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 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 🍫
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?
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
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
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.
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
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.
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 .
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.
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.
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 ?
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
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.
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
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!)
This can actually be solved without complex numbers Place the parameter in the argument of sin; then differentiate You'll then have to solve the integral of cos(t•ln(x)), which looks pretty intimidating but it can be solved with a u-sub and then integration by parts You then get I'(t)=1/(1+t²) I(t)=arctan(t)+c; but I(0)=0 so I(t)=arctan(t) I(1)=π/4
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.
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.
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.
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
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.
I solved it by changing the variables: t = -ln(x). I got the integral from 0 to +inf of sin(t)/t*e^(-t). Then I used the series definition of sin(t) and swapped integration and summation (bcs I can :)). The improper integral was equal to (2n)! which I very liked. In the end, the sum was exactly the arctan(1) by the series definition. The steps (in latex code - feel free to paste into desmos for readability): integral = \int_{0}^{+\infty}\frac{1}{t}\sum_{n=0}^{\infty}\frac{\left(-1 ight)^{n}}{\left(2n+1 ight)!}t^{2n+1}e^{-t}dt = \sum_{n=0}^{\infty}\frac{\left(-1 ight)^{n}}{\left(2n+1 ight)!}\int_{0}^{+\infty}t^{2n}e^{-t}dt = \sum_{n=0}^{\infty}\frac{\left(-1 ight)^{n}}{2n+1} = \arctan1 = \pi/4
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.
A much easier solution is to set u equal to ln(x), then set y=-u, then you get the integral from 0 to infinity of e^-xsin(x)/x and you realize we're done because this just requires the same feynman trick that gets us the solution to the dirichlet integral. define I(t)=e^-xt*sin(x)/x, I'(t)=-1/(1+t^2), I(t)=-arctan(x)+C, I(infty)=0, thus C=pi/2, plug in t=1 and pi/2-arctan(1)=pi/2-pi/4=pi/4, and ya done.
why not uosing u substituting right away ? strat whit u=lnx and then the same method but no need for complex numbers. the Integrand will be (e^bu*sinu)/u
No need to go into the complex world. Use Leibneiz-Feymann technique with the function I(b), where I(b) is the integral from 0 to 1 of sin(b*ln(x))/ln(x) in dx (our initial integral is I(1)). d(I(b))/db is the integral from 0 to 1 of cos(b*lnx) in dx, which is equal to the integral from -inf to 0 of cos(b*u)*(e^u) in du. (lnx=u) Integrating by parts twice we get that d(I(b))/db is 1/(1+b^2) (how cool is that). So I(b) is arctan(b)+c, where c must be 0 because I(0) is clearly 0 and arctan(0) is 0. So I(1) is arctan(1)= π/4
USE ?F(a) = integral(0 to 1 ) sin(aln(x)/ln(x) dx and Feynman diff under integral cos (a(ln(x)) = re (exp(i a ln(x))) = x ^(i a) integrate to x(i a +1) / (1+ i a) and put x = 1 and zero sp re part of (1/)1+i a) is the differential of the target Feynman integral. This is 1/(1+a^2) so the F(a) = arctan (a) + c put a = 0 and we seee c = 0 put a = 1 and the final integral is arctan(1)
I don't understand how you can plug in -1 for b. Didn't you have earlier ...x^(i+bi) ? You did plug in x=0 and said the result was 0, but with b=-1 you have 0^0. Please explain. Love your videos
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 🙏🙏🙏
Because it's the simplest value that satisfy the question. You could choose another value, or even all of then if you want, but there is no need since they're all kind of equivalent
Dear Blackpenredpen I'm one of your followers, and i noticed that you love complications lol, there's a more much easier way to do it look 1st let lnx=-t the integral become from 0 to infinity of sint*exp(-t)/t dt 2nd paramertrizing the integral by introduicing alpha in the sinus 3rd Derivate to alpha we obtain an easy integral (the laplacien of cos(at) The rest is a child game Bye
Dear Fares BERARMA If you work out *all* the steps from your outlines, then it will take about the same amount of time/steps to *make it clear to the viewers* who haven't seen this kind of things before....... Bye
Why can't you do the obvious of making u=-log(x), to get the integral of exp(-u)sin(u)/u from 0 to infinity? Then consider the integrand exp(-au)sin(u)/u, and then differentiate w.r.t a to get an integrand of -exp(-u)sin(u), which can be easily solved using repeated integration by parts.
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.
IS THIS RIGHT? Using your complex definition of a sin, I took the e^i common and rewrote i as e^(iπ/2), obtaining sin z = (e^ln z - e^ -ln z)/2e^(π/2) If z = ln x, We get sin ln x = x/e^(π/2) _(Which is like that shouldn't be right ???????)_
One thing that I don't understand about this derivation is: after you introduced b, you had x^(bi)-x^i in the numerator. Then you took the derivative and the x^i went away. HOWEVER any second term not involving b would have also gone away. So if you had started with any x^(bi) - f(x) would you have still gotten the same answer? Is that troublesome?
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?
very nice. this is a nice technique called complexification. but you should explain why we can take the derivative inside the integral sign. math subject is like that. we need to justify everything.
I remember this integral back in college several decades ago...indef int of [ ln (x^2)/(1+x^2) ] dx....never got the correct answer. Would love to see what this is.
this is one of those problems where it's so complex (figuratively and literally) that you could just enter this into a calculator, see the decimal approximation, which you recognize to be the same decimal approximation for pi/4. like how on earth would someone think to go to the complex world, make it a function, and find the derivative of that function, all to finally integrate it and somehow escape the complex world to get an answer like pi/4?
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
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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
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 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
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
U world is not powerful enough, but b world solves a problem. : ) YAY!
Yup
Good morning sir
How to become member ?
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!
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
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!!!
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.
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 .
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.
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)
Hey BPRP I really enjoyed that. It is very satisfying when complex maths leads to a simple result.
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.
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
I(1) can be converted to arctan(1)
This integral can also be calculated with Laplace transform
Calculate L(sin(t)/t) and plug in s = 1
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 ?
Very very good solution of this problem
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
Some great techniques used to find a very satisfying answer. So good :)
Yay!!
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?
@Hassan Akhtar i wasnt being,salty..I,asked an intelligent question...to,see how to actually solve,this.why cant you see that..
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.
That's incredible, never seen that before, feynmann was a legend.
If u know laplace transform, then proceed this way
Put - ln(X) =t
How does that help? Laplace transform I mean.
@@abdullaalmosalami you will an integral of form f(t) /t for which we have a formula. Then substitute s=1
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 creative problem solving process you used on this integral!
#YAY OMG I love how insane integrals ends with simple answers like pi/4 lol
yay!
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!)
Yay, I love these 20-minute integral videos!
Yay!!!
Again tesla for turing.
Video
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
This can be also be done with F(a)= sin(alnx)/lnx with F(1)= I pretty easily.
But complex world looks amazing.
This can actually be solved without complex numbers
Place the parameter in the argument of sin; then differentiate
You'll then have to solve the integral of cos(t•ln(x)), which looks pretty intimidating but it can be solved with a u-sub and then integration by parts
You then get
I'(t)=1/(1+t²)
I(t)=arctan(t)+c; but I(0)=0
so
I(t)=arctan(t)
I(1)=π/4
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.
ZOMG! That was a ride.
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.
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
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
That was very long and a very beautiful integral.
Great video !!
Good luck jaime for further maths
why we not using laplace properties L[f(x)/x]=integration of phi(x) from o to infinity.
Math for its own sake is beautiful. Thanks blackpenredpen
Winter Summers yay!!!
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.
Oh my Gosh! This was really awesome!
Brazilian congrats! #YAY
Thanks!!!
I solved it by changing the variables: t = -ln(x). I got the integral from 0 to +inf of sin(t)/t*e^(-t).
Then I used the series definition of sin(t) and swapped integration and summation (bcs I can :)). The improper integral was equal to (2n)! which I very liked. In the end, the sum was exactly the arctan(1) by the series definition.
The steps (in latex code - feel free to paste into desmos for readability):
integral = \int_{0}^{+\infty}\frac{1}{t}\sum_{n=0}^{\infty}\frac{\left(-1
ight)^{n}}{\left(2n+1
ight)!}t^{2n+1}e^{-t}dt
= \sum_{n=0}^{\infty}\frac{\left(-1
ight)^{n}}{\left(2n+1
ight)!}\int_{0}^{+\infty}t^{2n}e^{-t}dt
= \sum_{n=0}^{\infty}\frac{\left(-1
ight)^{n}}{2n+1}
= \arctan1
= \pi/4
Yes, this can be converted into the Laplace transform of (sin(t)/t), and the result = π/2 - arctan(S) with S=1, i.e. π/4.
Awesome integral! Thanks :D
YAY
Beautiful solution!
The convergence and continuity of the function sin(ln x)/ln x at x=0 need to be discussed.
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.
A much easier solution is to set u equal to ln(x), then set y=-u, then you get the integral from 0 to infinity of e^-xsin(x)/x and you realize we're done because this just requires the same feynman trick that gets us the solution to the dirichlet integral. define I(t)=e^-xt*sin(x)/x, I'(t)=-1/(1+t^2), I(t)=-arctan(x)+C, I(infty)=0, thus C=pi/2, plug in t=1 and pi/2-arctan(1)=pi/2-pi/4=pi/4, and ya done.
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?
Wolframalpha told me 0^i is undefined. Can you do a video on that and maybe other complex limits?
Really cool integral and a very nice explanation
why not uosing u substituting right away ?
strat whit u=lnx and then the same method but no need for complex numbers.
the Integrand will be (e^bu*sinu)/u
Easy by Substitution
U=lnx. and by gamma function
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
this was fun! but in the end how do you know that log i yields π/2, and not something like 5π/2 ?
No need to go into the complex world.
Use Leibneiz-Feymann technique with the function I(b), where I(b) is the integral from 0 to 1 of sin(b*ln(x))/ln(x) in dx
(our initial integral is I(1)).
d(I(b))/db is the integral from 0 to 1 of cos(b*lnx) in dx, which is equal to the integral from -inf to 0 of cos(b*u)*(e^u) in du. (lnx=u)
Integrating by parts twice we get that d(I(b))/db is 1/(1+b^2) (how cool is that).
So I(b) is arctan(b)+c, where c must be 0 because I(0) is clearly 0 and arctan(0) is 0.
So I(1) is arctan(1)= π/4
Wow, but Phew! I'm exhausted after watching that marathon.
: )
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.
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.
USE ?F(a) = integral(0 to 1 ) sin(aln(x)/ln(x) dx and Feynman diff under integral cos (a(ln(x)) = re (exp(i a ln(x))) = x ^(i a) integrate to x(i a +1) / (1+ i a) and put x = 1 and zero sp re part of (1/)1+i a) is the differential of the target Feynman integral. This is 1/(1+a^2) so the F(a) = arctan (a) + c put a = 0 and we seee c = 0 put a = 1 and the final integral is arctan(1)
I don't understand how you can plug in -1 for b. Didn't you have earlier ...x^(i+bi) ? You did plug in x=0 and said the result was 0, but with b=-1 you have 0^0. Please explain. Love your videos
Is it wrong for me to take parameter b for both the x variables in the numerator of I(b) such as _x^(ib)-x^(-ib)_ ?
I think it would be better to use the archangent and not complexes just for simplicity and speed, thanks.
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 🙏🙏🙏
1 power (1+bi) is equal to e power (- 2b pi).
I'm moved!! I don't know, at the last part of the video, why we can choose principal value of (ln i = πi/2 + 2nπi).
Because it's the simplest value that satisfy the question. You could choose another value, or even all of then if you want, but there is no need since they're all kind of equivalent
Dear Blackpenredpen
I'm one of your followers, and i noticed that you love complications lol, there's a more much easier way to do it look
1st let lnx=-t the integral become from 0 to infinity of sint*exp(-t)/t dt
2nd paramertrizing the integral by introduicing alpha in the sinus
3rd Derivate to alpha we obtain an easy integral (the laplacien of cos(at)
The rest is a child game
Bye
Dear Fares BERARMA
If you work out *all* the steps from your outlines, then it will take about the same amount of time/steps to *make it clear to the viewers* who haven't seen this kind of things before.......
Bye
What a wonderful first question on my exam.
May you please make a video on how to solve for x=y^2+x^2y y=x^2+y^2x
Where does 3^t.ln(3) come from? Don't you just have the exponent times whatever is in the power differentiated? So you'd have 3^t.(1) = 3^t.
Why can't you do the obvious of making u=-log(x), to get the integral of exp(-u)sin(u)/u from 0 to infinity? Then consider the integrand exp(-au)sin(u)/u, and then differentiate w.r.t a to get an integrand of -exp(-u)sin(u), which can be easily solved using repeated integration by parts.
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.
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
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?
Hello blackpenredpen plz solve the MIT intgration bee Question plz sir
That was a wild ride from beginning to end
can you prove the continuity of y=x^(1/x) pls ?
IS THIS RIGHT?
Using your complex definition of a sin, I took the e^i common and rewrote i as e^(iπ/2), obtaining
sin z = (e^ln z - e^ -ln z)/2e^(π/2)
If z = ln x,
We get sin ln x = x/e^(π/2)
_(Which is like that shouldn't be right ???????)_
Supreme integral? More like "Super good video!" 👍
One thing that I don't understand about this derivation is: after you introduced b, you had x^(bi)-x^i in the numerator. Then you took the derivative and the x^i went away. HOWEVER any second term not involving b would have also gone away. So if you had started with any x^(bi) - f(x) would you have still gotten the same answer? Is that troublesome?
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?
very nice. this is a nice technique called complexification. but you should explain why we can take the derivative inside the integral sign. math subject is like that. we need to justify everything.
Sooooo.... who knows the restriction for b? I understand series convergence but I dont think im familiar with integral convergence.
Do the integral of e^x*sin(ln x) from x=0 to infinity ?
Inverse Laplace transform to a double integral and changing the order of integration is how I evaluated it before watching the video.
Please solve for x in x= tan^-1(i)
I remember this integral back in college several decades ago...indef int of [ ln (x^2)/(1+x^2) ] dx....never got the correct answer. Would love to see what this is.
Just plug it into the integral calculator www.integral-calculator.com and show steps
this is one of those problems where it's so complex (figuratively and literally) that you could just enter this into a calculator, see the decimal approximation, which you recognize to be the same decimal approximation for pi/4. like how on earth would someone think to go to the complex world, make it a function, and find the derivative of that function, all to finally integrate it and somehow escape the complex world to get an answer like pi/4?
Integral(0 to 1) of (0/(1-x))Dx isn't exactly solid, considering 0/0 isn't easily dismissable
Substitute -u=lnx and you get an integral we have seen before
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
that was one heck of an adventure
: )
Really cool integral. Can someone please explain why we use lni=πi/2 and not any of the other values?
Ln e^i•п/2=п/2•i•lne= п/2•i
wouldn't Laplace transform be much shorter?
That was actually pretty cool.
hy blackpenredpen this integral will be amazing in a vidieo: integral of (1/cos^n(x)) n natural
this can be done more easily by using feynmen's trick in the first step itself
taking I(b)=integration(sin(blnx)/lnx)
using that
we won't have to deal with complex numbers, they will cancel out and just give arctan(b) as I(b)
Easier to start with I(t)=integral(dx sin(t lnx)/lnx,0,1) and apply Feinmann integration to that.
That supreme jacket tho