)). You can express this as a terminating series: f(n) = n - 9floor(n/10) - 9floor(floor(n/10)/10) - ... = n - 9(floor(n/10)+floor(n/10^2)+floor(n/10^3)+...).

[@natealbatros3848]

@@angelmendez-rivera351 S(n) = sum(j = 1 to (1+floor(log_10(n)))) of floor((n-((10^j)*floor(n/10^j)))/10^(j-1))

[@iambic-kilometer]

Nice work, especially with zeta(3) appearing as a surprise at the end! Minor stylistic quibble: In analysis courses, epsilon typically stands for very small positive values. I would have made the upper limit (1 - epsilon) rather than epsilon when discussing the interchange of integral and summation.

[@user-en5vj6vr2u]

Can you do calculus of variations like in the action principle

[@djvalentedochp]

I LOVED THIS SHIT

[@ilirdemiri2974]

Can you tell me reference of this problem?

[@dirichlettt]

I tried using Feynman's trick twice, seeing that cos(x)^t * ln(cos(x)) = partial(cos(x^t)) (I did the same with 1/sin(x)), but unfortunately it diverged at the very last few steps :(

[@gamingstars8956]

We are back to integrals at last

[@abhishekkp7121]

How many ways can you integrate this bad boi

[@Arnodorian6125]

Dude pls do a live Q&A!!!

[@sansamman4619]

this is our purpose in life.

[@tszhanglau5747]

I thought i won't see papa do integarahls again lel

[@stevencripe3987]

How much math do you generally need to know to solve the sorts of integrals on your channel? I am just starting Calculus III and I am completely lost haha. I of course can easily follow all your steps but oh lord I couldn't get there on my own.

[@rokarus7658]

None of the steps are hard in and of themselves; like proofs it's just experience working these types of problems to know which approaches to take.

[@sdparsons]

360 noscopes are too e4sy losers. Then I see (Ta)n→(Si)n holy cow that move too sick

[@valentinthieriot7506]

x=tan(t/2) vroum ?

[@tgeofrey]

Imperial mathematica

[@antoine5571]

PRO GAMER MOVEEEE

[@gym5959]

sinx=e^t simplifies it in someway

[@maqsudxorazm]

Itʼs great

[@maxwellsequation4887]

Yess, this is not a hahahaha video

[@Dr.1.]

niceeeee

[@epicmorphism2240]

Finally an integral… Edit: I‘ll post my way tomorrow

[@PapaFlammy69]

nice! :D

[@oofusmcdoofus]

*Whats an infinity boy*

[@RanEncounter]

7:30 he is refering to the infinite sum he is pointing. Just a meme form to say anything infinite related.

[@Alpha-gz6hk]

Allauakba🤣

[@carlosgiovanardi8197]

Nice integrahal!! And you are very red.

[@justacutepotato2945]

I'd like it when the answer was like a well known number.... that's probably cos i'm at like beginner level integration.

[@hoodedR]

Sin >> squiggly stuff

[@BoringExtrovert]

الله أكبر again

[@amirbannouri4642]

why u always write the congruent symbol instead of '=' , that's really annoying for arithmethicc lovers !

[@angelmendez-rivera351]

It is a multipurpose symbol in math, like most symbols. It shouldn't be annoying if you understand how it works. It means "identically equal."

[@KillianDefaoite]

It is often used to mean a definition. Like, "let q=sin(t)" is often written with a triple equal sign because q=sin(t) is not a property of q that we have discovered, rather it is it's definition that we have simply stated. Sometimes ":=" is used for this purpose. The triple equals is often used in modular arithmetic as well, personally I find however that "p mod q = x" is a much better notation than the common " p == x (mod q)". ["==" Is standing in place for triple equals]

[@futznation4230]

Should have written ln

[@youtuberdisguiser6075]

Wie ist das Wetter :P

[@PapaFlammy69]

viel zu fucking warm ;_;

[@youtuberdisguiser6075]

@@PapaFlammy69 feel you.

[@trigon7015]

Latural nog

[@Jeanstravels]

Damn iam 16 hahafjkl trying to understand this

[@drjohnsmith5282]

Division by zero = allahu akbar!

[@stefanoclaes1970]

Trig subs easy ??????

[@trifonmadas2215]

I am a bit perplexed watching this... People that can follow this are sitting "high" in the mountain of maths, and therefore will take little interest in this video People that are sitting "low" in the mountain of maths can hardly follow anything, and therefore also take little interest in this video So what is the target audience of this video? Any thoughts?