Supreme Integral with Feynman's Trick

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Supreme Integral with Feynman's Trick

Рет қаралды 209,915

6 жыл бұрын

We will do the integral of sin(ln(x))/ln(x) from 0 to 1 by using Feynman's Trick (aka differentiation under the integral sign). This is also closely related to the Dirichlet Integral, which is the integral of sin(x)/x from 0 to inf.
For the complex exponential definition of sine 👉 • Complex definitions of...
For another integral by Feynman's Technique: 👉 • Feynman technique: int...
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Пікірлер: 396

@jlxip 6 жыл бұрын

17:20 "So, this right here is pretty much the answer but what the heck in the world is this?" I'm crying 😂

@SteamPunkLV 6 жыл бұрын

from now on, I'll write that instead of pi/4 😂

@dijkstra4678 3 жыл бұрын

math video: so basically that's the answer. me: ok but what the heck in the world is this?

@JoJoJet100 6 жыл бұрын

I REALLY like it when you do improper integrals. It's so much more satisfying to get an answer that is an actual number instead of a bunch of math functions added together.

@blackpenredpen 6 жыл бұрын

: ) I actually like indefinite integral more tho

@ripjawsquad Жыл бұрын

@@blackpenredpen fr

@atheybengala5720 Жыл бұрын

@@blackpenredpenfr

@microscopicallysmall 10 ай бұрын

@@blackpenredpen fr

@ali-zl9ls 8 ай бұрын

@@blackpenredpenfr

@mortadhaalaa5907 6 жыл бұрын

I did it quite peacefully using feynman's trick with the parametrization: I(t) = sin(t lnx) / lnx I guess I've been watching too much flammable maths vids 😂 awesome video nonetheless 🍫

@hetsmiecht1029 4 жыл бұрын

I think your solution is more elegant, as it doesn't require complex numbers inside natural logs (which can have infinitely many values).

@euva209 3 жыл бұрын

Nice! It leads to dI/dt = (1/t) ∫e^(u/t)cos(u)du from -∞ to 0 = 1/(t²+1); After integrating you get I = arctan(t) + C; I(0) =0 = C ; I(1) = π/4

@GauravGupta-by1ml 3 жыл бұрын

Same bro.. And it doesn't involve complex numbers in any way so simple, but a little longer..

@k_wl Жыл бұрын

idk how it works, you would have to evaluate it from 0 to 1, you cant have 0 inside the ln tho even if u say t = 0 the integral has a 1/t after solving so can you explain please?

@user-hp2dr5qc8p Жыл бұрын

@@k_wl I = {0,1}∫ sin(ln x)/ ln x dx F(t) = {0,1}∫ sin(t*ln x) / ln x dx => F'(t) = {0,1}∫ ln x * cos(t*ln x) / ln x dx = {0,1}∫ cos(t*ln x) dx After solving (I spent like 15 minutes and couldn't figure it out tbh, so I just used wolfram) you get: F'(t) = 1 / (t^2 + 1) We see that F(0) = {0,1}∫ sin(0) / ln x dx = {0,1}∫ 0 dx = 0. Therefore: I = F(1) = F(1) - F(0) = {0,1}∫ F'(t) dt = {0,1}∫ 1 / (t^2 + 1) dt = arctan(1) - arctan(0) = π/4

@debrajbanerjee9276 6 жыл бұрын

You can more easily do this by substituting ln(x)=-y which will leads to ...... I=∫(sin(y)e^(y))/y dy from 0 to ∞ now breaking sin(y) into taylore series and pulling the sigma notation out from the integral the integral will be a gamma function of (2n)! At last dividing it by (2n+1)! You will get series of arctan(u) with u=1 which immediately says that I=π/4

@fengshengqin6993 5 жыл бұрын

yeah ! right ! Good!

@ianmoseley9910 5 жыл бұрын

"more easily" - 😳

@trace8617 5 жыл бұрын

Ian moseley easier as in doesnt require complex analysis and identities such as ln(i)

@BY-sh6gt 5 жыл бұрын

Anyway how can u write integral sign in the comnent? 😂

@FotisValasiadis 5 жыл бұрын

ik its a bit too late folks,but i solved it in 5 minutes.Set u=lnx dx=e^u du its now (sinu*e^u)/u,use feynman's method to get rid of u by writing the integral as (sinu*e^(uy))/u and solve.You will end up with a simple sinu*e^(uy).Use the DI method and by the end of the day you end up with a -1 over (y^2+1) so you just know its an inverse tangent.you get that the original is minus inverse tangent plus π/4 so if you replace y=1 you get π/2-π/4=π/4.without any complex numbers having to step in

@snejpu2508 6 жыл бұрын

U world is not powerful enough, but b world solves a problem. : ) YAY!

@blackpenredpen 6 жыл бұрын

Yup

@hervesergegbeto3352 3 жыл бұрын

Good morning sir How to become member ?

@blackpenredpen 6 жыл бұрын

It's 1:54am here. Good night!!!!!!!!!!!!

@CarDealersdotcom 6 жыл бұрын

Have a problem Mr D

@yoavcarmel1245 6 жыл бұрын

I solved it using I(t)=integral from 0 to 1 of sin(t*lnx)/lnx and got that right, maybe you could please upload a video using this method? If you would like, i can send you the picture of the solution somehow

@blackpenredpen 6 жыл бұрын

Flammy did that already... like 2 hrs after my upload, lolll

@yoavcarmel1245 6 жыл бұрын

blackpenredpen oh lol. Well done to him i guess :) will watch his video soon

@yoavcarmel1245 6 жыл бұрын

blackpenredpen no he did it different than me, i didnt use imaginary nums

@Jack_Callcott_AU 2 жыл бұрын

Hey BPRP I really enjoyed that. It is very satisfying when complex maths leads to a simple result.

@ryanhurst5096 4 жыл бұрын

Very creative problem solving process you used on this integral!

@mith_jain_here 2 жыл бұрын

I was wondering the whole time how can an integral of a real function have a complex answer, but at the end when the answer simplified I was so relieved 😂. Maths is indeed beautiful.

@CornishMiner 6 жыл бұрын

Some great techniques used to find a very satisfying answer. So good :)

@blackpenredpen 6 жыл бұрын

Yay!!

@axemenace6637 6 жыл бұрын

This integral is very similar to 0 to infinity of sinx/x after the substitution x=e^u and the substitution I(a)=integral from -inf to 0 of e^au(sinu)/u. We want I(1). Feynman's technique solves this for us.

@blackpenredpen 6 жыл бұрын

Yea!

@ariusmaximilian8291 6 жыл бұрын

Yay! This was so cool!! Thx for putting it up

@Tranbarsjuice 6 жыл бұрын

Really cool integral and a very nice explanation

@ayoubfenkouch5992 6 жыл бұрын

this is why i like you videos , even if you understand the lesson very well you always surprise us with some tricks , but i have a question ( to you and to whoever reads this and can anwer me ) : when to think of such a method ? how to know if taking an integral to the complexe world and B world will give a results ? is there some hints within the integral ?

@weerman44 6 жыл бұрын

Awesome integral! Thanks :D YAY

@galgrunfeld9954 6 жыл бұрын

Wow, that was so awesome! I haven't learned complex analysis, so I wouldn't think of expanding the scope to Complex numbers, that was clever! And when you zoomed in and I calculated the answer in my head, I was like "PFT, WHAT" and laughed, because the answer was so simple compared to how you solved it. One of the best videos of yours I've watched so far! :D

@blackpenredpen 6 жыл бұрын

Thank you!!! I am glad that you enjoy it!

@ChefSalad 5 жыл бұрын

You don't need complex analysis to learn about complex integration with exponentials. It's usually taught when doing differential equations. The reason is that it's way easier to do nonhomogeneous second order linear DE's using e^(ix) than with sin(x) and cos(x). To do ∫sin(x)dx, for example, you just do Im[∫e^(ix)dx] = Im[1/i*e(ix)] = Im[−i*e^(ix)] = −cos(x). It′s a bit overkill on a regular integral, but when doing nonhomogeneous second order DE′s, it′s a dream compared to the alternative method of undetermined coefficients. If you′re wondering what that looks like, I′ll give an example. Take x′′+2x′+x=sin(t). The characteristic equation is thus p(r)=r^2+2r+1=0, which means r=−1, twice. That′s makes the complementary solutions y₁=e^−t and y₂=t*e^−t. For the particular solution we can complexify the sin(t) as e^(it), thus α=i. We know that the particular solution has the form y*=e^(αt)/p(α), which means y*=e^(it)/(i^2+2i+1)=e^(it)/(2i)=−i/2*e^(it)=1/2*sin(t)−i/2*cos(t). Thus a particular solution is yₚ=−1/2*cos(t) and the whole solution is y=C₁e^−t+C₂te^−t−¹/₂cos(t). Finding the particular solution without using complex exponentials would involve solving a system of three equations or, even worse, a system of two equations with two integrals. This way just requires us to remember a simple rule.

@leif1075 4 жыл бұрын

Bit,can,younactually solve this without just knowing those formulas?

@hassanakhtar7874 4 жыл бұрын

@@leif1075 stop being salty, you are misled because you got high marks in baby level math.

@leif1075 4 жыл бұрын

@@hassanakhtar7874 i wasnt being,salty..I,asked an intelligent question...to,see how to actually solve,this.why cant you see that..

@colinjava8447 2 жыл бұрын

That's incredible, never seen that before, feynmann was a legend.

@jorgesponja3042 6 жыл бұрын

#YAY OMG I love how insane integrals ends with simple answers like pi/4 lol

@blackpenredpen 6 жыл бұрын

yay!

@srpenguinbr 6 жыл бұрын

First, I used u=ln(x) then used the feynman technique with I(t)= int from -inf to 0 of (sin(u)e^ut)/u

@blackpenredpen 6 жыл бұрын

Yup!!!

@omerangi4695 6 жыл бұрын

That was very long and a very beautiful integral.

@quidam3810 2 жыл бұрын

Great video !!

@TheBlueboyRuhan 6 жыл бұрын

Good luck jaime for further maths

@renesperb Жыл бұрын

A different approach is to set t = ln x . Then you get the Integral of sin t/t*Exp(-t) ,( limits zero and inf.). Setting I[a]= sin t/t *Exp[-a*t] you can use Feynman's trick now to find the result π/4 .

@iOhadRubin 6 жыл бұрын

That was actually pretty cool.

@varunsahni2128 4 жыл бұрын

The general answer would be n*pi + pi/4 where n is integer. Which also state that area of this curve can take variable values

@yusufmia 2 жыл бұрын

Hi. Your work is awesome

@silasrodrigues1446 6 жыл бұрын

Oh my Gosh! This was really awesome! Brazilian congrats! #YAY

@blackpenredpen 6 жыл бұрын

Thanks!!!

@Linkedblade 6 жыл бұрын

That was a wild ride from beginning to end

@alanturingtesla 6 жыл бұрын

Yay, I love these 20-minute integral videos!

@blackpenredpen 6 жыл бұрын

Yay!!!

@AmanSingh-rg7hk 4 жыл бұрын

Again tesla for turing.

@alfredod.cadionjr.7035 4 жыл бұрын

Video

@kunalbatra4166 6 жыл бұрын

loved this one..

@SVKODURU2008 3 жыл бұрын

put ln x =-y, then, 0 to inf ∫e^-y siny /y dy= lim s=1 , s to inf ∫1/(s^2 +1) ds = π/2 -π/4 =π/4 , (Laplace)

@The1RandomFool 3 жыл бұрын

It's true that the limit as x approaches zero of x^a is undefined if the real part of a is zero. However, the limit becomes 0 if the real part of a is greater than zero. This can be proven by showing the limit as x approaches 0 of | x^a | is 0. Therefore, his step of 0^(1+bi) = 0 is valid.

@thatpersononline 2 жыл бұрын

It's quite easy with Feynman's rule. I did it with g(t) = int sin(tlnx) dx/lnx evaluated at 0 to 1. Then evaluated g'(t) and complexified it. Pretty easy

@charlesnuett6674 6 жыл бұрын

Hey blackpenredpen, what books or bdf books do you propose when one want to study integration ( I mean calculus in general) to the fullest just like you. So he can Know quite a lot. Pls 🙏🙏🙏

@jasperh6618 6 жыл бұрын

that was one heck of an adventure

@blackpenredpen 6 жыл бұрын

: )

@SameerKumar-jf5mi 4 жыл бұрын

this was fun! but in the end how do you know that log i yields π/2, and not something like 5π/2 ?

@Saki630 5 жыл бұрын

What a wonderful first question on my exam.

@Hobbit183 6 жыл бұрын

More multivariable calculus videos would be neat 🤙

@mihaipuiu6231 Жыл бұрын

Beautiful solution!

@gagadaddy8713 6 жыл бұрын

Master Cao, u r way too clever ... Thanks a lot!

@General12th 6 жыл бұрын

Very nice!

@user-zo1mk3ek6m 4 жыл бұрын

This is so perfect

@jpradeesh3800 5 жыл бұрын

If u know laplace transform, then proceed this way Put - ln(X) =t

@abdullahalmosalami2373 5 жыл бұрын

How does that help? Laplace transform I mean.

@jpradeesh3800 5 жыл бұрын

@@abdullahalmosalami2373 you will an integral of form f(t) /t for which we have a formula. Then substitute s=1

@richtw 6 жыл бұрын

Awesome!

@wintersummers3085 6 жыл бұрын

Math for its own sake is beautiful. Thanks blackpenredpen

@blackpenredpen 6 жыл бұрын

Winter Summers yay!!!

@dgrandlapinblanc 5 жыл бұрын

Great. Thanks

@marcioamaral7511 6 жыл бұрын

Great journey !

@ikaros4425 6 жыл бұрын

this is the kind of content I love to see, also why are you up so late???

@qbetech4764 5 жыл бұрын

This can be also be done with F(a)= sin(alnx)/lnx with F(1)= I pretty easily. But complex world looks amazing.

@skeletonrowdie1768 5 жыл бұрын

I can only conclude it converges because at x=0 the integral is 0 (lnx>x for 0 to 1 domain). And finite everywhere else till x=1.

@skeletonrowdie1768 5 жыл бұрын

lol i already watched this video

@EMorgensztern 6 жыл бұрын

can you prove the continuity of y=x^(1/x) pls ?

@gnikola2013 6 жыл бұрын

Are there any restrictions for differentiating under the integral sign or can it be done each and every single time?

@user-wu8yq1rb9t 2 жыл бұрын

*Super Duper Cool* Thank you so

@6612770 5 жыл бұрын

Wow, but Phew! I'm exhausted after watching that marathon.

@blackpenredpen 5 жыл бұрын

: )

@kmac5912 6 жыл бұрын

May you please make a video on how to solve for x=y^2+x^2y y=x^2+y^2x

@rezanasiri7353 6 жыл бұрын

hello bro, your vids are awesome! i have a question, integrate of 1/(sin^2(x).cos^4(x)) thank you, plz answer this i know this is simple af, but somehow in the world i couldn't figure it out and didn't find a solution for it, thanks, you are awesome!#yay

@spudhead169 Жыл бұрын

ZOMG! That was a ride.

@dimosthenisvallis3555 6 жыл бұрын

That was pretty awesome. Hey, could you tackle this one--> 4(x^2) + x + 1 = ((2x - 1)^(1/2))((x+1)^(1/4)). i was able to prove there r no real solutions. but what about complex solutions. I think this could be an intresting video. Keep up the awesomeness

@srpenguinbr 6 жыл бұрын

Wolframalpha told me 0^i is undefined. Can you do a video on that and maybe other complex limits?

@azmath2059 6 жыл бұрын

Sensational

@bijalshah9113 4 жыл бұрын

I loved the way you solved this but I guess my method is easier... You can directly introduce a new variable: sin( 'b' lnx)/lnx, and then proceed with the same method. Finally you'll get: I(b) = arctan(b), where we want b=1, hence we get π/4. I hope that was helpful.

@arolimarcellinus8541 10 ай бұрын

Why suddenly become arctan?? We don't know the definition of arctan though

@NurHadi-qf9kl Жыл бұрын

Misal ln x=y maka dy=dx/x atau dx= x dy=e^ydy |=|e^y sin y dy= |sin y d(e^y) Lalu integral parsial.

@MrBoubource 6 жыл бұрын

That supreme jacket tho

@MarcLisevich 2 жыл бұрын

Thanks!

@blackpenredpen 2 жыл бұрын

Thank you, too.

@kutuboxbayzan5967 5 жыл бұрын

I think more easy way is I (b)=integral sin (blnx)/x 0 to 1 I'(b)=integral cos (blnx) 0 to 1 I' (b)=1/(1+b^2) I (b)=tan^-1 (b)+c And I (0)=0 =》c=0 I (b)=tan^-1 (b) I (1)=pi/4

@attepiltonen6607 6 жыл бұрын

Nice

@dimitris892000 6 жыл бұрын

very good bprp, i suggest you try the integral from 0 to 2π of e^(cosx)* cos(sin(x)) dx #YAY

@blackpenredpen 6 жыл бұрын

Hmmm, I can try

@aashsyed1277 2 жыл бұрын

for some reason i found the integral of sin(x)/x dx from 0 to a , a any constant. i calculated it to be -cos(a)/a +1/a by letting G(t)=integral from 0 to a of sin(xt)/x differentiated, got an easy integral, integrated found the constant , and finally got -cos(a)/a +1/a. please explain what happened.

@TheMiningProbe 6 жыл бұрын

This was an extremely clever method, you have my applause

@blackpenredpen 6 жыл бұрын

Yay!!! Thanks to Jamie tho! : )

@2thetutions153 2 жыл бұрын

why we not using laplace properties L[f(x)/x]=integration of phi(x) from o to infinity.

@sirmac6726 6 жыл бұрын

Aircraft trayectory: y = k / x k = 1 sqr km from: x1 = 0.5 km (y1 = 2 km) to: x2 = 2 km (y2 = 0.5 km) Velocity: V = const = 1000 km/h Max acceleration recommended a = 4 g a) Is the aircraft in danger? b) t=? time from x1 to x2.

@alegian7934 6 жыл бұрын

Really cool integral. Can someone please explain why we use lni=πi/2 and not any of the other values?

@alexanderpanov2326 Жыл бұрын

Ln e^i•п/2=п/2•i•lne= п/2•i

@leviszhou711 3 жыл бұрын

amazing！

@TheBetterVersion 6 жыл бұрын

I solved it differently. I used u=ln(x), got the integral from -∞ to 0 of sin(u)*e^(u)/u, used I(t) = the integral but with e^(tu), diffrentiated, integrated by parts, and got that I'(t) = 1/(t²+1), which leads to I(t) = arctan(t). Using t=1 we get that the integral is arctan(1) = π/4

@PunmasterSTP Жыл бұрын

Supreme integral? More like "Super good video!" 👍

@howdoi_yt Жыл бұрын

4:45 is there a way you can know where exactly to put the 2nd variable? or do you just keep trying to find the correct place?

@jimmylu7954 6 жыл бұрын

Satisfying

@EHTom 6 жыл бұрын

In math for fun could you do a non elementary integral but add something to it to make it elementary and thus make it solvable like the Gaussian integral? Would be a nice exercise I believe.

@angelmendez-rivera351 5 жыл бұрын

Do the integral of e^x*sin(ln x) from x=0 to infinity ?

@Mathmagician73 5 жыл бұрын

Great sir 😁😁😁😁👌👌👌👌😳😳u r doing very great work but i have another method to solve this integral can i share u?

@hendrixgryspeerdt2085 3 жыл бұрын

How do you know to take the principal value of ln(i)? Since there are infinite possible values for ln(i). i(Pi/2 + 2(Pi)n), n is an integer.

@user-jq6zr1dy1r 6 жыл бұрын

Thats beautiful

@jschnei3 2 жыл бұрын

The moment you plugged in b=-1 to solve for C, I slapped the table and shouted "You sneaky sonuvagun, you did it!!" That was an amazing moment

@luisafda6623 6 жыл бұрын

Sorprendente!

@joryjones6808 5 жыл бұрын

Now I’d buy an expensive t-shirt with that on it.

@sy-py 4 жыл бұрын

I solved it without complex number. I just used a Feynman's Trick to integrate sin(alnx)/lnx. For a=0 we get I(0)=0. I(1) is out integral in question. Now, I'(a)=integral of cos(alnx)dx from 0 to 1 which is 1/(a^2+1) (check for yourselves!)

@ethanchandler3934 3 жыл бұрын

If you use other values of ln(i), what would this mean intuitively? I believe if we add 2pi, it ends up being complex. What do complex values of integrals mean? Is this the area of the function but on a different slice of the plane?

@davidepeccioli4431 2 жыл бұрын

I may be wrong, but I assume that the ln function still refers to the inverse of the e^x function, knowing that e^(iπ)+1=0 (Euler identity)

@TonyStark-30001 3 жыл бұрын

Hello blackpenredpen plz solve the MIT intgration bee Question plz sir

@codegurt5165 6 жыл бұрын

You must've been out of breath by the end lmao

@vishruth.n11 5 жыл бұрын

how to integrate the same integral from 0 to 2,3...etc

@-james-8343 6 жыл бұрын

Hey awesome video, but you spelt the Jamie wrong in the title (you spelt it Jaime). Great video nevertheless, and keep it up!

@blackpenredpen 6 жыл бұрын

-James- thanks!! I just fixed.

@-james-8343 6 жыл бұрын

blackpenredpen no problem!

@MrCuteguylol 5 жыл бұрын

@@blackpenredpen jaime lannister?

@wiwaxiasilver827 3 жыл бұрын

There actually is another way in the final step. Using the definition that a+bi = re^(i*theta), with r being sqrt(a^2+b^2) and theta being arctan(b/a), or just the angle that forms on the Cartesian when the points are graphed with the x-axis are real and y-axis as imaginary, we get that ln(1+/-I) = ln(sqrt(2))+/-pi/4 (ln(r) + i*theta by logarithmic product rule and cancellation with e) but because it’s (1/(2i))(ln(1+I) - ln(1-i)), the parts with ln(sqrt(2)) cancel and we get pi/4 ultimately. Meanwhile, I guess this could be a cheat explanation but I think we can consider the 2pi*n of both thetas to mutually cancel in my way of calculation through subtraction.

@muneebahmad7729 6 жыл бұрын

Blackpenredpen yay❤️❤️

@slahenejjari5334 2 жыл бұрын

hy blackpenredpen this integral will be amazing in a vidieo: integral of (1/cos^n(x)) n natural

@shanmugasundaram9688 6 жыл бұрын

The convergence and continuity of the function sin(ln x)/ln x at x=0 need to be discussed.

@1412-kaito 6 жыл бұрын

Love it

@blackpenredpen 6 жыл бұрын

Thanks!

@AnuragKumar-io2sb 5 жыл бұрын

So good

@gilbertmiya4199 10 ай бұрын

Tantalizing indeed

@ivanpogildiakov1948 6 жыл бұрын

Nice idea.

@blackpenredpen 6 жыл бұрын

: )

@EAtheatreguy 3 жыл бұрын

I did this integral without bringing in complex numbers at all, quick u-sub u=lnx turns it into int(e^(u)sin(u)/u)du from negative infinity to 0, parametrize with I(b) = (e^(bu)sin(u)/u)du from negative infinity to 0, proceed. I got the wrong answer a few times because when solving for the constant at the end, I let b go to negative infinity instead of positive infinity, which is wrong because u is always negative, so letting b go to negative infinity actually causes divergence.