Complex Analysis: Fancy Branch Cuts

[@girianshiido]

Saying that a logarithm of a product of complex numbers is the sum of the logarithms of its factors is of course incorrect. It is true in some cases but it’s not true in general. This is one of the difficulty of dealing with logarithms of complex numbers. The chosen branch cut of log(z^4+1) is not equal to the natural logarithm of z^4+1 when z is real (you have +8iπ). The arguments have to be chosen carefully in order to cancel themselves.

[@qncubed3]

You are correct, completely overlooked that. Should be fixed if we adjust the argument of the lower two branch cuts by subtracting 2pi :)

[@Pyroguy92]

Greeeeeeeetings everybody...lol im trained to hear that every time I click on one of your videos. I'm doing my phd in physics right now and I want to thank you so much for all your help on complex analysis. By far, this channel has been the reason I've been able to handle a lot of the mathematical physics problems they throw at me on exams...

[@manstuckinabox3679]

I love how intimidating 6:23 the product sign is, they're like these "I'm a tough guy" mini people you can easly destroy with a log. (in both cases ;p).

[@kummer45]

A 3 dimensional picture of this surface illustrates WHY such multi branched functions requires SUCH contour. Contouring these functions is the main task because it reveals the nature of such multi branch functions that cuts with itself or surfaces that clip with each other. Would be great if we could have 3D images of what is going on through the process of integration. I know that such videos are intended to be aimed at scholars with the intention of finding the Riemann sum or a closed form of such sum. But there's a lot of more happening when such contour is decided in that way around the POLES. For me this is my daily morning. Classical complex integration IS a theme on itself. There should be an ENTIRE book dedicated only to this fascinating topic.

[@captainkim7738]

Been looking for your new uploads. Thanks for the video. P.S. I’m a former VCE student. :))

[@TheHellBoy05]

Sick video. Now do this again, but switch the quartic with the quadratic, and vice versa. I tried it and I messed up. And so, I turn to you. Love from India🔥🔥

[@Decrupt]

New video, hype!

[@Decrupt]

And these truly are some fancy branch cuts.

[@The1RandomFool]

I've evaluated this before, but I actually used Feynman's technique to differentiate the logarithm under the integral sign. I then used contour integration on that. It's interesting to see how to evaluate it the way it is.

[@ahmetleventtakr7625]

I tried Feynman’s trick too but got stuck after the partial fraction decomposition. How should I proceed?

[@The1RandomFool]

@@ahmetleventtakr7625 I didn't use partial fraction decomposition after Feynman's technique. I used the residue theorem to evaluate the resulting contour integral. But in your case, I imagine you have to factor the quartics in the denominators into a product of two irreducible quadratic polynomials.

[@violintegral]

@@ahmetleventtakr7625 if you factor x^4 + 1 into (x^2 + sqrt(2)*x + 1)(x^2 - sqrt(2)*x + 1), you can use log(ab) = log(a) + log(b) to split the integral into two separate, but very similar integrals. Using the substitution u = -x, it becomes clear that these two integrals are equivalent, leaving us with only one integral to solve. From there, you can use Feynman's trick to evaluate the integral of the parameterized function log(x^2 + tx + 1)/(x^2 + 1) w.r.t. x from -inf to inf. After taking the partial derivative of the integrand w.r.t. t, what follows is just simple calculus integration techniques. To find the initial condition, set t = 0 in your parameterized integral, and employ the substitution x = tan(u). Here we run into the integral from 0 to pi/2 of log(cos(u))du, which is a famous integral, commonly solved using the symmetry of the integrand. This method requires only the use of real-analytic techniques, no complex analysis.

[@zestyorangez]

studying for my complex prelim. ty for the resources ^-^

[@lordstevenson9619]

Starting the holidays with a juicy integral. This is a throw away thought, since I haven’t done a complex analysis course my knowledge is based on random videos, after seeing this example it’s made me think. My question is, are branch cuts only linear intervals or could they be polynomial or some other function? Like a function not being defined along some nonlinear branch cut.

[@qncubed3]

On the riemann sphere, a branch cut is a simple curve connecting two branch points. For example, log(z) has a branch point at 0 and infinity, and a branch cut for log(z) can be any nice enough curve from 0 out to infinity. So the answer is no, branch cuts do not have to be linear, however we usually work with straight line branch cuts since they are easier to handle.

[@lordstevenson9619]

@@qncubed3 That’s awesome thanks for the explanation. ❤

[@neelmodi6693]

Subscribed for the Frederic Schuller clouds lol

[@pl412]

Chair intro hype!

[@Dominikbeck12]

Try this: Integral (ln^2(x))/(x^4 + 1) from 0 to infty! (To solve this alternatively, I used the Feynman's method, differentiating the integral wrt a cleverly inserted parameter)

[@trannam2185]

In 20:14 how to know which side of omega should we approach for each psi? Why psi 1 need to spin around whereas psi was just approach directly?

[@qncubed3]

For Omega 1, the argument of psi1 can only approach pi/4 + 2pi since that is the argument range defined

[@trannam2185]

Is that psi1 takes the negative pi/4 value, but it needs to turn around since it go opposite with the direction of Omega1 and due to the angle definition, as well?

[@qncubed3]

@@trannam2185 The path orientation has nothing to do with the argument the we are approaching. We only have access to the angles on the interval (pi/4,9pi/4). Psi 2 approaches an angle of pi/4 and psi1 must approach 9pi/4 since it is coming form the other direction.

[@Dominikbeck12]

I found another solution. Denote our integral as I. Second, denote N = int 4ln|x|/(1+x^2) dx from -infty to infty. This integral vanishes (as can be shown simply by x=1/y substitution). Then, you can perform by parts on I - N, giving you an expression I = 4 int (arctanx)/(x (x^4+1)) dx from -infty to infty. Let J = int_C ln (1+z)/(z(z^4+1)) dz, where ln has branch cut from -infty to -1. The contour C is C_1 U C_2, where C_1: z = it and C_2 = R e^(it) with t from -pi/2 to pi/2

[@vascomanteigas9433]

The result are the same given by Mathematica 4.1, runing on my phone using the modded ExaGear Windows Emulator. It requires to use FullSimplify after ComplexExpand and FunctionExpand, once done the integral.

[@superkiller5800]

integrating from 0 up to infinity of ((ln(x^(2)+x+1))/(x^(2)+1)) dx. I would really appreciate it if you make a video about it. I managed to solve it from -inf to inf but i have no clue how to solve from 0 to inf

[@vascomanteigas9433]

The integral of exp(-x^2)/(x^2+1) along all real line equals to %e*%pi*erfc(1) and not %e*%pi as sugested by Residue Theorem due to essential singularity of exp(-x^2) at x=Infinity*%i. It can be bypassed, or the Feynmann technique are the only way ?

[@lordstevenson9619]

When your evaluating integrals over separate curves Γ, ψ1,ψ2. Is there a common way of knowing which integrals contribute 0, before computing them, is this intuition or a theorem or just knowing beforehand?

[@theelk801]

usually if it’s over a circular arc then you can show that it goes to 0 in the limit

[@theelk801]

yessssss you took my request

[@qncubed3]

:D

[@theelk801]

thanks so much for making this man it was driving me crazy trying to solve it but I’m glad to see I did the same things you did

[@codyriceandothers]

Is it possible to apply the same techniques for integrating log(1+x^6)/(1+x^2)? I tried the techniques used in this video and since there is a branch point at z = i in my case, I ended up integrating 1/(1+z^2) from i to i(infinity), which diverges.

[@maximebree4360]

In that case I guess that you can't do it but you can integrate log(z⁶+1)/z²

[@karomusaelyan338]

Hello can you slove this integeal 0to infinity ln(x²+1)/(x²+1)

[@karimshariff7379]

Nice video. Please write larger though. Thanks.

[@DJe451]

❤️❤️❤️

[@pascal3458]

🔥🔥🔥🔥

[@taterpun6211]

You can also solve this with real analysis (I just realized this was not the integral, but hopefully still a cool solution) I = (-∞,∞)∫log(x^4 + 1)/x^2dx = 2(0,∞)∫log((x^2 + 1)x^2)/x^2dx = 2(0,∞)∫log(x^2 + 1/x^2)/x^2dx + 2(0,∞)∫log(x^2 )/x^2dx = 2(0,∞)∫log(x^2 + 1/x^2)/x^2dx - 2(0,∞)∫log(1/x^2 )/x^2dx define I(t) = 2(0,∞)∫log(tx^2 + 1/x^2 )/x^2dx , t>= 0 note I = I(1) - I(0) I'(t) = 2(0,∞)∫1/(tx^2 + 1/x^2)dx u = 1/x -1/u^2du = dx 2(0,∞)∫1/(x^4 + t)dx t^1/4u = x t^1/4du = dx 2t^1/4(0,∞)∫1/(tx^4 + t)dx = 2t^-3/4(0,∞)∫1/(x^4 + 1)dx integrate now from 0 to 1 or solve unparameterized integral 2t^-3/4(0,∞)∫(1/x^2)/(x^2 + 1/x^2)dx = 2t^-3/4(0,∞)∫(1/x^2)/(x^2 + 1/x^2)dx = t^-3/4(0,∞)∫(2/x^2 + 1 - 1)/(x^2 + 1/x^2)dx = t^-3/4[(0,∞)∫(1 + 1/x^2)/((x - 1/x)^2 + 2)dx - (0,∞)∫(1 - 1/x^2)/((x + 1/x)^2 - 2)dx] let's find indefinite integrals with c = 0 to avoid interval problems u = x + 1/x du = (1-1/x^2)dx v = x - 1/x dv = (1 + 1/x^2)dx ∫1/(v^2 + 2)dv - ∫1/(u^2 - 2)du 2^1/2r = v 2^1/2dr = dv 2^1/2s = u 2^1/2ds = du 2^-1/2[∫1/(r^2 + 1)dr - ∫1/(s^2 - 1)ds] = 2^-1/2[arctan(r) - 1/2∫1/(s - 1) - 1/(s + 1)ds] = 2^-1/2arctan(2^-1/2(x - 1/x)) - 2^-3/2(log(2^-1/2(x + 1/x) + 1) - log(2^-1/2(x + 1/x) - 1) = 2^-1/2arctan(2^-1/2(x - 1/x)) - 2^-3/2log((2^-1/2(x + 1/x) + 1)/(2^-1/2(x + 1/x) - 1) now we can use result in original integral t^-3/4[2^-1/2arctan(2^-1/2(x - 1/x))|(0,∞) - 2^-3/2log((2^-1/2(x + 1/x) + 1)/(2^-1/2(x + 1/x) - 1) |(0,∞)] I'(t) = 2^-1/2t^-3/4 integrate from 0 to 1 (0,1)∫I'(t)dt = (0,1)∫2^-1/2t^-3/4dt I(1) - I(0) = 2^3/2πt^1/4|(0,1) note I(1) - I(0) is our original integral I = 2^3/2π

[@stephyjose9533]

Can you help with the integral $\int_ {- \infty} ^ {\infty} dk\frac {e^ {-a\sqrt {s+\iota k}} e^ {-a\sqrt {s-\iota k}}} {(s+\iota k) (s-\iota k)} $ ?

[@Carnifindion]

ξ is pronounced like 'hexi' but without the 'he'