Note: from the 18:30 mark onwards the right hand side includes arctan(t/2) and not arctan(1/2) untill we actually plugged in t=1. Apologies for the error, I subconsciously drifted to the target case of I(1) while carrying out all the math correctly 😂
Here's how I'd do it with the Fourier transform. After the first u sub, you have an exponent times sin(u)/u squared, so the integral I is half the integral of e^-|u| times sin(u)/u squared, so you use Plancherel theorem to turn it in the integral of a triangle windows function times 1/(1+x^2), which is already easier
Parameterising it as e^(-u)sin^2(tu)/u^2 actually makes the ensuing working out much nicer. You get that I''(t)=2/(1+4t^2) with I(0)=I'(0)=0 and directly end up with the nicer looking solution of arctan(2)-1/4ln5
Also works really nicely if you define the parameter inside the sine and just immediately differentiate twice with respect to the parameter. You end up integrating 2/x^(2)cos(2a ln(x)) which can be done by parts, where [a] was my parameter.
In your final step you could have written -¼ ln5 which is slightly more elegant. Another great video! Just keep control of your script and editing...you got a bit nervous in the middle there and started rambling 😜
I don't like using scripts....it makes things too organized and is not the optimal way to carry out the reverse cowgirl formulation of integral calculus😂😂😂
@@maths_505 I support this motion :) I love the quirky format. If I wanted a dry integral calculus recitation, there’s plenty of those around. Dry works for some people, quirky works for others. Me quirky. Me likes.
It's funny that I got the exact same answer, differently. I put the alpha in the sin function and got the second derivative of it instead of changing variables first. ( which actually involved some really cool tricks that I learned from this channel as well). So anyway thanks a lot for your great content and I'd love to see more(especially if you would upload some tutorials about differential equations)
theres an easier way to go about this integral IMO, first we start by setting ln(x) to x getting us the integral from 0 to infty of sin^2(x)e^-x/x^2, then we actually define our integral function by putting the parameter t inside of the sin term instead of the exponential term, note for later that I(0)=0, for I'(t) we end up getting sin(2tx) inside so note again that I'(0)=0, we will have no worries over constants in this scheme. Then we differentiate one more time and we get the integral from 0 to infty of 2cos(2tx)e^-x which is equal to 2/(1+4t^2), then we integrate twice and plug in one. edit: my solution is no better, it's just what I came up with before I watched your solution. You literally taught me every advanced integration technqiue I know, you're the GOAT.
I am a 56 years old electrical engineer and this channel gives me more pleasure than having sex...duh! His choice of integrals and diff equations are top notch. Keep up the good work.
Where and how could I learn those ways to solve integrals?? I mean how to learn gamma function and beta function and Feynman technique and laplace transformation...etc??
Awesome vids, dude! What I especially like is you approach, balanced between regorousity and fun. Don't stop! Can you crack professor Penn's 'members only' recent sum?
@@maths_505 Neither can I find it. I saw it yesterday in my TH-cam feed and I don't see it anymore. It involved summation of Reinmann zeta of n , all over pi^n, and the RHS was something ridiculously simple like 1\(exp(2)-1). I'm not sure about the details though.
It's just easy to see , you compare the growing rates of your functions! Any power of t will dictate its behavior over the behavior of an ln function. So the product t*ln(t²/t²+4) , inside the ln , when t approches 0 , the whole thing converges to 0 , then we have ln(0) , so it goes to -infinity but the whole thing is multiplied by a power of t that actually wins there cuz "it goes to 0 way faster than the ln function". And you should be able to verify this tesult using l'Hospital rule
You can split it up as 2tln(t)-tln(t^2+4). Using the fact that the limit as t approaches 0^+, tlnt approaches 0 gets you to the answer. That fact can be proven using L'Hopital's Rule
At the minute 18:07, the factor of 8 outside the integral should be divided by 4 not by 2. So, What about using Lobachevsky integral instead of Feynman technique from the beginning. Anyway, this is very interesting integral and you have a talent to solve such complex integrals.
@@maths_505 the form of 1/(1+t^2) is the derivative of arctan. Thus, if we took 4 as a common factor from the denominator, then we divide 8 by 4. Thanks
@@MrWael1970 factor out 4 from the denominator and you get 1/(1+t²/4). Write t²/4 as (t/2)². Now you need (1/2)dt as the differential element so insert the ½ in the numerator and to balance that out we need to multiply outside the integral operator by 2. That gives you 2/4 outside the integral operator which is ½. Multiply that by the factor of 8 we already have then we get the factor of 4. You're right about factoring the 4 from the denominator but we also need to address the form of the differential element.
@@MrWael1970 it's cool....the signs of great mathematicians include forgetting constant factors cuz they're completely hypnotized by the beautiful structure 😂
If there were problems with boundedness or convergence. Dirichlet's convergence theorem for integrals is a pretty good criteria to test whether or not the integral converges or not.
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Note: from the 18:30 mark onwards the right hand side includes arctan(t/2) and not arctan(1/2) untill we actually plugged in t=1.
Apologies for the error, I subconsciously drifted to the target case of I(1) while carrying out all the math correctly 😂
"We are gathered here today..." LMAO THE CULT OF FEYNMANN INTEGRATION I LOVE IT
Towards the end, it is written inverse tan of 1/2. However, this should be inverse tan of t/2.
Final result stays the same, as t=1.
Thanks!
ja that 's not a arctg he forgot the 1/t cause he cant do the math
SO many smart people and no one pointed out the mistakes he made ...... wow
Nice. I fear I may never be skilled enough to know when to apply Feynman's technique. But it is really cool.
whenever you have a monomial in the denominator
Here's how I'd do it with the Fourier transform. After the first u sub, you have an exponent times sin(u)/u squared, so the integral I is half the integral of e^-|u| times sin(u)/u squared, so you use Plancherel theorem to turn it in the integral of a triangle windows function times 1/(1+x^2), which is already easier
Excellent video, thanks. I would mention that those I’’’ integrals are Laplace transforms. Never a bad time to point out the power of those things.
Parameterising it as e^(-u)sin^2(tu)/u^2 actually makes the ensuing working out much nicer. You get that I''(t)=2/(1+4t^2) with I(0)=I'(0)=0 and directly end up with the nicer looking solution of arctan(2)-1/4ln5
lmao, I did the same thing
2:30 put t into sin^2tx and you'll get a quicker answer
Also works really nicely if you define the parameter inside the sine and just immediately differentiate twice with respect to the parameter. You end up integrating 2/x^(2)cos(2a ln(x)) which can be done by parts, where [a] was my parameter.
I was scared to integrate cos(2nlnx) and then integrate that more times
love the channel, keep on keeping on!
Year 2050: "Feynman's technique is still working 😆"
Wow ,really crazy integral !
Excellent. A very surprising answer.
In your final step you could have written -¼ ln5 which is slightly more elegant.
Another great video! Just keep control of your script and editing...you got a bit nervous in the middle there and started rambling 😜
I don't like using scripts....it makes things too organized and is not the optimal way to carry out the reverse cowgirl formulation of integral calculus😂😂😂
@@maths_505 I support this motion :) I love the quirky format. If I wanted a dry integral calculus recitation, there’s plenty of those around. Dry works for some people, quirky works for others. Me quirky. Me likes.
y = ln(x)
Integration by parts with
dv=1/y^2dy and u=e^{-u}sin^2(u)
Express sin^2(y) as cos(2y)
Calculate Laplace transform and evaluate it at s=1
It's funny that I got the exact same answer, differently. I put the alpha in the sin function and got the second derivative of it instead of changing variables first. ( which actually involved some really cool tricks that I learned from this channel as well). So anyway thanks a lot for your great content and I'd love to see more(especially if you would upload some tutorials about differential equations)
From 2:40 , you could probably solve it using the Laplace transform and its properties.
But is it not just the same procedure to evaluate the Laplace transform?
theres an easier way to go about this integral IMO, first we start by setting ln(x) to x getting us the integral from 0 to infty of sin^2(x)e^-x/x^2, then we actually define our integral function by putting the parameter t inside of the sin term instead of the exponential term, note for later that I(0)=0, for I'(t) we end up getting sin(2tx) inside so note again that I'(0)=0, we will have no worries over constants in this scheme. Then we differentiate one more time and we get the integral from 0 to infty of 2cos(2tx)e^-x which is equal to 2/(1+4t^2), then we integrate twice and plug in one.
edit: my solution is no better, it's just what I came up with before I watched your solution. You literally taught me every advanced integration technqiue I know, you're the GOAT.
I am a 56 years old electrical engineer and this channel gives me more pleasure than having sex...duh! His choice of integrals and diff equations are top notch. Keep up the good work.
brilliant solution. thanks man !
I think you have an error at about the 15:25 point...You should have 0=1/4(1)+c, which implies that c=-1/4. Although, I could be wrong.
Where and how could I learn those ways to solve integrals??
I mean how to learn gamma function and beta function and Feynman technique and laplace transformation...etc??
"You have come to the right place my child"
Just stay tuned to this channel I guess? 😂
Sir I didn't understand the part when you found the value of c please explain
Awesome vids, dude!
What I especially like is you approach, balanced between regorousity and fun.
Don't stop!
Can you crack professor Penn's 'members only' recent sum?
I'll have a look at it
Could you send me a link?
I can't seem to find the members content part of his page
@@maths_505 Neither can I find it. I saw it yesterday in my TH-cam feed and I don't see it anymore. It involved summation of Reinmann zeta of n , all over pi^n, and the RHS was something ridiculously simple like 1\(exp(2)-1). I'm not sure about the details though.
12:44 Is Dominated Convergence justified?
mistake: the answer is actually arctan(2)-ln(5)/4 not arctan(2)+ln(5)/4
Can we do IBP with e-u and sin2u/u2 then the term containing sin is a Dirichlet integral.
Uso la funzione integranda I(a) =sin(alnx) ^2.... Derivo 2 volte... I=arctg2-ln5/4 se non ho fatto errori di calcolo...
It can easily be done by using Laplace transform.....
I failed to solve by L Hospital rule that limit t approaches to zero tlnt^2/(t^2+4) is equal to zero.
It's just easy to see , you compare the growing rates of your functions! Any power of t will dictate its behavior over the behavior of an ln function.
So the product t*ln(t²/t²+4) , inside the ln , when t approches 0 , the whole thing converges to 0 , then we have ln(0) , so it goes to -infinity but the whole thing is multiplied by a power of t that actually wins there cuz "it goes to 0 way faster than the ln function". And you should be able to verify this tesult using l'Hospital rule
You can split it up as 2tln(t)-tln(t^2+4). Using the fact that the limit as t approaches 0^+, tlnt approaches 0 gets you to the answer. That fact can be proven using L'Hopital's Rule
Writing z for ln(x) one gets
Integration (0, INF)
( sin^2(z)/( z exp(2z)) dz
At the minute 18:07, the factor of 8 outside the integral should be divided by 4 not by 2.
So, What about using Lobachevsky integral instead of Feynman technique from the beginning.
Anyway, this is very interesting integral and you have a talent to solve such complex integrals.
The table of anti derivatives says that we need to divide by a=2 and not a²=4
@@maths_505 the form of 1/(1+t^2) is the derivative of arctan. Thus, if we took 4 as a common factor from the denominator, then we divide 8 by 4. Thanks
@@MrWael1970 factor out 4 from the denominator and you get 1/(1+t²/4). Write t²/4 as (t/2)². Now you need (1/2)dt as the differential element so insert the ½ in the numerator and to balance that out we need to multiply outside the integral operator by 2. That gives you 2/4 outside the integral operator which is ½. Multiply that by the factor of 8 we already have then we get the factor of 4. You're right about factoring the 4 from the denominator but we also need to address the form of the differential element.
@@maths_505 thanks, I got confused, but now it is clear. Regards.
@@MrWael1970 it's cool....the signs of great mathematicians include forgetting constant factors cuz they're completely hypnotized by the beautiful structure 😂
ok, this is cool
In which cases wouldn’t it be possible to switch the differential operator with the integral one?
If there were problems with boundedness or convergence. Dirichlet's convergence theorem for integrals is a pretty good criteria to test whether or not the integral converges or not.
Nice
a great channel to satisfy my integration fetish
Woooooow
You're going so fast, I can't well understand what you're doing ☹️☹️
Plz tell me, whether Feynman's technique is essentially solving an integration using Leibniz's rule?
It is the same. Just Feynman made it famous.