We could use the Laplace transform after introducing Feynman's technic I'(a)= - L(sin(bt))=-b/(a^2 +b^2) , with b=1, we can easily reach the solution faster.
Integral of sin(x) / x dx from 0 to infinity is a classic. Here's an algebraic approach. It does extend into the patterns of series, binomial theorem - identities, as well as the complex plane: This may not be a complete proof or solution, but it illustrates the point. I find this to be another decent approach towards evaluating or trying to solve it. Setup and a few basic common principles: Not all of them may be directly used but are good and useful to keep in mind. Slope-Intercept form of a line y = mx + b. Slope formula: m = (y2-y1)/(x2-x1) = deltaY/deltaX = sin(t)/cos(t) = tan(t) where t is the angle theta between the line y = mx+b and the +x-axis. Initial Conditions: m = 1, b = 0. Constraints: b will always be 0. Simplification: y = 1*x + 0 y = x. Substitutions: y = mx == y = sin(t)/cos(t) * x == x * tan(t). We can write this as sin(t) / t. The thing to recognize here is that the integration here is in relation to the angle, as opposed to the x - dx form. We know that 90 degrees or PI/2 radians is a Right Angle. We know that, multiplying by the imaginary unit i vector has the same exact effect of rotating by 90 degrees and by multiplying any value by i^(4*N) where N is an +Integer is the same as multiplying by 1 since it rotates it by 360 degrees or 2PI radians. Taking the graph of this function and looking for the area under the curve can be broken down into intervals based on the properties and relationships between PI/2 and i within the context of the summation of their series that converges to PI/2 or 90 Degrees. The Series: n=0 |--> +infinity of: (2n)!! / (2n+1)!! * (1/2)^n = PI/2 The double factorial (!!) is define by 0!! = 1!! and n!! = n(n-2)!! Then: f(t) = Series: n=0 |--> +infinity of: (-1)^n / (2n+1) * t^n Note that f(1) = PI/4. We can take the Euler Transform of the series: 1/(1-t) * f( t / (1-t) = OuterSum: n=0 |--> +infinity { InnerSum: k=0 |-->n ( n : choose k) ( -1)^k / (2k + 1) } * t^n Then: Sum: k = 0 |--> n (n: choose k) (-1)^k/(2k+1) = (2n)!! / (2n + 1)!! Proving the above just refer to proving a binomial sum identity. We can see that: The Integral from 0 to infinity of Sin(x)/x dx is equal to: The Series: n=0 |--> +infinity {(2n)!!/(2n+1)!!}*(1/2)^n = PI/2. Forgive me if there's any typos in the math... "Y.T." isn't very friendly with their parsing of comments. Here's a link for the above Series: math.stackexchange.com/a/14116/405427
I wanted to stress test Feynman's method, so I used sin(ax) instead of introducing exp(-ax). What happens is, you get the integral of cos(ax) over [0,∞) which is undefined. But if you regularize it by introducing the regulator exp(-tx), then solve that equation, you find the regularized integral is atan(a/t) which goes to 𝜋/2 for all a>0. That regularization is basically the exact thing you did right from the beginning, just with an extra step. So, it seems introducing exp(-ax) is the "canonically correct" way to use Feynman's method here. Still feels kinda interesting to me two different choices for where to put the Feynman parameter end up giving similar results, if you grant the use of regularizing divergent terms as a tool.
is the function you used exp(-ax) just e^(-ax)? i've seen it used before but couldn't find what it was but when i did this same method i used e^(-ax) and got the same thing
I was upset about the dx until you fixed it! I found this derivation in high school, and loved it, and only now as an undergrad can I appreciate this technique's similarity to the Laplace or Fourier transform.
Sin x = 1/1! x - 1/3! x³ + 1/5! x⁵ - .... So sin x /x = 1/1! - 1/3! x² + 1/5! x⁴ - .... And the I tergal is just : ∫ Sin x/x . dx = 1/1! x - 1/3! x³/3 + 1/5! x⁵/5 - .... + (-1)^n 1/n!. x^n /n +... You can also integrate many functions including the naughty function f(x)= e^x² which is otherwise does not have an explicit formula of integral. Off course with a condition of being an analytic function.
@@thedocotrL Off course you can only write a limited number of terms of the series. This seems to be useless , but in computer programming since the values they give have limited lengths that means they have a margin of error. and since that the Taylor series are convergent then you can take only a limited amount of terms that don't exceed a certain error, so you can define the sin x , ln x , e^x, ch x, functions etc by a limited number of terms of their Taylor series.
I think this is a two-part trick method. First trick part comes from the need to “temperate” the integrand, which is swinging wildly. (Btw, we need to state, right off the bat, that a is positive.) This way we get a function, the temperated integrand, whose integral on [0,infinity) is finite. The second trick part is to use differentiation (under integral) wrt the parameter of the “temperator”.
Thanks for your interesting video. Area under a curve is often equivalent to energy. Buckling of an otherwise flat field shows a very rapid growth of this area to a point. If my model applies, it may show how the universe’s energy naturally developed from the inherent behavior of fields. Your subscribers might want to see this 1:29 minutes video showing under the right conditions, the quantization of a field is easily produced. The ground state energy is induced via Euler’s contain column analysis. Containing the column must come in to play before over buckling, or the effect will not work. The sheet of elastic material “system”response in a quantized manor when force is applied in the perpendicular direction. Bonding at the points of highest probabilities and maximum duration( ie peeks and troughs) of the fields “sheet” produced a stable structure when the undulations are bonded to a flat sheet that is placed above and below the core material. Some say this model is no different than plucking guitar strings. You can not make structures with vibrating guitar strings or harmonic oscillators. ruclips.net/video/wrBsqiE0vG4/видео.htmlsi=waT8lY2iX-wJdjO3 At this time in my research, I have been trying to describe the “U” shape formed that is produced before phase change. In the model, “U” shape waves are produced as the loading increases and just before the wave-like function shifts to the next higher energy level. Over-lapping all frequencies together using Fournier Transforms, can produce a “U” shape or square wave form. Wondering if Feynman Path Integrals for all possible wave functions could be applicable here too? If this model has merit, seeing the sawtooth load verse deflection graph produced could give some real insight in what happened during the quantum jumps between energy levels. The mechanical description and white paper that goes with the video can be found on my LinkedIn and RUclips pages. You can reproduce my results using a sheet of Mylar* ( the clear plastic found in some school essay folders. Seeing it first hand is worth the effort!
@@98danielray Feynman found it in a textbook, 'Advanced Calculus', by Frederick Woods, loaned him by his middle-school(!) physics teacher to keep him out of trouble. See 'Surely you're joking, Mr. Feynman.'
There is a much easier way to evaluate that integral. Since sin(x)/x is an even function, the integral in question is 1/2 the integral from -infinity to + infinity, which can easily be evaluated using Cauchy's residue theorem.
Good to know, although there are many ways of solving most problems, I specifically chose to solve this via feynmans method, also doesn't require too much knowledge beyond standard integration techniques, therfore, in my opinion, making it more digestible to the average viewer who has some sort of background in maths
To exchange the derivative with the integral, you need to show that the function respects the hypotheses of the theorem of Dominated Convergence: in this case, they are; but the Feymann trick is usually seen as this integral-killing machine because no one ever wants to check if these hypothesis are actually true. You can see that proving all the things that are required by the various theorems related to the change of the order of derivation and integration requires at least 10 more minutes of video
I did exactly this in my x^x Bernoulli integral video. Because yes you're right, it is important but honestly it's so tedious and probably very boring to the average viewer
@@Jagoalexander @Jagoalexander sorry if my comment could be seen as a criticism, but I am just a poor student traumatized by the uniform convergence and the endless list of theorems that I had to study because my professors are crazy people who like working with integrals with all the properties that Lebesgue’s measure possesses even if this means to explain to the first-year undergraduate Physics student that with *30* lemmas per theorem it is possible to obtain a weaker version of all theorems which can be proved by Lebesgue’s measure and, even more comical, they asked them during the oral examination. It was more a spontaneous reflex to write that you had to do it (caused by my PTSD) than a criticism
a=0 has a problem. At 6:50, you show 1/a as a factor in the formula. So, if you repeat all the steps with "a" having a fixed value of zero, the whole thing breaks down at that formula..
A can be any number you choose, however when we go ahead and say a=0 is our integral, it’s more the limit as a-> 0. You’re right it could not be a=0 it should be a limit, sorry if I didn’t make that clear.
@@Jagoalexander yes, I agree, but tan(x) is not bijective. We usually use the main branch for arctan(), namely the one that goes from -pi/2 to pi/2. My question is how can I know that the main branch is the right o e to consider.
"Feynman Technique" is RUclips's favorite moniker for "The method of integration by-parts". Just open up any elementary calculus book written before Feynman was born.
@@Cow.cool. again this "trick" was known way before Feyman. Only physics enthusiasts with insufficient math background would refer to it as "Feynman technique", Feynman himself never claimed over such a thing. Some Feyman's pupils may have referred to it as Feyman's teqnique because they did not see this method prior to taking Feynman's lecture. He's a great scientist and had great contributions to the field of QFT, but this is not his "technique".
Actually not. There he tried to find the values when 'a' approaches infinity. Then the Left side of the equation would result into zero while the right side would be -π/2 + c. That just gives us the equation c -π/2 = 0 or c = π/2. Zero is just equal to zero and he has actually integrated zero as zero.
@@giorgibliadze1151 Yes, you’re right that the integral of 0 is a constant. When dealing with definite integrals, like the one in the video, we evaluate the integral over a specific range. This process gives a specific value, so we don’t need to add a constant. It’s similar to differentiation: when you differentiate a function at a specific point, you get a single value, not a function
We could use the Laplace transform after introducing Feynman's technic I'(a)= - L(sin(bt))=-b/(a^2 +b^2) , with b=1, we can easily reach the solution faster.
Excellent idea
14:35 He ended up with the Laplace transform.
@@xtr3m385 yes but later!
The integral is actually the laplace transform of sin(x)/x evaluated at 0
L(sin(x)/x)(0)
Integral of sin(x) / x dx from 0 to infinity is a classic.
Here's an algebraic approach. It does extend into the patterns of series, binomial theorem - identities, as well as the complex plane:
This may not be a complete proof or solution, but it illustrates the point.
I find this to be another decent approach towards evaluating or trying to solve it.
Setup and a few basic common principles: Not all of them may be directly used but are good and useful to keep in mind.
Slope-Intercept form of a line y = mx + b.
Slope formula: m = (y2-y1)/(x2-x1) = deltaY/deltaX = sin(t)/cos(t) = tan(t) where t is the angle theta between the line y = mx+b and the +x-axis.
Initial Conditions: m = 1, b = 0.
Constraints: b will always be 0.
Simplification: y = 1*x + 0 y = x.
Substitutions: y = mx == y = sin(t)/cos(t) * x == x * tan(t).
We can write this as sin(t) / t. The thing to recognize here is that the integration here is in relation to the angle, as opposed to the x - dx form.
We know that 90 degrees or PI/2 radians is a Right Angle. We know that, multiplying by the imaginary unit i vector has the same exact effect of rotating by 90 degrees and by multiplying any value by i^(4*N) where N is an +Integer is the same as multiplying by 1 since it rotates it by 360 degrees or 2PI radians.
Taking the graph of this function and looking for the area under the curve can be broken down into intervals based on the properties and relationships between PI/2 and i within the context of the summation of their series that converges to PI/2 or 90 Degrees.
The Series: n=0 |--> +infinity of: (2n)!! / (2n+1)!! * (1/2)^n = PI/2
The double factorial (!!) is define by 0!! = 1!! and n!! = n(n-2)!!
Then: f(t) = Series: n=0 |--> +infinity of: (-1)^n / (2n+1) * t^n
Note that f(1) = PI/4. We can take the Euler Transform of the series:
1/(1-t) * f( t / (1-t) = OuterSum: n=0 |--> +infinity { InnerSum: k=0 |-->n ( n : choose k) ( -1)^k / (2k + 1) } * t^n
Then:
Sum: k = 0 |--> n (n: choose k) (-1)^k/(2k+1) = (2n)!! / (2n + 1)!!
Proving the above just refer to proving a binomial sum identity.
We can see that:
The Integral from 0 to infinity of Sin(x)/x dx is equal to:
The Series: n=0 |--> +infinity {(2n)!!/(2n+1)!!}*(1/2)^n = PI/2.
Forgive me if there's any typos in the math... "Y.T." isn't very friendly with their parsing of comments.
Here's a link for the above Series: math.stackexchange.com/a/14116/405427
Absolutely brilliant
I wanted to stress test Feynman's method, so I used sin(ax) instead of introducing exp(-ax). What happens is, you get the integral of cos(ax) over [0,∞) which is undefined. But if you regularize it by introducing the regulator exp(-tx), then solve that equation, you find the regularized integral is atan(a/t) which goes to 𝜋/2 for all a>0.
That regularization is basically the exact thing you did right from the beginning, just with an extra step. So, it seems introducing exp(-ax) is the "canonically correct" way to use Feynman's method here.
Still feels kinda interesting to me two different choices for where to put the Feynman parameter end up giving similar results, if you grant the use of regularizing divergent terms as a tool.
is the function you used exp(-ax) just e^(-ax)? i've seen it used before but couldn't find what it was but when i did this same method i used e^(-ax) and got the same thing
Yes exp(x) is e^x
Interesting. My first time here. I look forward to more.
I was upset about the dx until you fixed it! I found this derivation in high school, and loved it, and only now as an undergrad can I appreciate this technique's similarity to the Laplace or Fourier transform.
Sin x = 1/1! x - 1/3! x³ + 1/5! x⁵ - ....
So sin x /x = 1/1! - 1/3! x² + 1/5! x⁴ - ....
And the I tergal is just :
∫ Sin x/x . dx = 1/1! x - 1/3! x³/3 + 1/5! x⁵/5 - .... + (-1)^n 1/n!. x^n /n +...
You can also integrate many functions including the naughty function f(x)= e^x² which is otherwise does not have an explicit formula of integral. Off course with a condition of being an analytic function.
So how do you handle the infinity limit in your series solution?
@@thedocotrL
Off course you can only write a limited number of terms of the series. This seems to be useless , but in computer programming since the values they give have limited lengths that means they have a margin of error. and since that the Taylor series are convergent then you can take only a limited amount of terms that don't exceed a certain error, so you can define the sin x , ln x , e^x, ch x, functions etc by a limited number of terms of their Taylor series.
I think this is a two-part trick method. First trick part comes from the need to “temperate” the integrand, which is swinging wildly. (Btw, we need to state, right off the bat, that a is positive.) This way we get a function, the temperated integrand, whose integral on [0,infinity) is finite. The second trick part is to use differentiation (under integral) wrt the parameter of the “temperator”.
Thanks for your interesting video.
Area under a curve is often equivalent to energy. Buckling of an otherwise flat field shows a very rapid growth of this area to a point. If my model applies, it may show how the universe’s energy naturally developed from the inherent behavior of fields.
Your subscribers might want to see this 1:29 minutes video showing under the right conditions, the quantization of a field is easily produced.
The ground state energy is induced via Euler’s contain column analysis. Containing the column must come in to play before over buckling, or the effect will not work. The sheet of elastic material “system”response in a quantized manor when force is applied in the perpendicular direction.
Bonding at the points of highest probabilities and maximum duration( ie peeks and troughs) of the fields “sheet” produced a stable structure when the undulations are bonded to a flat sheet that is placed above and below the core material.
Some say this model is no different than plucking guitar strings. You can not make structures with vibrating guitar strings or harmonic oscillators.
ruclips.net/video/wrBsqiE0vG4/видео.htmlsi=waT8lY2iX-wJdjO3
At this time in my research, I have been trying to describe the “U” shape formed that is produced before phase change.
In the model, “U” shape waves are produced as the loading increases and just before the wave-like function shifts to the next higher energy level.
Over-lapping all frequencies together using Fournier Transforms, can produce a “U” shape or square wave form.
Wondering if Feynman Path Integrals for all possible wave functions could be applicable here too?
If this model has merit, seeing the sawtooth load verse deflection graph produced could give some real insight in what happened during the quantum jumps between energy levels.
The mechanical description and white paper that goes with the video can be found on my LinkedIn and RUclips pages.
You can reproduce my results using a sheet of Mylar* ( the clear plastic found in some school essay folders.
Seeing it first hand is worth the effort!
@@ronney6121 drag due to Turbulence is usually a big factor in air especially water.
@@ronney6121 I so sorry that I can not help with your work. My interest in mathematics well exceeds my ability’s in the subject.
Wonderfully explained. Loved It!
You know a guy is genius when he invents a new way to integrate!
What’s funny is Feynman had an average IQ the guy just developed extraordinary thinking techniques. So it’s possible for you and I as well!
he popularized it, clearly not invented. would not be surprised if even Euler used it
@@98danielray Feynman found it in a textbook, 'Advanced Calculus', by Frederick Woods, loaned him by his middle-school(!) physics teacher to keep him out of trouble. See 'Surely you're joking, Mr. Feynman.'
That was a joy to watch.
So differentiating under integral sign was introduced only recently by Feynman? Anyway I wanted to learn it.
Excellent video.
Differentiating under the integral was found by Leibniz iirc, but Feynman popularised it, so people started calling it Feynman's technique
There is a much easier way to evaluate that integral. Since sin(x)/x is an even function, the integral in question is 1/2 the integral from -infinity to + infinity, which can easily be evaluated using Cauchy's residue theorem.
Good to know, although there are many ways of solving most problems, I specifically chose to solve this via feynmans method, also doesn't require too much knowledge beyond standard integration techniques, therfore, in my opinion, making it more digestible to the average viewer who has some sort of background in maths
Loved the explanation
To exchange the derivative with the integral, you need to show that the function respects the hypotheses of the theorem of Dominated Convergence: in this case, they are; but the Feymann trick is usually seen as this integral-killing machine because no one ever wants to check if these hypothesis are actually true. You can see that proving all the things that are required by the various theorems related to the change of the order of derivation and integration requires at least 10 more minutes of video
I did exactly this in my x^x Bernoulli integral video. Because yes you're right, it is important but honestly it's so tedious and probably very boring to the average viewer
@@Jagoalexander @Jagoalexander sorry if my comment could be seen as a criticism, but I am just a poor student traumatized by the uniform convergence and the endless list of theorems that I had to study because my professors are crazy people who like working with integrals with all the properties that Lebesgue’s measure possesses even if this means to explain to the first-year undergraduate Physics student that with *30* lemmas per theorem it is possible to obtain a weaker version of all theorems which can be proved by Lebesgue’s measure and, even more comical, they asked them during the oral examination. It was more a spontaneous reflex to write that you had to do it (caused by my PTSD) than a criticism
Thank you bro loved it
great vid mate
Wow. That’s a great video.
At 13:30 you argued incorrectly about the limit at infinity. You should divide by “a^2” first, then take the limit.
Love the bow
Use cos(x)+isin(x) = exp(ix) to avoid the integration by parts.
I was always taught , in the absence of x or ln, that e would be chosen as u in integration by parts. But it gives the same result.
a=0 has a problem. At 6:50, you show 1/a as a factor in the formula. So, if you repeat all the steps with "a" having a fixed value of zero, the whole thing breaks down at that formula..
A can be any number you choose, however when we go ahead and say a=0 is our integral, it’s more the limit as a-> 0. You’re right it could not be a=0 it should be a limit, sorry if I didn’t make that clear.
Love the « in the a world »
Wonderful.
It's just the laplace transform of sin(x)/x evaluated at 0
Could you explain why we switch to a partial derivative once inside the integral sign?
At 17:55 you say that arctan(infinity)=pi/2, but someone could argue that arctan(infinity) is also equal to 3pi/2, or 5pi/2, or 7pi/2, …
No they couldn’t, look at the arctan graph
@@Jagoalexander yes, I agree, but tan(x) is not bijective. We usually use the main branch for arctan(), namely the one that goes from -pi/2 to pi/2.
My question is how can I know that the main branch is the right o e to consider.
Very interesting 🤔 so what does the f without the dash mean?
Which software are you using for writing this ?? Please inform me
Goodnotes on iPad
Very cool.
Solved easily with Laplace transform
Any engineer would know that the integral of
sin(x)/x = x
what app is this?? the one u are writing on??
Goodnotes on IOS on my ipad
Gangsters record when they have
Bro what 😭
@@Jagoalexander u is the skibidi top G andrew tate
@@Jasturtle thank you man
@@Jagoalexander no problemo my ohio skibidi rizzler
Бесконечность
t can not approche ∞ because ∞ is not a finite number 🌵🌵🌵
That is precisely why the term "approach" is used instead of equal.
I love obi wan teaching me calculus!
Haha what?
"Feynman Technique" is RUclips's favorite moniker for "The method of integration by-parts". Just open up any elementary calculus book written before Feynman was born.
incorrect, this is a different technique involving the introduction of a completely new variable
@@Cow.cool. again this "trick" was known way before Feyman. Only physics enthusiasts with insufficient math background would refer to it as "Feynman technique", Feynman himself never claimed over such a thing. Some Feyman's pupils may have referred to it as Feyman's teqnique because they did not see this method prior to taking Feynman's lecture. He's a great scientist and had great contributions to the field of QFT, but this is not his "technique".
Thaks a lot, however , I lost you at 17:39, integral of 0 is not 0, is it? Its some contstant.
Actually not. There he tried to find the values when 'a' approaches infinity. Then the Left side of the equation would result into zero while the right side would be -π/2 + c. That just gives us the equation c -π/2 = 0 or c = π/2. Zero is just equal to zero and he has actually integrated zero as zero.
Thank you, All I know is that int of 0 is constant as diff of constant is zero.
@@giorgibliadze1151 Yes, you’re right that the integral of 0 is a constant. When dealing with definite integrals, like the one in the video, we evaluate the integral over a specific range. This process gives a specific value, so we don’t need to add a constant. It’s similar to differentiation: when you differentiate a function at a specific point, you get a single value, not a function
You took so much time in just applying by parts your way of doing is such a waste of time
Who asked
Vulgar abuse contributes nothing to the discussion. If you can't make a useful contribution please refrain from antisocial comments.
well how do you do it then genius
You might consider it to be a waste of time, but the presenter of this video didn't...
Kudos: respect and appreciation to the Content Creator!
@@robert-skibeloit’s his video…….if you don’t like it, scroll away you fossil 🦖
❤❤
§°°21 -••(1.43218(---)Limpa