Integral with oscillation
ฝัง
- เผยแพร่เมื่อ 21 ก.ย. 2020
- Integral of (1-cos(x))/x^2 from 0 to infinity
In this neat video, I calculate the integral from 0 to infinity of 1-cos(x)/x^2 using complex analysis. For this I choose a very clever contour that jumps over the singularity at 0. This problem is taken from the book Complex analysis by Stein and Shakarchi
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You and BlackPen RedPen are my favorite Math TH-camrs :)
I wrote some JUNK on my last math test.
Long story short, let's say my teacher wasn't impressed.
Complex analysis
I don't know much complex analysis, but this was still clear and well explained. Lots of steps seem hard to come up with though!
I had a course on complex analysis back in January at Uni, the ideas are really elegant and beautiful! Thanks for making this video!
Thank you for thIs video! Watching you do a complex analysis problem makes me feel more comfortable giving this a try as someone who is learning on their own 🙏🏽🥳
Watching it with my morning coffee. Thanks.
So cool) Complex analysis is just amazing
اسطورتي دكتور بيان
3:55 You erased a white board with your fingers and I wanted to scream at my monitor.
I like how you explain. Hello from Rep.Dom.
It’s a beautiful integral keep up bro
Beautiful integral
Great vid, I love me some integrals!
Love it!
Dr peyam you can solve it using Jordan lemmas for the circular contour
Physicist method: 1-cos^2(x) is approximately x^2/2. So the integrand = 1/2 and so the integral diverges
I hope in grad school I can do some cool work in complex analysis. I already took the basic course in my undergrad and want to see what I missed out on.
Wow I love compex analysis so much. It’s a shame they don’t do this for engineers, in my case atlteast
Compkex analysis also known as Funktionetheorie in german
I think funktionentheorie is functional analysis
@@drpeyam No, funktionentheorie is complex analysis. I took that course. functional analysis is called funktionalanalysis in german.
Actually the dominated convergence theorem does not apply for the integral of [1-e^(it) ]/t^2 over [ \varepsilon, R], as \varepsilon ->0. The same goes for the integral over [-R,-\varepsilon]. The problem is that the imaginary part of the integrand given by -sin(t)/t^2 is not integrable in a nbh of 0. However, the dominated convergence does apply for the real-part of the integral, so I would suggest that you take the real-part first and then the limit in \varepsilon->0+.
Bravo
Big and small "circle" @ 0:59. Growing up, those used to have constant radii. 😂😂😂
The smol circle was in the pool. xD
Can you do it ( without C.A(
I separated this integral in two single integrals, calculated integral of some beautiful exponent using Euler Puasson integral, find Im of all that and got sqrt(pi/2). Help me understand where is the mistake? May be we can’t use Euler Puasson integral with complex alfa?
who is Puasson?
It is simpler to set (1-exp(i z))/z^2=lim_(eps->0)(1+exp(iz))/(z^2+eps^2) and then use the theorem of residues.
That’s kind of what I’m doing
Ah yes, complex analysis to evaluate real integrals. Very cool technique but it can be shortened using the Leibnitz Rule. Let's say I=our desired integral and I(t)=the same but with cos(tx) instead and t>0. Now notice that I(0)=0 and I(1)=I. Now differentiate with respect to t to arrive at integral from 0 to infinity of sin(tx)/x dx which is pi/2 by the Fresnel integrals. You have I'(t)=pi/2.
Now just integrate with respect to t from 0 to 1 to get integral from 0 to 1 of I'(t) dt =I(1)-I(0)=I-0=I=pi/2. Would be like 3 lines or so. But you way certainly shows the power of complex analysis^^
No that is super not rigorous
@@drpeyam It might have been rather rigorous to me due to being into applied mathematics.^^ I see your point that the switch of integral and partial derivative with respect to t is not justified. But I gave a reason for everything else. And the switch can be justified that (1-cos(tx))/x^2 and sin(tx)/x are continuous on the open region
(x,t) from (0,inf)x(0,1). If that's wrong please correct me.
No because it blows up near 0 so it’s dangerous to differentiate. Better to do it with complex analysis
@@drpeyam That's why I went for the open region and excluded 0 and not the closed xD
you're not making it easy to wait for next semester. I'll have complex analysis :D
All the best. It is going to be hard but really interesting
according to the same calculation, if we replace cos with sin, we get 0. This is nonsense because we have a singularity at x=0. I think it would be better to show that your function is well-defined around 0 using the L'H, for example.
Unless you have already taken a complex analysis course, this is black magic. And if you have, you have probably already solved this problem, or something similar.
In both cases it’s fascinating
You could have juste done integration by parts and you find sin(t)/t and it's п/2
But then why is sin(t)/t equal pi/2, you’d have to prove that
@@drpeyam Well that's a famous result we can do it by Feynman method by putting a e^(-tx) term in the integral
Feynman method is not rigorous, you’d have to do it with complex analysis like in this video
You could also parametrize this integral as-
I(a)=Int(0,infty){ (1-cos(ax))/x^2 }
That would give I'(a)=Int(0,infty){sin(ax)/x}
Which we can convert into the usual sin(u)/u form by u-subbing u=ax (bounds remain same)
This we know to be π/2 and therefore I'(a)=π/2
=>I(a)=πa/2+C ------(1)
Now we have to find C which we do by trying to find the Feynmanian special case which gives the value of the integral for a specific value of a by using alternative methods.
Here we notice I(0) has (1-cos(0x)/x^2 as its integrand which is just (1-1)/x^2=0 and so the integral is 0 too
Using the results we found in (1), I(0)=(π/2)*0+C which results in the equation 0=0+C => C=0 for all a because it is a constant
Therefore I(a)=πa/2 and our integral is I(1) which then is π/2
It's interesting to see how we both got the same integral (sin(u)/u) slightly differently
We got the sin part the same way by differentiating (1-cos(ax)) which although was wrt x in your integral and wrt a in mine, gave the same result as the numerator is symmetric in a and x
The denominator is a more interesting case as you eliminated an x by integrating 1/x^2 and I got rid of it differentiating cos(ax) which popped out an x due to the chain rule. A fun exercise, this was!
It's pretty easy by integration by parts but whatever
Not really, you’d have to know sin(x)/x