Wow I have gotta say the result was damn clean. Didn't think an integral would be explained so nicely. Thanks for sharing all these nice integrals with us . 😄
I love how intuitive all of this felt, I said it once but I'm starting to get the hang of solving these problems! keepin the summer cool my dude keep it up!
The nice expansions for cotangents, cosecants etc at the heart of this method are closely related to Weierstrass factorizations, which can be shown efficiently using complex analysis. So there probably are more miraculous looking results along these lines.
Thank you for you video! I found your method interesting. I solved that integral using complex analysis, just calculated a sum of residues of f(z) = [1-e^(i*tan(z))]/[z^2]. To be clear, I= Int_0^{infty} {sin^2(tanx)/x^2} dx= 1/2*Int_{-infty}^{infty} {sin^2(tanx)/x^2} dx= 1/4*Int_{-infty}^{infty} {[1-cos2(tanx)]/x^2} dx =Re(1/4*Int_{-infty}^{infty} {[1-exp(i*2*tanx)]/x^2} dx) = Re(1/4*2*pi*i*Sum(res(f)) Then i found residues. tan(z) = z+z^3/3+o(z^3) -> exp(i*tan(z)) = exp(i*[z+z^3/3+o(z^3)]) = 1+i*z+i*z^3/3+[(-1)*z^2]/2+[(-i)*z^3]/6 +o(z^3) = 1+i*z-z^2/2+i*z^3/6+o(z^3) -> f(z) = -[i*z-z^2/2+i*z^3/6+o(z^3)]/z^2 = -i/z+1/2-i*z/6+o(z) -> res(f(z))_{z=0} = -i -> I = Re(1/4*2pi*i*(-i)) = pi/2. :^)
As a weekly thing could you do another one of those “Every way to solve …(insert interesting integral here”” ? I loved the one about sin(x) / x and never saw a video on integral calculus that instructive
Do you come up with these solutions on your own?! Because they are absolutely elegant as hell! Okay, maybe not hell, but you get my point. Amazing video!
you do integration by parts first and then Lobachevsky formula (or breaking up into segments of length pi, which is the same as the proof to Lobachevsky) and you're done. You wouldn't need the series expansion of cot(z) either.
Look at the integrand: There are no problems regarding boundedness or convergence so you can switch up the order of the operators. Basically it all boils down to dominated convergence which is trivial in this case.
By the fundamental theorem of engineering tanx = x sinx = x Hence Sin^2(tanx) = sin^2(x) = x^2 infinity is just a very large number for all practicality, so the integral is Int from 0 to N dx Hence we get N I'm sorry I couldn't stop myself It hurt me to type this too
I tried to employ the Feynman integration, but inserting free variable $t$ as $sin^{2}(tan tx)$ or $sin^{2}(t tan x)$ did not lead to something familiar to me. Any idea on the regard?
Another cool way to arrive at the result is through the Fourier Transform. The Fourier Transform of a rectangle function from -1 to 1 is sin(w)/w, so you can just use that an Parseval's Identity to instead integrate the rectangle function multiplied by itself.
I solved an integral similar to this one a while back. I found that one on math stackexchange. So I just modified that integral to get the monster integral in the video
Omg... How on Earth does a nested-trig beast like that have a closed form? Amazing.
Nature surprises. And Maths 505 surprises every time 😊
Wow I have gotta say the result was damn
clean. Didn't think an integral would be explained so nicely.
Thanks for sharing all these nice integrals with us . 😄
I love how intuitive all of this felt, I said it once but I'm starting to get the hang of solving these problems! keepin the summer cool my dude keep it up!
Your integrals are just getting more daunting day by day 😄
Very interesting integral. Very smart substitutions. Thanks you very much.👍
Holy shit that was amazing
The solution is marvelous. Thanks for the insights.
The nice expansions for cotangents, cosecants etc at the heart of this method are closely related to Weierstrass factorizations, which can be shown efficiently using complex analysis. So there probably are more miraculous looking results along these lines.
Excelente trabalho 🇧🇷🇧🇷🇧🇷🇧🇷 excelent work
How did you come up with this integral? Amazing!!!!!💯💯
Thank you for you video! I found your method interesting.
I solved that integral using complex analysis, just calculated a sum of residues of f(z) = [1-e^(i*tan(z))]/[z^2]. To be clear, I= Int_0^{infty} {sin^2(tanx)/x^2} dx= 1/2*Int_{-infty}^{infty} {sin^2(tanx)/x^2} dx= 1/4*Int_{-infty}^{infty} {[1-cos2(tanx)]/x^2} dx =Re(1/4*Int_{-infty}^{infty} {[1-exp(i*2*tanx)]/x^2} dx) = Re(1/4*2*pi*i*Sum(res(f))
Then i found residues. tan(z) = z+z^3/3+o(z^3) -> exp(i*tan(z)) = exp(i*[z+z^3/3+o(z^3)]) = 1+i*z+i*z^3/3+[(-1)*z^2]/2+[(-i)*z^3]/6 +o(z^3) = 1+i*z-z^2/2+i*z^3/6+o(z^3) -> f(z) = -[i*z-z^2/2+i*z^3/6+o(z^3)]/z^2 = -i/z+1/2-i*z/6+o(z) -> res(f(z))_{z=0} = -i -> I = Re(1/4*2pi*i*(-i)) = pi/2. :^)
Beautiful solution
Hello Maths 505, I love watching your videos even though I am just a beginner in calculus
May I ask what drawing program you are using? Very nice results. :-)
As a weekly thing could you do another one of those
“Every way to solve …(insert interesting integral here”” ?
I loved the one about sin(x) / x and never saw a video on integral calculus that instructive
Do you come up with these solutions on your own?! Because they are absolutely elegant as hell! Okay, maybe not hell, but you get my point. Amazing video!
Mostly yes....or I just modify solutions I learn from others
@@maths_505 Awesome
you do integration by parts first and then Lobachevsky formula (or breaking up into segments of length pi, which is the same as the proof to Lobachevsky) and you're done. You wouldn't need the series expansion of cot(z) either.
Hello. May I ask, where do you get these integrals? Do you do them yourself? Or do you invent them yourself?
You can find most of em on the internet and some of em I just mess around and create them
hey can you explain how u switched the summation and the integral I didn't know that was allowed
Look at the integrand:
There are no problems regarding boundedness or convergence so you can switch up the order of the operators.
Basically it all boils down to dominated convergence which is trivial in this case.
By the fundamental theorem of engineering
tanx = x
sinx = x
Hence
Sin^2(tanx) = sin^2(x) = x^2
infinity is just a very large number for all practicality, so the integral is
Int from 0 to N dx
Hence we get N
I'm sorry I couldn't stop myself
It hurt me to type this too
You joke, but that approximation for the tangent does yield the correct result.
QED😂
How Van you prove the switchability of infinite sum and integral?
It all boils down to dominated convergence which is pretty trivial in this case.
I tried to employ the Feynman integration, but inserting free variable $t$ as $sin^{2}(tan tx)$ or $sin^{2}(t tan x)$ did not lead to something familiar to me. Any idea on the regard?
I don't think Feynman's trick will be useful here.
May I request you to solve Integral from 0 to infinity of ln (1+x)/ (1+x^2). note this is not 0 to 1 (rather 0 to infinity)
Integrate 1/(1+x^2) then add do x/(1+x^2) by integration by parts. Add the results of the 2 integrals and you will have the results.
int[sin^2(u)/u^2,u] from -inf to inf= pi
how?
can u have a video on that?
Yup
You can find it in the Feynman integration playlist
Another cool way to arrive at the result is through the Fourier Transform. The Fourier Transform of a rectangle function from -1 to 1 is sin(w)/w, so you can just use that an Parseval's Identity to instead integrate the rectangle function multiplied by itself.
Very nice ! By the way :MATHEMATiCA can't do it .
Where do you find these integrals? Your nightmares or what?
I solved an integral similar to this one a while back.
I found that one on math stackexchange.
So I just modified that integral to get the monster integral in the video
You're basically trolling me RN. Just admit it 505!
Nah man I'm just integrating wild stuff😂😂😂
Wow
You go too fast !!! And you don't explain some of your steps involving sign changes you made several times .
Your caligraphy is almost illegible. Are you a medic?
No he is a teacher.
@@thomasblackwell9507 You really didn't get the joke...