The residue Theorem is so powerful in this case. I only had to look at the integral and calculate it in my head, I didn’t have to write a single line down. I could SEE what the answer would be. Such a beautiful, powerful tool.
Another aprroach is to use Fourier -transforms:Consider the integral of Exp[i x y]/(1+x^2) ,which corresponds to √(2*π) * Fourier transform of 1/(x^2+1). The FT of 1/(x^2+1) is √(π/2)*Exp[- | y |]. For y=1 you get π/e . To see that the Fourier transform of 1/(x^2+1) is given by the term written before , one can just use the inverse FT of Exp[- | y |] which is easy to calculate.
Using the famous engineering equation cosx=1, we can transform this integral into int(-inf, inf) 1/(x^2 + 1) dx, which is simply the arctan integral evaluated from -inf to inf, giving us pi. A truly beautiful result, round of applause
The technique used is by no means simple. I would assume that someone who is able to do that also knows residue theory .Use that you may take the real part of Exp[ i x] and use that 1/(x^2+1)= 1/2 i (1/(x-i)-1/(x+ i) ,which leads to π * i *(-i *Exp[-1] = π/e
Not exactly The feynman technique is based on differentiating a parameter The laplace transform is more of an integration w.r.t a parameter along with the introduction of an exponential term involving that parameter
Let's say I integrate from -infty to +infty the function x, wrt x. Can I use the fact that x is an odd function in order to say that the I definite integral is zero? The same way here, how do we know the parity on an indefinite integral applies ? Because the function you're integrating is bounded maybe? In that case why is it a valid argument?
You can integrate x from -a to +a and you'll get zero. Taking the limit to infinity of something that's zero is zero so yes the argument is valid in this case Nd yes part of the reason why the parity applies is that the function we're integrating is bounded. However that's part of the bigger picture where we need boundedness to actually get a solution that's valid.
I'm not very sure on 0:30 passage: function is even ok, but doubling the integral and behalfing the integration domain in case of integrand even function works only when extremes of integration are finite. If extremes are not finite we don't know how each extreme approaches its infinity (+inf and -inf respectively) and so we don't necessarily have a symmetric integration area. I think the passage is true only in the principal value of the integral sense.
Beautiful interplay of the two techniques! This result was marvelous.
Indeed
So far I haven't seen an integral that gives a similar result
nice. however there is an elementary solution in one line using the residue theorem applied to the function Re(e^(iz)/(z^2+1)).
Yes indeed but I just love using real methods wherever possible
The residue Theorem is so powerful in this case. I only had to look at the integral and calculate it in my head, I didn’t have to write a single line down. I could SEE what the answer would be. Such a beautiful, powerful tool.
Very nice method! Really liked it. Thank you🙏🙏
Another aprroach is to use Fourier -transforms:Consider the integral of Exp[i x y]/(1+x^2) ,which corresponds to √(2*π) * Fourier transform of 1/(x^2+1).
The FT of 1/(x^2+1) is √(π/2)*Exp[- | y |]. For y=1 you get π/e . To see that the Fourier transform of 1/(x^2+1) is given by the term written before , one can just use the inverse FT of Exp[- | y |] which is easy to calculate.
Absolutely gorgeous one ❤️🔥
Using the famous engineering equation cosx=1, we can transform this integral into int(-inf, inf) 1/(x^2 + 1) dx, which is simply the arctan integral evaluated from -inf to inf, giving us pi. A truly beautiful result, round of applause
such an underrrated math channel
Wow! Great result and techniques for solving! Thank you👍
wow... very nice, stunning result
The technique used is by no means simple.
I would assume that someone who is able to do that also knows residue theory .Use that you may take the real part of Exp[ i x] and use that
1/(x^2+1)= 1/2 i (1/(x-i)-1/(x+ i) ,which leads to π * i *(-i *Exp[-1] = π/e
Indeed but I just love using real techniques whenever possible
Nice video along a touristic route of mathematical techniques. Contour integration in the complex plane makes this a one-liner.
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Beautiful interplay of techniques!!
beautiful solution.
Very nice.
Thank you very much!
which software you use to write like these?
Love from India 👍
Beaufiful result. Can also be accomplished using complex analysis.
Yep that was my first reaction
Can somebody tell me how all of this is used in the real world
In engineering
Beautiful! This is a combination of the Feynman Technique and the Laplace Transform right?
Not exactly
The feynman technique is based on differentiating a parameter
The laplace transform is more of an integration w.r.t a parameter along with the introduction of an exponential term involving that parameter
@@maths_505 --Thank you, sir.
@@maths_505 is special case of integration with respect to a parameter
Why should I use such a long method? There are much shorter methods
But what if replace the cos(x) with sin(x) ?
It surely converge (seems converge to 0.64...). , but I can't work out. I get it equal to infnity*0
It gives 0 because the function is odd
@@blblblblblbl7505 sorry ！i forgot to rewrite the integrated boundary from 0 to plus infinity .
It is converged. Converge to 0.6xxxxx
Nice!
awesome
Let's say I integrate from -infty to +infty the function x, wrt x. Can I use the fact that x is an odd function in order to say that the I definite integral is zero?
The same way here, how do we know the parity on an indefinite integral applies ? Because the function you're integrating is bounded maybe? In that case why is it a valid argument?
You can integrate x from -a to +a and you'll get zero. Taking the limit to infinity of something that's zero is zero so yes the argument is valid in this case
Nd yes part of the reason why the parity applies is that the function we're integrating is bounded. However that's part of the bigger picture where we need boundedness to actually get a solution that's valid.
No need to delete my response really, it's okay if you don't have the response to my question...
I haven't deleted it
I responded an hour ago
@@maths_505 I was talking about the 3 other comments that were deleted in a row ...
Deleted?
I haven't deleted anything
Perhaps they're being held in review
Let me check
I'm not very sure on 0:30 passage: function is even ok, but doubling the integral and behalfing the integration domain in case of integrand even function works only when extremes of integration are finite. If extremes are not finite we don't know how each extreme approaches its infinity (+inf and -inf respectively) and so we don't necessarily have a symmetric integration area. I think the passage is true only in the principal value of the integral sense.
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Are U kidding us?
Hi
You skipped too many core concept steps
:(