The

[@thomasblackwell9507]

Thank you for going extra slow; it makes me feel less of a dumby.

[@maths_505]

My friend you're not dumb by any means You display a passion and discipline for learning and that is the core of intelligence.

[@sophiophile]

100%. I'm sure it's frustrating for math wizards, but this is one of the few channels tackling these problems that I don't need to incessantly pause (except for his poor writing sometimes)

[@renesperb]

I used another approach: substitute y = ln x. Then one gets in the next step the integral I of sin y/y *Exp[y]. Define I(a) = ∫ sin y/y *Exp[-a y] ( limits 0 and inf). Then one gets after simple steps I[a]= arctan a. Hence I = π/4

[@maths_505]

Already made a video on this integral using this approach.

[@MrWael1970]

Very cool integral. Thanks.

[@snow5064]

dude what a coincidence, i just solved this yesterday, our teacher gave this as a challenge, im super happy that i was able to do it

[@Charky32]

what year are you?

[@snow5064]

@@Charky32 im in high school but im preparing for the jee exam. i didnt use the feynmen technique however. i used the series expansion of e^(ilnx) to get terms. then i evaluated all the lnx terms by getting a generalized formula for int (lnx)^n (which turns out to be (-1)^n *n!), then it becomes an expansion for arctan.

[@Charky32]

@@snow5064 you are very smart, I am only in year 4 primary school, but this type of calculus is still very far away from my knowledge level, I certainly applaud you my friend.

[@nicknice7839]

@@snow5064 good luck for your JEE preparation fellow aspirant

[@forestreee]

@@Charky32 Damn that you know what this stuff even is while being in primary school is impressive!

[@holyshit922]

My approach is t = -ln(x) -t = ln(x) x = exp(-t) dx = -exp(-t)dt Int(-sin(-t)/(-t)exp(-t),t=infinity..0) Int(sin(t)/t*exp(-t),t=0..infinity) Calculate Laplace transform of sin(t)/t and plug in s = 1 To calculate L(sin(t)/t) we probably need double integral unless we know formula for L(f(t)/t) I know that purpose of this video is presentation of Leibniz rule but I have read something about differential equations lately and I can see that some integrals can be calculated the same way as Laplace transform which is defined as integral (definite and improper)

[@manstuckinabox3679]

I SMELLED COMPLEX ANALYSIS THE MOMENT YOU SAID "DIFFERENT APPROACH"!

[@maths_505]

Yeah I decided we should try being complex men instead of real men for a change😂😂😂

[@manstuckinabox3679]

@@maths_505 It's the time of the year to apply a morbius transformation on our manlyhoods .

[@maths_505]

@@manstuckinabox3679 you mean mobius right cuz I don't think there's a morbin time in math😂

[@manstuckinabox3679]

@@maths_505 lol yeah it was a "my dad is a math professor and trying to relate to the young people" kind of Dad joke.

[@ThePiMan0903]

Nice Integral Maths 505

[@maths_505]

Thanks bro Happy holidays

[@mikeoffthebox]

The Bob Ross of calculus!

[@nicogehren6566]

very nice question

[@maths_505]

Thanks Nico Happy holidays

[@nicogehren6566]

Happy hollydays !!!!!

[@ErickPolar8607]

I first use feynman technique to cancel out the ln(x), so that I made a substitution (y=lnx), after that I used euler's formula, I also got pi/4 By the way, thank you for the job you have done in your chanel, I haven't finished school yet, but I'm learning calculus by my own and videos like yours or bprp's are really helpful Sorry if my english were bad, I'm still learning

[@1-.-.-.-..-.-.--.]

My aproach : I use the series expension of sinx so that the two ln(x) cancel out then we notice that we can switch the integral and sum sign and we have : Int [0, 1] (ln(x)^2k) dx Using Feymann technique we prove that this integral is equal to (n)! if n is even since 2k is even we now have : Sum [0, infinity] ( [(-1)^k * (2k)!]/(2k+1)! the (2k)! and (2k+1)! cancel out to give us 2k+1 This sum we have is the Dirichlet beta function evaluated at 1 which is pi/4 So this integral is equal to pi/4

[@RizwanKhan-ib9xh]

We can even use Laplace Transform.

[@jorgedeadejesus9339]

Please solve Integral of Cos(lnx)dx

[@maths_505]

That's the real part of the integral of x^I You can evaluate that using the power rule

[@memeboy1057]

Here’s my solution - Perform a substitution : u = - ln x Then the integrand becomes, (sin u)/(u) * e^(-u) And the bounds go from 0 to infinity. This can be done using simple Feynman’s technique, just take the parametrisation as e^(-αu) and then the answer is pretty obvious. It comes out to π/4 (I am wroting this before watching the video so I hope I am right, otherwise, have a good laugh at how I embarrassed myself :) )

[@maths_505]

You're correct.....this is actually the 2nd video I made on this integral....the first video involved exactly the technique you've described.

[@jieyuenlee1758]

8:48 the Immaginary part of a real number is 0

[@renesperb]

Sorry,I haven't seen that video.

[@solitonacademy]

🤓