Supreme Integral with Feynman's Trick
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- Опубликовано: 25 июн 2018
- 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.
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17:20 "So, this right here is pretty much the answer but what the heck in the world is this?" I'm crying 😂
from now on, I'll write that instead of pi/4 😂
math video: so basically that's the answer.
me: ok but what the heck in the world is this?
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.
: )
I actually like indefinite integral more tho
@@blackpenredpen fr
@@blackpenredpenfr
@@blackpenredpen fr
@@blackpenredpenfr
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 🍫
I think your solution is more elegant, as it doesn't require complex numbers inside natural logs (which can have infinitely many values).
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
Same bro.. And it doesn't involve complex numbers in any way so simple, but a little longer..
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?
@@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
U world is not powerful enough, but b world solves a problem. : ) YAY!
Yup
Good morning sir
How to become member ?
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
yeah ! right ! Good!
"more easily" - 😳
Ian moseley easier as in doesnt require complex analysis and identities such as ln(i)
Anyway how can u write integral sign in the comnent? 😂
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
Hey BPRP I really enjoyed that. It is very satisfying when complex maths leads to a simple result.
Very creative problem solving process you used on this integral!
It's 1:54am here. Good night!!!!!!!!!!!!
Have a problem Mr D
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
Flammy did that already... like 2 hrs after my upload, lolll
blackpenredpen oh lol. Well done to him i guess :) will watch his video soon
blackpenredpen no he did it different than me, i didnt use imaginary nums
Awesome integral! Thanks :D
YAY
Yay! This was so cool!!
Thx for putting it up
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
Thank you!!! I am glad that you enjoy it!
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.
Bit,can,younactually solve this without just knowing those formulas?
@@leif1075 stop being salty, you are misled because you got high marks in baby level math.
@@hassanakhtar7874 i wasnt being,salty..I,asked an intelligent question...to,see how to actually solve,this.why cant you see that..
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.
Yea!
Some great techniques used to find a very satisfying answer. So good :)
Yay!!
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.
Really cool integral and a very nice explanation
#YAY OMG I love how insane integrals ends with simple answers like pi/4 lol
yay!
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 ?
Oh my Gosh! This was really awesome!
Brazilian congrats! #YAY
Thanks!!!
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
That's incredible, never seen that before, feynmann was a legend.
Great video !!
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
Yup!!!
Good luck jaime for further maths
That was very long and a very beautiful integral.
Yay, I love these 20-minute integral videos!
Yay!!!
Again tesla for turing.
Video
That was a wild ride from beginning to end
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 .
Hi. Your work is awesome
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.
loved this one..
this was fun! but in the end how do you know that log i yields π/2, and not something like 5π/2 ?
That was actually pretty cool.
that was one heck of an adventure
: )
Awesome!
If u know laplace transform, then proceed this way
Put - ln(X) =t
How does that help? Laplace transform I mean.
@@abdullahalmosalami2373 you will an integral of form f(t) /t for which we have a formula. Then substitute s=1
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.
lol i already watched this video
Very nice!
Math for its own sake is beautiful. Thanks blackpenredpen
Winter Summers yay!!!
Great journey !
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 🙏🙏🙏
Beautiful solution!
This is so perfect
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
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
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?
More multivariable calculus videos would be neat 🤙
Master Cao, u r way too clever ... Thanks a lot!
Sensational
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!)
Are there any restrictions for differentiating under the integral sign or can it be done each and every single time?
This can be also be done with F(a)= sin(alnx)/lnx with F(1)= I pretty easily.
But complex world looks amazing.
ZOMG! That was a ride.
Great. Thanks
Refer to video about lim(x->0) {sin(ln x)/ln x} = DNE.
Can we observe the convergence of this integral to determine lim(x->0) {sin(ln x)/ln x}? I am thinking: If the area under {sin(ln x)/ln x} is finite (for 0
this is the kind of content I love to see, also why are you up so late???
What a wonderful first question on my exam.
Wow, but Phew! I'm exhausted after watching that marathon.
: )
amazing！
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.
Euler evaluated this integral centuries ago by focusing on sin(ln(x)) first expanding it into an infinite series of sin(y) ie y-y^3/3!+y^5/5!-.......then you substitute y=ln(x)....ln(x) can be factored out and cancelled with the ln(x) in the denominator. Then it’s a simple ln(x) to a power evaluated term wise by Bernoulli first. Then you get the Leibniz series. Pi/4. Simples.
Nice
*Super Duper Cool*
Thank you so
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)
Thanks!
Thank you, too.
Supreme integral? More like "Super good video!" 👍
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
very good bprp, i suggest you try the integral from 0 to 2π of e^(cosx)* cos(sin(x)) dx #YAY
Hmmm, I can try
Sorprendente!
But isn't ln(i)=iπ/2+2kπ for integer k? But this integral clearly cannot have more than one answer. Am I missing something?
Let me know if you the answer
Because x is between 0 and 1
Hey awesome video, but you spelt the Jamie wrong in the title (you spelt it Jaime).
Great video nevertheless, and keep it up!
-James- thanks!! I just fixed.
blackpenredpen no problem!
@@blackpenredpen jaime lannister?
Satisfying
can you prove the continuity of y=x^(1/x) pls ?
Love it
Thanks!
blackpenredpen becomes blackpenredpenbluepen
Thats beautiful
Blackpenredpen yay❤️❤️
*if you watched a video in the past*
me who watched a video about feynman's technique 5 minutes ago: damn what kind of distant past is this
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.
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.
Why suddenly become arctan?? We don't know the definition of arctan though
That supreme jacket tho
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.
A non-complex, perhaps less exciting approach:
Begin with u = lnx
u = lnx
x = e^u
dx = e^u du
Now we have: ∫ (sinu / u * e^u) du. Let's define a function I(a) such that I(a) = ∫ (sinu / u * e^(au)) du. Notice that I(1) is equal to the original definite integral at hand. Let's differentiate both sides with respect to a:
I(a) = ∫ (sinu / u * e^(au)) du
I'(a) = ∫ (sinu / u * ue^(au)) du
I'(a) = ∫ (sinu * e^(au)) du
Perform integration by parts so that the integrand repeats. This yields the following equation:
(1 + a^2)∫ (sinu / u * e^(au)) du = -1
∫ (sinu / u * e^(au)) du = -1/(1 + a^2)
I'(a) = -1/(1 + a^2)
Time to integrate both sides with respect to a. It is a well known result that ∫ 1/(1 + x^2) dx = tan^(-1)(x) + C. Let's use that piece of information to our advantage:
∫ I'(a) da = ∫ (-1/(1 + a^2)) da
I(a) = -tan^(-1)(a) + C
Recall that I(a) is defined as I(a) = ∫ (sinu / u * e^(au)) du. If we plug in a = 0, we get the definite integral of sinu / u from negative infinity to zero which is essentially the same thing as going from zero to positive infinity. This is because sinx/x is an even function. The value of this definite integral is known very well to be equal to π/2. We could also let a go to negative infinity and carry on from there but things get a little akwards that way. I believe that also works but it is much simpler this way. Let's continue with a = 0:
I(0) = -tan^(-1)(0) + C
π/2 = C
C = π/2
Now that we know another representation for I(a) besides it's original definition, let us finally plug in a = 1. If you remember, this is equal to the definite integral going from 0 to 1 of sin(lnx)/lnx. That's what we are trying to solve.
I(1) = -tan^(-1)(1) + π/2 = -π/4 + π/2 = π/4
In conclusion, I(1) = ∫ (sin(lnx) / lnx) dx (from 0 to 1) = π/4.
Amazing 😍
Amazing :)
Wolframalpha told me 0^i is undefined. Can you do a video on that and maybe other complex limits?
omg that was awesome
This was an extremely clever method, you have my applause
Yay!!! Thanks to Jamie tho! : )
Guys, this in America is not calc I or Calc II OR calc III this a non elementary function and is fourth or fifth semester Calc that is junior or senior level analysis (All calc is a form of analysis). So unless you are a math major, engineer or physics major don't sweat this stuff. You will not need it.
Exactly, even a lot of engineers don't need this stuff. Nor programmers etc. Fun to watch but not anything that most of those people involved in the technical or science fields need to worry about.
Do the integral of e^x*sin(ln x) from x=0 to infinity ?
So good
hy blackpenredpen this integral will be amazing in a vidieo: integral of (1/cos^n(x)) n natural
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
Tantalizing indeed
why we not using laplace properties L[f(x)/x]=integration of phi(x) from o to infinity.
Hello blackpenredpen plz solve the MIT intgration bee Question plz sir
This would have been much simpler if you made ln(x) = -v and then use Feynman's technique! I(a) = Integral of [sin(v)/v]*e^-av between 0 and infinity.
Now I’d buy an expensive t-shirt with that on it.
May you please make a video on how to solve for x=y^2+x^2y y=x^2+y^2x
how to integrate the same integral from 0 to 2,3...etc
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.