This integral looks crazy
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- Опубликовано: 30 июл 2022
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With your development of the problem you overcame certain inconsistencies that fortunately did not affect the final result. First in the interval from 0 to pi there is pi/2 in the middle which is a discontinuity of tan(x). In that case we have to divide the integral into two pieces.
Second, the complex function you arrive at to integrate has two essential singularities on the real axis, z=-1 and z=1, that must be avoided. Fortunately the integral over small semicircles tends to zero at these singularities and does not affect the final result.
Hello there! I just have a question about the World outside of my country: Is it normal that my teachers expect me to solve problems like this, during my third year at University? I am doing Rocket Engineering btw.
P.s.: Good point about singularities. I also wanted to point them out.😉
@@HavaN5rus We need more info to answer this; I know many upper-level engineering courses make use of complex analysis for techniques such as Fourier/Laplace transforms and computing integrals, but this is a super general statement
That being said, a random integral like this which is for show for RUclips? Probably not
@@angelmendez-rivera351 unfortunately it's still not undefined at pi/2.
@@angelmendez-rivera351 but the integration is still over [0, pi). He just expressed tan x in terms of tan x/2.
Love how such diverse types of problems are covered on this channel. Your videos are truly appreciated, please never stop creating them!
A simple binary thumbs-up/thumbs-down rating system is insufficient. Michael Penn clearly tops 95 out of 100.
I don't see why we need the decomposition to cos and sin. With a bit of hand-waving we can change the limits of integration to +/- pi/2. Then substitute
z = tan x
The limits are +/- real infinity. The integrand is exp(i pi/2 z)/(z^2+1) dz and we proceed with contour integration as shown
Clever. I think it wouldn't be handwaving to say the integral from -pi/2 to 0 is the same as from pi/2 to pi. The integrand is a periodic function such that f(x) = f(x+pi) just like tan(x). I think you can say that rigorously.
@@Cancellator5000 Yes, quite true
I remember studying the theoretical foundations of the link between residuals and integrals around 15 years ago... but it still strikes me like a miracle. Math is beautiful.
Cauchy in “Sur Les Integrales Definees” gives the integral of F(x)/(1+x^2) from 0 to inf as pi/2*F(i) iff F(x) can be written as an even infinite series here you get
Pi/2*cosh(pi/2) saving the lots of work.
The part showing that the imaginary part of the integrand doesn't contribute to the integral was fairly useless since you use the full exponential function for your contour integral anyway
The integral of the full exp could be different than its real part
I didn't get the meaning of it.
You can see that cos(pi/2*tan(x)) is symmetric on the interval 0 to pi, so the integral can be rewritten as 2 times the integral from 0 to pi/2. Sin(pi/2*tan(x)) is also symmetric, but oppositely. By the Cauchy principle value, this integral goes to 0. The tangent substitution will work without the half angle formula on the interval 0 to pi/2 resulting in a simpler contour integral. Otherwise I evaluated this problem the same way. Interestingly, SageMath says this integral is divergent. WolframAlpha got it right.
Nice Lynel shirt, @Michael Penn!
Amazing!!! thanks
Are we sure that the integral from 0 to pi of cos(pi/2*tan(x)) is even well-defined? tan(x) has a divergence at x=pi/2 and, looking at the graph of y=cos(pi/2*tan(x)), the function approaches infinite frequency at x=pi/2, which gives me some concern that the Riemann integral is not well-defined over the given interval. Maybe under a different measure it is?
@ 15:06 The exponent of the complex exponential inside the integrand needs to have a pi.
One can use the new variable y= tan x . Then one has to integrate the real part of exp[i*Pi/2*y ] from - inf. to + inf. An easy application of the Cauchy
integral-theorem gives the the final result Pi * Exp[- Pi/2] .
At 4:54 the two negative signs can only "cancel" each other because sin is an odd function so sin(-x) = -sin(x)
@7:53, where did that 2 out front come from?
I also got lost there a bit, but it comes from solving from the substitution for dx = 2du/(1+u^2)
I said “Weierstrass!” out loud when you set u = tan(x/2)
A splendid 'problem' video!
I'm not quite sure that the contour function is defined at ±1, and that integrating over it won't cause problems.
In this case it works, I tried it with semicircles around it and it was fine.
You can also use a different *u*-substitution to get rid of those two singularities entirely - at 5:50 shift the integration domain to make it symmetric, then use *u := tan(x)* .
Yeah I used that sub!
16:02
It may be a stupid question, but at the beginning, what happens with x= pi/2?
And in the integral, I don't remember what caution should be taken with z= 1 and z= -1
excellent problem. starts with complex numbers, then becomes a real integral that ends up being solved using complex contour integration. beautifully done
Bravo Prof.
I wonder, what is the simplest integrand requiring every trick in the book to be integrated?
*every* trick?
The integrand is not defined at a single point x=π/2, but this is not essential for a certain integral.
Although, when approaching this point, the argument of the integrand increases indefinitely, but the integrand itself is limited:
-1≤cos ((π/2)*tan(x))≤1.
Therefore, the value of a certain integral is also limited (let it be equal to Int):
∫(from 0 to π) (-1)*dx≤Int≤∫(from 0 to π) 1*dx, -π≤Int≤π.
At the same time, the contribution to the integral of a small δ-neighborhood near the point x =π/2 is small.
Consider the contribution "from the left": F(δ)= ∫(from π/2-δ to π/2) cos ((π/2)*tan(x))dx =
= lim(ε→0) [∫(from π/2-δ to π/2-ε) cos ((π/2)*tan(x))dx] .
lim(ε→0) [∫(from π/2- δ to π/2-ε) (-1)*dx]≤ F(δ)≤lim(ε→0)[ ∫(from π/2-δ to π/2-ε) 1*dx],
lim(ε→0) (-δ+ε)≤ F(δ)≤lim(ε→0)(δ-ε), -δ≤ F(δ)≤δ and at δ →0 : F(δ)→0.
Unfortunately, I cannot estimate the possibility of applying calculation methods from the theory of functions of a complex variable to an integral with such a function.
Hi Dr.!
I guess since i^i=e^-pi/2, then the final answer could also be written as pi*i^i
I think I should start from full course of mathmajor lol.
What program do you use for thumbnails?
Doesn’t only look crazy
Is it okay to just say that on the period from 0 to pi/2 the function sin(pi/2*tanx) is like a mirror image of itself on the period from pi/2 to pi?
Why bother using Euler's formula to go to sin and cos when you are going to use it a second time to go back to a complex exponential?
It was at this point that my perfect score in high school math failed me. Contour integrals are cool though, always thought about trying to integrate over the complex plane but never knew the notation or how to do it. I think it’s time I stop watching math videos expecting to osmosis it all and actually do a silly little math course.
1. Intro.
2. Magic.
3. "And that's a good place to stop".
u = tan(x) : easier than u = tan(x/2) ==> x = Arctan(u) ==> dx = du / (1+u²)
f(z) = exp(i.PI.z/2)/(1+z²): process easier
What if we take a i=e^(i*5pi/2) or in general (2k+1/2)pi
@5:20 - The function sin(pi/2 tan(x)) has an essential singularity at x = pi/2. How do you justify integrating past it?
In general, you would want to split the integral up at every singularity and compute each part as a limit approaching that singularity in the endpoints of definite integrals. I don't know if that can be done here. What is done here (though he doesn't mention it) is the Cauchy Principle Value. The idea is that you still split up the integral at every singularity, only instead of computing each limit independently, you bind them together so that you approach each singularity at the same rate (and similarly each infinity at the same rate). He ends up doing this with the substitution u=pi-x.
@@SlipperyTeeth I suspect it would have to be the Cauchy principal value, because I don't believe that the integral on either side of the singularity converges, although I freely admit I have not actually tried to compute it.
@@jimschneider799 The integral does indeed converge on either side of the singularity at *x = pi / 2* .
Consider the graph on one side of the singularity: It consists of alternating positive and negative "bumps".
The area of those "bumps" goes to zero monotonically as you approach the singularity, so the integral over one side of the singularity turns into a convergent sum via _Leipniz-Criterium_ !
I wonder where an integral like that would ever come about?
why substitute u=tan x/2 when you can just u=tan x? then you get the cos(pi/2 u) /(1+u^2) and that's just Laplace integral
Even regarding discontinuity, you can split the integral and shift one part to make it from -pi/2 to pi/2
Michael's disregarding of the singular poles -1;+1 bothers me. Does the integral vanish around them? Or cancel between those two points?
Each integral around a small semicircle vanish, if I'm not mistaken. Inside the exponential is an expression with a negative real part.
WARNING: tan(pi/2) IN INTEGRAL DETECTED--
Why do we have only 1 singularity? Inside the exp there is 1-z^2 in the denominator, that gives us 2 more singularities (-1, 1)
I think he meant within the contour. You could also choose the contour in the lower half plain and evaluate the residual of the function at -i.
As far as I understood , because the 'i' that he considered is a principal root of sqrt(-1) , which is only 'i' (but not -i), add he said at the beginning.
Но могу ошибаться.
You're right, you do have to worry about singularities on the contour as well. You could feasibly take a small semicircle around 1 and -1 and set their radius to 0 and hopefully that part of the integral goes to 0.
I think I misread. Since the exp is imaginary, it's bounded on the real line i.e. the limit as x->infinity of e^(i*x) is not infinity. It is indeterminable.
You can further simplify the integral and get rid of the singularities at *⨦1* !
-----------------------------------------
5:50 The integrand *f(x) := cos(𝞹/2 * tan(x))* is *𝞹*-periodic so we may shift the integration domain to a symmetric interval. The integral becomes
*I := \int_0^𝞹 f(x) dx = \int_{-𝞹 / 2}^{𝞹 / 2} f(x) dx*
With this symmetric integration domain we may use the simpler substitution *u := tan(x)* to get
*I = \int_ℝ cos(𝞹 / 2 * u) / (1 + u^2) du*
The contour integration remains the same, but the exponent is a lot simpler - no singularities at *⨦1* anymore, and it is easier to check the upper part of the contour integral really vanishes via _Jordan's Lemma_ .
Would have been nice to have the link to that (tanx)^i integral (EDIT: Had to search through videos: ruclips.net/video/_6hxhUZM_G8/видео.html ). I can't find it in the playlist "interesting integrals". Funny that I can't find this video in there either.
I like your channel. I view the math content on RUclips as continuum on an axis, with one end of the axis being like your channel (adroit, skillful manipulation of symbols as done in math classes) and the other end of the axis being like 3Blue1Brown (peering underneath the symbol manipulation to actually understand what is going on). While I like your channel, it seems to me that the 3Blue1Brown approach is somehow 'better'. Here's why. While I have a PhD in physics and have gone through a LOT math I now realize that I did not really understanding DEEPLY what is going on in many cases, but I got by. Having more emphasis on concepts rather than technique would have been more useful to me. I'm not trying to trash your channel, just interested in your thoughts. For example, to what end (other than passing high level math classes) does having the ability to do these manipulations lead to. What insight is gained? Again, just interested in your thoughts
Weierstrass substitution nice nice
Can we ignore singular points z=1,-1?
I think it should be adressed, but if you work out the expansions of exp(pi*z/(1-z^2))/(1+z^2) at z = 1, -1 they seem to both lack residual terms 1/(z-1), 1/(z+1) respectively according to WolframAlpha. Thus, there is no contrubition if we avoid the singularities by passing them using small semicircles.
@@mansklingspor2032 That's not the case. You have to evaluate the integral around the small semicircle directly. It may not be directly related to the residue at those singularities if the singular points are not simple.
@@replicaacliper You are right, I stand corrected.
What is the difference between Newton and Leibniz calculus?
Newton's calculus is about functions.
Leibniz's calculus is about relations defined by constraints.
In Newton's calculus, there is (what would now be called) a limit built into every operation.
In Leibniz's calculus, the limit is a separate operation.
Weierstrass
I typically love your work but this one not so much. Your evaluation of an integral using the Residue Theorem (at time ruclips.net/video/MLKMJs4JvpE/видео.html ) needed further justification. In particular, you need to show that the integral along the half circle of radius R in the upper half plane, centred at the origin, approaches zero as R approaches infinity. Or, at least state your assumption of this fact.
I think substitution v = tan(x) may be more simple:
∫[0,π] i^tan(x)dx ( with log i = πi/2 )
= ∫[0,π] exp( πi/2 tan(x) ) dx
= re∫[0,π] exp( πi/2 tan(x) ) dx
( ∵ imag. part vanishes with king property )
= re∫[-π/2,π/2] exp( πi/2 tan(x) ) dx
= re∫[-π/2,π/2] exp( πi/2 v )/( 1 + v² )dv
= re 2πi res( exp( πi/2 v )/( 1 + v² ), v=i )
= re 2πi ev( exp( πi/2 v )/( v+i ), v=i )
= re 2πi exp( -π/2 )/( 2i )
= π exp( -π/2 )
wtf
I could mostly follow this until the “integral of a contour that I’ll describe in a picture”, introduced with “complex analysis involving residue”. No idea what happened there! I’d love to understand this ring-integral thing some day (I mean: ∮).
Check his other channel "MathMajor". He has a full course on Complex Analysis there.
@@luisaleman9512 cool! Tx! I'll check it out