Another ridiculously awesome integral with a beautiful result

This integral combines 2 of my favourite integration techniques so I had a wonderful time solving it

[@taterpun6211]

Favourite constants throughout a lifetime boys: π teenagers: e men: γ

[@kamalsaleh6497]

As a teenager, my favorite number is the Euler Mascheroni constant.

[@trelosyiaellinika]

You are a true MIRZA(DEH) of integration! I deeply enjoy your videos. Even though I watch and take notes from the videos and lectures of various people on the net, often comparing different approaches, yours have a special place. You are a true teacher... Bahat shukriya!

[@joshuaiosevich3727]

I’m using your videos as lectures to get better at math, you do great work man!

[@archinsoni1254]

This was a very high level integral.

[@davidblauyoutube]

8:45 Or you could just note that this is the Laplace transform of sin(x), which you can look up in a table or recall from memory. 😁

[@maths_505]

"You dare use my own spell against me Potter"😂 The idea is extremely efficient but this was fun

[@MrWael1970]

Awesome Integral and awesome solution. Good Luck

[@redroach401]

3:14 (pi) - Why not proceed by getting imaginary part of integral and then ix=-u to get gamma function and from there you can immediately get -pi/2 * gamma

[@javiergilvidal1558]

The integral at 9:08 will diverge unless s < 2. You can suppose that, because you are interested in (I)´(1), so the end computations will involve specialising s at 1, but it´s important that you say so explicitly. Therefore, in the exposition we must specify 1 less than or equal to s < 2.

[@herbertdiazmoraga7258]

this is video is epicc!!

[@margozamora5573]

Saludos desde mexicooooo, me encanta tu contenido, sigue así🤩😊

[@ayoubnasmi6794]

You could use Ramanujan master theorem

[@slavinojunepri7648]

Excellent

[@manstuckinabox3679]

EYOO its the integral we did together

[@MaalikJohannes]

what..? you're memorising specific integrals? you cant be serious......become a priest or something lmao

[@manstuckinabox3679]

@@MaalikJohannes LOL! it's funny cuz I actually wanted to be a priest, in all seriousness I do believe I have somewhat of the human capability of evaluating the quality of certain decisions, while deciding to memorize specific integrals could be observed (trivially)as a blunder of one's time, comprehending the process and memorizing the logical evaluation of the method used to solve said integral, would indeed aid in quickly noticing similar integrals that fall under the same "umbrella" (I'd consider the general way of seeing this is that this integral falls under the family of derivatives with respect to s of the integral(f(x)/x^s) where f(x) is the function denoted in Ramanujan's Master theorem)). in summary, Lol no I'm not serious but I do have good memory so I automatically memorized the Ramanuj-variation of the solution because it's the one I used.

[@MGoebel-c8e]

10:20 I can’t wrap my head around this integral representation of the reflection formula. This seems to diverge for all positive exponents. Did you forget to mention a restriction on z? Would be helpful to clear this up a bit since the formula itself seems to be very helpful. Thanks!

[@Circuito28]

Should it be possible to with feynman choosing ln(ax), differentiating for a we obtain pi/2a for I'(a), now I'm blocked because I was trying to obtain I(1) as it's the solution be integrating I' between infinity and 1, on the left hand side I obtain I(inf) - I(1) = pi/2 (lim a->inf of ln(a)) which is ALMOST the euler mascheroni😂 I miss the zeta(1) term which Im sure it's cointained in I(infinity) but that I cannot evaluate, any ideas?

[@minhnguyen1338]

elegant

[@nguyenquangtuyen6626]

I love it

[@robertlittlejohn8666]

I think I've got a derivation that's more straightforward. At least it doesn't involve so many technical gamma function results. Let me know if you want to see it. Basic idea (Feynman's method etc) is the same however.

[@unknownhero6187]

From where does the rule about partial derivatives come? Why if the integral converges we can switch the operators?

[@insouciantFox]

Dominated convergence and Fubini's theorem

[@samirhamdi1557]

What pen and tablet are you using for your pencast video?

[@puceno]

wow

[@FilippoCutaia]

Using lobochevsky integral?

[@maths_505]

Nope

[@FilippoCutaia]

@@maths_505 onsari9

[@bilkishchowdhury8318]

Lobochevsky kaisen

[@NurHadi-qf9kl]

|sin x {d(ln x)^2}/2= =(1/2)sin x( ln x)^2+ -(1/2)|(ln x)^2cos x dx= =...-(1/2)|(sin x){(ln x)^2} + +(1/2)|sin x d{(ln x)^2} Jadi |=(1/4)(ln x)^2(cos x+sin x)+ +C Lalu dimasukkan batas2nya.

[@giuseppemalaguti435]

Con lnx e infinito ci rinuncio subito

[@FilippoCutaia]

Socio,puoi usare semplicemente l'integrale di lobachevsky, dove integrale da 0 a infinito di f(t) per sinx/x dx è uguale a integrale da 0 a π/2 di f(t) dx. in questo caso diventa integrale da 0 a π/2 di lnx dx, che è xlnx+x, facciamo le sostituzioni appropriate e otteniamo che il risultato è π/2 ln(π/2) + π/2 - 0ln0 + 0, che è semplicemente {π/2 ln(π/2) + π/2 + C}

[@abi3135]

Found a way to do it using laplace transform Using, ℒ (ln(x)) = -(γ + ln(x))/x ⇒ ℒ (-γ - ln(x)) = ln(x)/x ∫ sin(x)ln(x)/x → ∫ sin(x)ℒ (-γ - ln(x)) → ∫ ℒ (sin(x))(-γ-ln(x)) → ∫ (-γ-ln(x))/(1+x²) → -γ ∫ 1/(1+x²) = -γπ/2

[@Impossiblegend]

Tools he used: Laplace transform, Euler's reflection formula, Feynman's technique/ differentiation under integral, more gamma and digamma related identities, change of order, and substitution. Tools you used: Laplace transform, Catalan's constant identity, arctan formula. I think it's safe to say yours is better :P

[@pardeepgarg2640]

Suggestion: Integral from 0 to infinity (ln(tanx))^2 dx Result : (π/2)^3 or (π^3)/8

[@eroi1963]

LETSGOOOO

[@finnr3472]

Cool video but whats not beautiful that you put parentheses at sin but not at ln. Either both or none.

[@maths_505]

Got it boss

[@finnr3472]

@@maths_505 sry for the useless comment ;)

[@maths_505]

No comment is ever useless from any one of you. All of you are amazing and you have no idea how much the support of each and everyone here means to me

[@Mario_Altare]

My attempt/try: sin x = [e^(ix) - e^(-ix)]/2i f(t)=\int_0^infty [e^(ix) - e^(-ix)](ln x)/x dx, f(-i)=0, [f(i)]/2i = I f'(t)=\int_0^infty [e^(-tx) ln x] dx tx = u dx = du/t f'(t)= - γ/t - (ln t)/t f(t) = -γ ln t - (ln^2 t)/2 + C f(-i) = 0 --> C = - πγi/2 + π^2/8 f(t)= -γ ln⁡ t - (ln^2 ⁡t)/2 -πγ/2 i+ π^2/8 I= [f(i)]/2i = (-γπi)/2i = -γπ/2