Thanks so much! I'm really glad you enjoyed - feel free to let me know if there's any other topics you'd like to see or any feedback you have on my other videos too!
I managed to "solve" it this way : First, using trig identities, it can be shown that cos(2x) / cos^4(x) can be rewritten as (tan(x) - tan^3(x)) * 2 / sin(2x). This is good because the derivative of ln(tan(x)) is 1/(cos(x)sin(x)) = 2 / sin(2x). Set u = ln(tan(x)) => du = 2 / sin(2x) dx. This also means that tan(x) = e^u. The bounds become -inf to 0. So you get the integral from -inf to 0 of (e^u/u - e^(3u) / u) du. I believe that these can be solved using infinite series but Feynman's trick seems to work for I(a) = integral from -inf to 0 of e^(au) / u du (I). I'(a) = 1/a => I(a) = ln|a| + C (II). This is where it got messy. if you plug I(0) into (I), you get ln(0). Plug it into (II), and you get ln(0) + C, so C = 0. The problem is that ln(0) is undefined, and even if you take the limit, you get -inf = -inf + C, which is inconclusive for C. But if you ignore that, and set C = 0, then you get I(a) = ln|a|, and so the integral becomes : I(1) - I(3) = - ln(3), the expected answer. It seems to work but I don't know how to make Feynman's trick work rigorously... Any ideas on how to make it work ? Edit : lim x->0 ln(x) = lim x->0 ln(x) + C => C = 0. It works fine if you take the limit.
I did it by taking the limit as a approaches negative infinity - that seems to suggest c is 0. Really nice that Feynman's trick could be used here even though you did it differently.
When learning Feynman's trick I remember running into this exact conundrum at one point, so this is right up my alley :D The issue you're having is to do with the fact that the integral from -inf to 0 of e^(au) / u du, for a>0, DOES NOT CONVERGE. However, the integral from -inf to 0 of [ e^(au) - e^(bu) ] / u du, for a>0, DOES converge (you can see both of these things by doing some integral comparison tests). So the way to proceed with feynman's from here is: 1. Define I(a,b) = integral from -inf to 0 of [ e^(au) - e^(bu) ] / u du, for a,b>0. Note that if you set a=b, I(b,b) = 0 [since the exponential terms in the integrand cancel each other out] 2. dI/da = integral from -inf to 0 of [ u*e^(au) - 0 ] / u du = integral from -inf to 0 of e^(au) du [Differentiating partially with respect to a on both sides] 3. dI/da = e^(au)/a, evaluated from -inf to 0 4. dI/da = 1/a --> I = ln(a) + C. [Integrate both sides with respect to a] 5. [Since when a=b, I = 0] 0 = ln(b) + C --> C = -ln(b) 6. Therefore I(a,b) = ln(a) - ln(b) = ln(a/b). 7. For our question, plug in (a,b) = (1,3), and we get that I = ln(1/3) = -ln(3), which is the answer presented in the vid. The main issue was to do with the fact that the integral you were trying to solve doesn't converge. Check out this book: galoisian.wordpress.com/wp-content/uploads/2018/11/undergraduate-lecture-notes-in-physics-paul-j-nahin-inside-interesting-integrals-2015-springer-1.pdf , check out Chapter 3.3 pg 84 (pg 106 in the pdf I've sent). This section gives you an overview of a generalisation to the integral. This book is also fascinating, I would highly recommend going through the rest of it if you want to learn more about integrals!
@@GGBOYZ583bro we can also do a double integral by converting (e^au-e^bu)/u into int from b to a e^ut dt hence by switching the order of integration we get I= int from -int to 0 int from b to a e^ut du dt then integrating with respect to u we get I= int from b to a 1/t dt which is equal to I = ln (a)- ln(b) giving the same result u found As I = ln (a/b) 😊
To learn calc 1 and 2 what TH-cam channels/books would you recommend? I saw the comment where u recommended black pen red pen but is there anyone else who you would recommend? Thanks and nice video.
@@THE_LAST_STYLEBENDER For the basics maybe check out three blue one brown's calculus series? Paul's online maths notes is also a great website to use. Hope this helps!
@OscgrMaths i think i just need to stop skipping concepts and take things slow. I keep trying to learn the whole of calc when i havent even started calc 1. Math is just too interesting. I cant help but dive head first into it.
@@duckyoutube6318 bro I had exactly the same thing - most of the stuff I know is self taught and I started trying to do some super hard stuff but I always had to take a step back and get foundations first cause otherwise it's so frustrating!! Definitely get through calc 1 and calc 2 - black pen red pen has some great content you could watch as you go through them!
Yeah this is well known. It became known as "Feynman's trick" even while Feynman was still alive and he disagreed with the name, stating that he found it in an old calculus text as a demonstration of how to use the Leibniz rule. He became famous for it while in school as he could evaluate difficult integrals without contour integration.
Damnnnnnn that was clean
Thanks for the amazing integral in the first place!!
One of the best intros to feynmans method imo; shows the whole process without getting bogged down in the details
I always thought the same! Thanks for the comment
bro i'm addicted to your channel, keep going u are doing a great work.
Thanks so much! Really appreciate it
Very good video and I will definitely be watching more of your videos!
Thanks so much! I'm really glad you enjoyed - feel free to let me know if there's any other topics you'd like to see or any feedback you have on my other videos too!
I managed to "solve" it this way :
First, using trig identities, it can be shown that cos(2x) / cos^4(x) can be rewritten as (tan(x) - tan^3(x)) * 2 / sin(2x). This is good because the derivative of ln(tan(x)) is 1/(cos(x)sin(x)) = 2 / sin(2x).
Set u = ln(tan(x)) => du = 2 / sin(2x) dx. This also means that tan(x) = e^u. The bounds become -inf to 0.
So you get the integral from -inf to 0 of (e^u/u - e^(3u) / u) du. I believe that these can be solved using infinite series but Feynman's trick seems to work for I(a) = integral from -inf to 0 of e^(au) / u du (I).
I'(a) = 1/a => I(a) = ln|a| + C (II). This is where it got messy. if you plug I(0) into (I), you get ln(0). Plug it into (II), and you get ln(0) + C, so C = 0. The problem is that ln(0) is undefined, and even if you take the limit, you get -inf = -inf + C, which is inconclusive for C. But if you ignore that, and set C = 0, then you get I(a) = ln|a|, and so the integral becomes :
I(1) - I(3) = - ln(3), the expected answer.
It seems to work but I don't know how to make Feynman's trick work rigorously... Any ideas on how to make it work ?
Edit : lim x->0 ln(x) = lim x->0 ln(x) + C => C = 0. It works fine if you take the limit.
I did it by taking the limit as a approaches negative infinity - that seems to suggest c is 0. Really nice that Feynman's trick could be used here even though you did it differently.
When learning Feynman's trick I remember running into this exact conundrum at one point, so this is right up my alley :D
The issue you're having is to do with the fact that the integral from -inf to 0 of e^(au) / u du, for a>0, DOES NOT CONVERGE.
However, the integral from -inf to 0 of [ e^(au) - e^(bu) ] / u du, for a>0, DOES converge (you can see both of these things by doing some integral comparison tests). So the way to proceed with feynman's from here is:
1. Define I(a,b) = integral from -inf to 0 of [ e^(au) - e^(bu) ] / u du, for a,b>0. Note that if you set a=b, I(b,b) = 0 [since the exponential terms in the integrand cancel each other out]
2. dI/da = integral from -inf to 0 of [ u*e^(au) - 0 ] / u du = integral from -inf to 0 of e^(au) du [Differentiating partially with respect to a on both sides]
3. dI/da = e^(au)/a, evaluated from -inf to 0
4. dI/da = 1/a --> I = ln(a) + C. [Integrate both sides with respect to a]
5. [Since when a=b, I = 0] 0 = ln(b) + C --> C = -ln(b)
6. Therefore I(a,b) = ln(a) - ln(b) = ln(a/b).
7. For our question, plug in (a,b) = (1,3), and we get that I = ln(1/3) = -ln(3), which is the answer presented in the vid.
The main issue was to do with the fact that the integral you were trying to solve doesn't converge. Check out this book: galoisian.wordpress.com/wp-content/uploads/2018/11/undergraduate-lecture-notes-in-physics-paul-j-nahin-inside-interesting-integrals-2015-springer-1.pdf , check out Chapter 3.3 pg 84 (pg 106 in the pdf I've sent). This section gives you an overview of a generalisation to the integral. This book is also fascinating, I would highly recommend going through the rest of it if you want to learn more about integrals!
@@GGBOYZ583 Wow!!!! Thanks so much for this amazing comment! I'll definitely check this book out.
@@GGBOYZ583bro we can also do a double integral by converting
(e^au-e^bu)/u
into
int from b to a e^ut dt
hence by switching the order of integration we get
I= int from -int to 0 int from b to a e^ut du dt
then integrating with respect to u we get
I= int from b to a 1/t dt
which is equal to
I = ln (a)- ln(b)
giving the same result u found
As I = ln (a/b)
😊
@@aravindakannank.s. NEAT! That’s also a nice method.
Awesome
Glad you enjoyed it!
Nice and clean
Glad you enjoyed!!
To learn calc 1 and 2 what TH-cam channels/books would you recommend? I saw the comment where u recommended black pen red pen but is there anyone else who you would recommend? Thanks and nice video.
@@THE_LAST_STYLEBENDER For the basics maybe check out three blue one brown's calculus series? Paul's online maths notes is also a great website to use. Hope this helps!
@@OscgrMaths thanks, will check these out
I just can seem to master solving these. Hope this video will help.
It's all practice - I can do some more videos like this if you'd like!
@OscgrMaths i think i just need to stop skipping concepts and take things slow.
I keep trying to learn the whole of calc when i havent even started calc 1.
Math is just too interesting. I cant help but dive head first into it.
@@duckyoutube6318 bro I had exactly the same thing - most of the stuff I know is self taught and I started trying to do some super hard stuff but I always had to take a step back and get foundations first cause otherwise it's so frustrating!! Definitely get through calc 1 and calc 2 - black pen red pen has some great content you could watch as you go through them!
Perfect!
Thanks! Glad you enjoyed.
@@OscgrMaths btw is converting the integration into a Contour integration work with this integral?
This trick is not Feynman's trick. It is a method to do integrals well known from long time!
Yeah this is well known. It became known as "Feynman's trick" even while Feynman was still alive and he disagreed with the name, stating that he found it in an old calculus text as a demonstration of how to use the Leibniz rule. He became famous for it while in school as he could evaluate difficult integrals without contour integration.
Don't forget me when you become famous
I won't mohammedmadani7277 🙏
Leibniz integral rule