15:32 Both integrals along the vertical sections should have an addition factor of i because the parameterisation z= \pm R + iy results in the differential being dz = i dy, naturally. Obviously this doesn’t affect the cancellation but it is a necessary nitpick as my first thought when seeing the initial u =tanx substitution carried out was to do a semicircular contour with a branch cut on the negative imaginary axis, making sure to go around the origin with another small semicircle (of radius epsilon) traversed in a clockwise direction. In that situation, the positive and negative real axis contours would have different parameterisations and the differential would therefore be important to get right.

@alainleclerc233

2 жыл бұрын

Correct !

I was really fascinated. Every time I solve Improper integrals without Definite function in this way, I feel like I'm doing magic.

Alternately Beta Function identities make mince meat of this problem. Also you can expand it as a series but that requires some work to express it in terms of sechx

Alternatively, at 2:20 you can actually do the standard half-circle contour in the upper half plane as the radius R goes to infinity to get the correct result. After watching the video I went and evaluated that myself. I think it is simpler, but not as interesting as the additional substitution he does which shows a contour I haven't seen before.

@alainleclerc233

2 жыл бұрын

No, since the number of singularities in the half-disc increases to infinity with R while Michael’s rectangular contour contains only 1 singularity.

@The1RandomFool

2 жыл бұрын

​@@alainleclerc233 Taking f(z) = z^i/(z^2 + 1) the denominator has only two poles, at i and -i. Only i would be in the enclosed region. Did you not click on the timestamp I linked?

@alainleclerc233

2 жыл бұрын

No I did’n’t click on your timestamp because I thought you referred to the intégral from -infinity to infinity which was consistant with your half-circle proposal. Such a function icos(t)/(2cosh(t) has an infinity of poles in your contour

Wolfram Alpha gives; integrate (tan(x))^(i) from x=0 to x=pi/2 = π sinh(π/2) csch(π)≈0.62602 which is simply equiv. π /[e^( π/2)+e^(- π/2)] as Michael given at end.

for a bit shorter notation we can use the hyperbolic cosine: (exp(x)+exp(-x))/2=cosh(x). it's easy to see that cos(ix)=cosh(x)

@alainleclerc233

2 жыл бұрын

Agree ! It’s better only because it’s nicer. The intégration of the vertical segments is much easier to handle with the shorter notation as only half of the expressions disappear while the other half does not but has à zéro limit while R goes to infinity.

4:32 I think, you should have mentioned that it works only if integral converges. For example, somebody may think that integration of sinx from -∞ to +∞ can be done in such manner due to symmetric limits and odd integrand (no, this integral diverges and is not equal to zero)

@mcqueen424

Жыл бұрын

It still “doesn’t work” even if it converges; f(x)=x integrated from -inf to inf has a Cauchy principle value of zero, but it (the integral) doesn’t actually exist.

You can solve the integral very easily using the beta function and the euler reflection furmula

The final answer should be pi*(e^pi/2+e^(-pi/2))/(2+e^(-pi)+e^pi).

@JM-us3fr

2 жыл бұрын

I’m pretty sure the final answer is actually pi/(e^(pi/2)-e^(-pi/2))

@alainleclerc233

2 жыл бұрын

@@JM-us3fr I confirm!

@koengroot3285

2 жыл бұрын

Ah, I needed a double take as well: your denominator is a perfect square: e^(-pi)+2+e^pi = (e^(-pi/2)+e^(pi/2))^2

I understand all the steps and it’s really fascinating. I just don’t know how people get so good at recognizing what substitutions to make.

19:19

A very nice and interesting example !

I wrote the numerator of the u integral as e^(i ln(u)), and used a keyhole contour with the ln branch cut on the positive real axis. The integral below the branch cut is -e^(-2 pi) times the integral (the original) above the cut. There are simple poles at +/- i.

Why does it say you need a full math major to understand. Teach us complex analysis, professor michael. I love you , and i will donate.

Nice T-Shirt, Paul !!

Even if the last 2 integrals didn't cancel, they would still vanish as R->infinity, right?

@RishaadKhan

Жыл бұрын

yes. we can also use integral inequalities to arrive at 0.

the result can be expressed further as pi/2 sech(pi/2)

couldn't you also argue that for the thin sides of the contour, in the limit as R goes to inffinity, the e^R in the denominator basically kills the entire thing?

@mingmiao364

2 жыл бұрын

No. Because the cos function for complex variables is also unbounded. So the numerator goes to infinity as well

@ayernee

2 жыл бұрын

@@mingmiao364 in the interval under consideration it is bounded

@uriaviad9617

2 жыл бұрын

Yes this also works

@alainleclerc233

2 жыл бұрын

@@mingmiao364 Here it works when you develop everything in trig and hyperbolic functions of real numbers

@slenderman478

Жыл бұрын

Why are ppl making those vertical integrals so complicated and handwavey? The denominator can be written as 2*cos(R+/-i*pi) and the whole integrands cancel out to i/2 (he missed a factor of i from the contour jacobian). The opposite bounds then eliminate the two.

Wolfram alpha calculates Laplace transform of cos(t)/cosh(t) with digamma function (derivative of logarithm of Gamma) If we plug in s=0 in Laplace transform of cos(t)/cosh(t) we will get the answer This digamma function has complex argument

You could have already solved it for general complex s as in I(s)=int_0^infinity dx x^s/(x^2+1) by considering the integral int_C z^s/(z^2+1) dz where C={-R+i0,R+i0} + {|z|=R,Im(z)>0} and R->infinity, choosing the cut of x^s to be (-infinity,0). The integral over the arc |z|=R vanishes in the limit R->infinity for Re(s) € (-1,1) and for those s also the integral over the semicircle of radius epsilon at z=0 vanishes as epsilon->0. The integral over (-infinity,0) is the same as the one over (0,infinity), except it picks up a phase factor of e^{i*Pi*s). On the other hand the residue is easily obtained to be e^{i*Pi/2*s}/(2i) for the pole at z=i. Hence (1+exp(i*Pi*s))*I(s)=2Pi*i*exp(i*Pi/2*s)/(2i)=Pi*exp(i*Pi/2*s) which gives I(s)=Pi/2*1/cos(Pi*s/2).

Is there a general method for deciding when to make the upper part of the contour a rectangle rather than a semi-circle? I usually see a semi-circle in these types of problems on KZread. Also, how are you inspired to use the e**t substitution?

Hey, Professor Penn. Would you please explain in full detail why the values of the integrals along the vertical parts of the contour are equal and opposite? Thank you.

@alainleclerc233

2 жыл бұрын

Only the real parts are equal and opposite. The imaginary parts are not : they are equal but have a zero-limit when R goes to infinity

@krisbrandenberger544

2 жыл бұрын

@@alainleclerc233 Thank you!

the plot of (tanx)^i scares me.

QNCUBED3 would do it much better!

Teacher: Today I've got a nice integral. The integral: 👁👄👁

how do you come up with the subs? that's amazing - i am crushing on the math

@xulq

2 жыл бұрын

you mean the first subs for tanx or the parametrizations?

@thandolwethu3430

2 жыл бұрын

@@xulq everything - all the subs and approach

@xulq

2 жыл бұрын

@@thandolwethu3430 okay i will give it a try but please keep in mind im not a native english speaker and i have only studied math in school until 12th grade, im self taught. i hope it still helps^^ well about the first substitutions: the original integrand has an i in the exponent which makes this integral hard in the first place. so the substitution u=tanx at least made this easier by not having to deal with trigonometric functions anymore and also when i first watched this video i honestly thought he'd use the beta function from that point on or go straight into complex analysis/residue theorem. but the second substitution is brilliant because by letting that u=e^t we now have the i composed into the exponential function and that means we can use eulers formula to split it into the real and imaginary parts. now we have to only deal with 2 real integrands and a factor of i which doesnt really bother us since it can be pulled outside the integral. im not sure how to explain the paramatrizations but you might notice that all those 4 path integrals are actually being parametrized since we are integrating over a path where 1 variable is changing. lets take a look at "our original integral" which goes from -R to R. here the changing variable is R while the imaginary part is constant at i*0 as we integrate along that path, so we could parametrize it like this: z=t+i*0 and dz=dt where t is an element of the interval [-R;R] and viola we parametrized it so that we can now integrate from -R to R since our variable t is an element of that interval (it feels like a substitution) now lets look at another example, the path integral on the right side of the rectangle: here you can see that the R stays constant while the imaginary part varies from 0 to pi as we "go" along the path so our parametrization would be: z=R+i*t and dz = i*dt where t is an element of the interval [0; pi] now we can integrate from 0 to pi since our function has been parametrized with the that variable t

@xulq

2 жыл бұрын

maybe someone else might explain it much better im sorry^^

@thandolwethu3430

2 жыл бұрын

@@xulq you explained well thank you

Mr Penn so you are a dune fan too? It's my favourite book! :D

pi/2 / cosh(pi/2) is elegant

@knivesoutcatchdamouse2137

2 жыл бұрын

I think (pi/2) sech(pi/2) looks better but yeah, it's nice. It's also (pi * e^(pi/2)) / (1 + e^π)

I don't see the algebraic equation in the final step. Why can't you just divide the residue result with the (1+cos(i*pi)) after being expanded?

@xizar0rg

2 жыл бұрын

I think he meant "algebraic" as a shorthand for "non-trigonometric" rather than an actual algebraic expression. As for why he didn't just divide by 1+cos(i pi), he *did* do that.

@christianbuzzio1002

2 жыл бұрын

@@xizar0rg The thing is I computed both results and they are different. The division of the residue and (1+cos(i*pi)) results in 0.68... while pi/(e^(pi/2)+e^(-pi/2)) is 0.626... I didn't do the simplification by hand, maybe he made a simple mistake, but I thought that maybe there is another calculation that I'm completely missing

@krisbrandenberger544

2 жыл бұрын

That does need to happen.

@christianbuzzio1002

2 жыл бұрын

@@krisbrandenberger544 what?

@krisbrandenberger544

2 жыл бұрын

@@christianbuzzio1002 The division of (1+cos(i*pi)) did not happen when it needed to.

Hi. We are used to taking the big semicircle connecting -R to R as part of the contour. I guess it wouldn't work in this case and the integral over it would not be zero. What is the reason?

@brandonwillnecker8060

2 жыл бұрын

The semicircle contour would include infinitely many more poles because of the periodicity of the exponential. You'd have all the poles along the imaginary axis in the upper plane. It's much easier using this rectangle because there's only one pole to deal with.

@erfanmohagheghian707

2 жыл бұрын

@@brandonwillnecker8060 that's not the main issue though. The function on the semicircle does not tend to zero as R tends to infinity. That wasn't clear to me in the first glance

Kinda glad I did physics ;)

Does Wolfram's Mathematica get this right?

Feynman found it problematic to derive integration of tangent. For QM involves the complex plane and not only the real.

Prof right at the very end you made a mistake of not including the +1 with cos(ipi) Because 1+cos(ipi) = 1 + (e^-pi+e^pi)/=(2+e^-pi+e^pi)/2 The final answer I got for that beautiful integral is pi(e^pi/2+e^-pi/2)/(2+e^pi+e^-pi)....

@lingzhao5719

Жыл бұрын

I factored it out.same answer

Could a semicircle be used as our contour?

@mingmiao364

2 жыл бұрын

Beware that as the radius of the semicircle goes to infinity, the contour will include infinitely many singularities. That means that in the residue theorem, the sum of residues becomes an infinite series, which is not easy to calculate. The rectangle contour only contains one singular for all values of R

@youknowwhatsreallysofunny

2 жыл бұрын

@@mingmiao364 Ah right, I forgot about the periodicity of the denominator's solutions. Thank you for pointing it out.

@threedee5831

2 жыл бұрын

You can use an upper half circle as the contour of integration if you don't do the second substitution and instead just look to integrate u^ i / (1+u^2) = e^(iln(u))/(1+u^2) from 0 to infinity. In this case the upper circle part goes to 0, and the radius from -R to 0 is just a multiple of the integral from 0 to R (which is the integral we want to solve for). The only reside is at i.

at around 5 minutes you cancel out the odd function, my question is how do we know the -infinity and positive infinity are of the same magnitude? does it matter?

@randomname7918

2 жыл бұрын

I would also like to know

@Minskeeeee

2 жыл бұрын

the -inf to inf interval is taken as the limit of symmetric intervals around zero. this means that at all values of R, the odd functions will cancel thus they do in the limit

@yoav613

2 жыл бұрын

The integral sinx/(e^x+e^(-x)) from 0 to inf converges ,so this integral from -inf to inf is 0 because the integrand is odd.

@randomname7918

2 жыл бұрын

@@yoav613 wait does it converge? I thought periodic trigonometric functions like sinx in the numerator didn't converge Maybe I'm thinking of something else Because like it would be just, like, snaking over and under around 0 wouldn't it? Oh but it will be getting close and closer to 0, right?

@yoav613

2 жыл бұрын

@@randomname7918 yes. You can also use : sinx/(e^x+e^-x)

In my knowledge, complex power function is multivalved, e.x, 1^i = e^(-2 k pi), for all integers k. Why is the result here singlevalued?

@maxhofman6879

2 жыл бұрын

x^i = e^ln(x)i and because x is a real number it's more natural to only consider the real natural logarithm, which was done here

@murongwangqing

2 жыл бұрын

@@maxhofman6879 number 1 is real. but 1^i = e^(-2 k pi). Why should other values than 0 be excluded?

@maxhofman6879

2 жыл бұрын

If you let x^i be multivalued for real x then eulers formula e^ix = cosx + isinx which was used in the video also doesn't hold anymore because you would get e^ix=e^2kpi * (cosx+isinx)

@murongwangqing

2 жыл бұрын

@@maxhofman6879 yes.why not. If k=0,,then we have Euler Formel. There is no restriction on k. Why not e^ix = e^2k pi(cos x +I sin x). If multi value is allowed, the result in this video may look like (…)e^(2pi k) for all integer k

@murongwangqing

2 жыл бұрын

@@maxhofman6879 kzread.info/dash/bejne/iIWc165wZNS0gaQ.html kzread.info/dash/bejne/g3inpZinj7PAmco.html

Can someone demonstrate how the very last step is correct? Is there a typo?

@alainleclerc233

2 жыл бұрын

Yes. The correct answer is half of the one shown on the board

@alainleclerc233

2 жыл бұрын

No typo. Cancel my comment

@reeeeeplease1178

2 жыл бұрын

@@alainleclerc233 i dont get your comments? Others seems to doubt the final step too I think it is incorrect in the way it is presented

Your comment that an integral of an odd function over (-inf,inf) is zero is not necessarily true, you need to show it. For example, f(x)=x integrated from -inf to inf does not exist.

@nucreation4484

Жыл бұрын

That was my thought as well. kzread.info/dash/bejne/YoeElMRpnqnWp6w.html

If x = arctan(u) then dx = du/(u^2 + 1) not dx = du / u^2 + 1.

I have been along until minute 18:29, then he made to many steps at once kicking me out ...

Playing a bit hard and fast with some convergences/cancellations, I think.

@mingmiao364

2 жыл бұрын

Cancellation of the integrals over the “thin sides” of the contour is elementary algebra. There is no convergence to show in this problem because the integral over the two “wide sides” do not vanish as R goes to infinity

asnwer=t isit

Please credit the discoverer of this powerful technique: Augustin Louis Cauchy (1789-1857)

@norajcarnaj9207

2 жыл бұрын

just say Cauchy bud everyone knows who he is by saying his name

@kostasbr51

2 жыл бұрын

Hats off to Mr. Cauchy. One of the greatest mathematicians of all times.

@norajcarnaj9207

2 жыл бұрын

@@kostasbr51 yes and a French one hahha France>>>all

@RozarSmacco

2 жыл бұрын

@@norajcarnaj9207 I disagree. For a great man like him you give his full name.

@norajcarnaj9207

2 жыл бұрын

@@RozarSmacco nobody cares dude even in France when we use one of his theorems just Cauchy is required and that's all

Is it just me or is this video really quiet lol