Supreme Integral with Feynman's Trick 

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 actually like indefinite integral more tho

@@blackpenredpen fr

@@blackpenredpenfr

@@blackpenredpen fr

@@blackpenredpenfr

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

Yup

Good morning sir How to become member ?

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

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

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..

Yea!

Yay!!

Thanks!!!

yay!

Yup!!!

Yay!!!

Again tesla for turing.

Video

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

Winter Summers yay!!!

-James- thanks!! I just fixed.

blackpenredpen no problem!

@@blackpenredpen jaime lannister?

Hmmm, I can try

: )

lol i already watched this video

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

: )

Yay!!! Thanks to Jamie tho! : )

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)

Let me know if you the answer

Because x is between 0 and 1

Ln e^i•п/2=п/2•i•lne= п/2•i

I suppose the answer is hidden somewhere in finding the "constant" C, which is not really a constant, but rather is a function of x, right? Perhaps that is some subtlety which needs to be discussed, that C is not a constant, but is only constant with respect to b?

Thank you, too.