integral of (-1)^x from 0 to 1

A complex integral of (-1)^x from 0 to 1, is it possible? What is the answer?
Euler's formula: • Euler's Formula (but i...
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[@AssemblyWizard]

You can eliminate the i and get a real number result: 2i/pi = 2/p

[@NoNameAtAll2]

XD

[@Raikaska]

Best comment

[@Tomaplen]

yes, this result is used in string theory

[@drdca8263]

p is imaginary though (it is about -3.1415i )

[@n484l3iehugtil]

angery react

[@Harlequin314159]

4:01 "i don't like to be on the bottom, i like to be on the top."

[@hikarifathan5143]

Lol

[@gorthorki]

hmm ok

[@azzanporter4377]

Lol i

[@sandipanplayz7282]

i noticed, i should have not

[@przemezio]

( ͡° ͜ʖ ͡°)

[@jerry3790]

This is one of the times where the calculations make sense, it’s just I have no idea what I’m actually calculating

[@raylee1222]

bluechalkredchalk

[@goodplacetostart9099]

No bluechalkredchalkwhitechalk

[@-minushyphen1two379]

Whitechalkredchalkbleuchalk

[@erincondron8105]

Redchalkwhitechalkbluechalk. AMERICA, F*CK YEAH!

[@tazking93]

YES, Please more on complex logarithms!

[@johnny_eth]

ln -x = i.pi.ln x

[@azzanporter4377]

This is integrals

[@bubbletea-ol4lr]

@@johnny_eth This was a year ago, but ln(-x) = ln(x) + ln(-1) = ln(x) *+* i(pi)

[@THOMAS_SHELBY434]

​@@johnny_eth but it's illegal to put negative number in log then how you put negative in log..........

[@attyfarbuckle]

OK, now integrate 1^x using the substitution 1 = e^42πi

[@angelmendez-rivera351]

attyfarbuckle Then the antiderivative is 1/(42πi) e^(42πix) in which case the integral from 0 to 1 is 0.

[@skylardeslypere9909]

Integrating 1^x dx is just integrating 1 dx Is just x :) From zero to one we just get 1 :)

[@gio5607]

@@skylardeslypere9909 whooooosh

[@ivornworrell]

afterfarbuckle, y u gadda complicate things? keep it simple farbucke!

[@ayushbudhiraja9056]

Why does this happen? What's the mistake in using euler form ?

[@bilz0r]

But what does it *mean* if an integral has many answers? I'm used to thinking of an integral as being the area under a curve or a sum of infinitely small chunks. How can either of those have multiple answers?

[@MrZkeggia]

bilz0r It is related to the fact that you can't extend so freely the logarithm to complex numbers (the integrand is not always a real number, for example (-1)^(0.5) = i). This means you have to extend the definitions of your functions to the complex plane. For the logarithm, one way is to say: log(z) = log( r e^(i theta)) = log(r) + i theta = log(|z|) + i arg(z). But the argument of z repeats itself every 2pi, so you must first decide what is the argument of z. This is up to you and eventually leads to the answer. Every time you choose an interval for arg(z) (geometrycally you are in a sense "turning around" the origin k times) your answer changes by 2pi k. Search for Rieemann sheets on Wikipedia to find more.

[@mike4ty4]

The curve itself is ambiguous - there is many possibilities (infinitely many) for the curve of y = (-1)^x. Each possible curve has one possible "area" under it (well not quite, because its complex numbers, so it doesn't directly represent an area, more like a corkscrew vector sum), and thus one possible integral. In particular, y = (-1)^x = e^(x log(-1)) but log(-1) is ambiguous with log(-1) = (2n + 1)pi i, n e Z. The reason for this is log is the inverse of exp (e^x), but exp is periodic with purely imaginary period 2pi i and so not injective and thus its inverse is not a function but a one-to-many relation instead (like arcsine and square root in perhaps more familiar real-analytic settings.). Thus so also must (-1)^x be ambiguous/one-to-many as well (at least at non-integer x; at integer x there is no ambiguity as that is just the sequence of powers flipping between -1 and +1.). If it is graphed, it looks like a bunch of helical threads wound around a cylinder of radius 1, where the "y-axis" is a complex plane (thus a real and imaginary axis), and the x-axis is just an axis perpendicular to that plane, and the cylinder is also so perpendicular. (You might want to imagine a spool of thread from a sewing kit.) The value of n controls the pitch and handedness of the helix, I believe positive n is a right-hand helix and negative n a left-hand helix (at least if your coordinate system is set up in the usual way). The integral looks like the "area" of a piece of screw-shaped sheet (like Archimedes' screw) between the x-axis and the curve (but as said the value of the integral is not the geometric area because the complex numbers have direction associated with them so will tend to cancel each other in various ways. It's like how that a negative part of a function cancels the area of the positive part, only more so, with more dimensions involved.), and there is one such sheet and so one such integral value for each of the different helical curves.

[@znhait]

This is a complex analysis problem. If you're or have only taken calculus where all functions are real, then this problem will not make sense. It only makes sense on the complex plane.

[@stranger0152]

Because Complex Functions have 4D space but integral is just 2D. So 4D spaces have infinitely many 2D spaces.

[@esse8407]

this is the black magic

[@lucazara9137]

Thank you very much blackpenredpen!! Very cool answer

[@ILGiullareDiCorte]

Luca Zara bellissima domanda! :D

[@paololeaer7838]

Cia

[@benjaminsanchez3735]

It's a false answer

[@benjaminsanchez3735]

You can't tell that [exp(i*pi)]^x=exp(i*pi*x) if x is not an integer For example, 1^x is always equal to 1 and 1=exp(2*i*pi) But [exp(2*i*pi)]^x is different from exp(2*i*pi*x)=cos(2*pi*x)+i*sin(2*pi*x) you can easily see that exp(2i*pi*x) is different than 1 when x is not an integer

[@simranakter007]

Ey are you still watching

[@angelmendez-rivera351]

Using the rule that Antiderivative(b^x) = b^x/Ln(b), this work just fine too.

[@manojbansal2798]

you are doing a great job through this channel. Keep going. It makes me happier

[@tommyrosendahl7238]

"ISN'T IT?"

[@blackpenredpen]

It is!

[@GeekTommy]

Really cool! Hope you'll make videos about other complex integrals in the future :)

[@zokalyx]

Keep the complex math coming!!! I love it

[@fizixx]

Love your videos. I already know these things, but I like your presentation and explanations.

[@blackpenredpen]

Thanks!!!!

[@koenth2359]

Very nice, keep up the good work! You explain every step so maticulously! Therefore, it would be so nice if you could work out some problems involving the residue theorem, which sometimes seems to look like a hattrick. For example the definite integral from 0 to pi of 1/(2+cos theta). That would be so great.

[@ZipplyZane]

Did Dr. Payam steal your markerboard?

[@sansamman4619]

ZipplyZane, OMG that's what I wanted to comment!

[@Egonkiller]

wonderful as always

[@HanhTangE]

I liked the video! It would be awesome if there is some geometric or in-depth explanation of what does it mean to integrate and get a complex answer!

[@kamoroso94]

Fun with imaginary numbers, I love it! Thank you bprp!

[@josephgrossenbacher7642]

i was glad & reliefed at the same time when you eventually "brought in" the "blue chalk" ... !!!

[@CossZt6]

This has so far been the most satisfying video I've seen in 2018

[@benjamindorsey2058]

Gorgeous! Well done mate!

[@cwldoc4958]

I enjoyed this video. That having been said, it is necessary to define specifically what is meant by the integral of a multiple-valued function. In fact what is being done is that for each integer, n, we are choosing a branch of (-1)^x and integrating that continuous, single-valued function. Just saying that we are integrating (-1)^x, without any further explanation is ambiguous, because we might try to integrate some function that is not one of the branches. For example, we could integrate the non-continuous function defined by e^[ i pi x] for 0

[@ivornworrell]

*Beautifully explained!*

[@Mtmtmtmtmtmtmtmtmtmtmt]

Fantastic video!

[@casa1420]

Parabéns meu amigo! Gostei muito do vídeo!

[@n0ita]

Such simple, but so cool answer !!

[@_DD_15]

I had no clue how to solve this one, as soon as I saw euler, i was like :that's a realllyyyyy smart way :)

[@allaincumming6313]

0:02 "Hello darkness my old friend..."

[@avtaras]

Best video so far !!! :)

[@steelguard]

Super cool - thank you!

[@erwinrojasarabia]

Hola, quería saludarte y felicitarte por tu trabajo. Te sigo desde los 1000 subcriptores :)

[@RITESHYADAV-se2ym]

Yes sir, i agree to your hyphysical solution. Thanks

[@mistervoldemort7540]

I like how random your problems are, but that you always find out the answers

[@luis96xd]

Amazing, this is so cool!

[@J7Handle]

I'm trying to find a decent approximation for the analytic continuation of f(x) = A^^x using a piecewise function. Because of the nature of the piecewise function (each segment "k

[@nicholasandrzejkiewicz]

That was a satisfying integral to solve!

[@drumervara]

I found out the solution of this by considering the integral as a sum of all complex numbers in a radius = 1 circle in the complex plane between 1 and -1. Then that sum will be 2 times the imaginary part of the same sum from 1 to i, which is an integration of sine function. Actually the same, just wanted to share. Love your videos mate! Always check my results to find out I missed all the solutions taking periodicity into account.

[@Re-lx1md]

Love the change to the chalkboard :)

[@SultanLaxeby]

but only if you choose a continous branch of the multivalued function (-1)^x. theoretically, you could switch branches while integrating and get any arbitrary result :)

[@mike4ty4]

If you want to go really far down that route you will need to dump Riemann integration in favor of Lebesgue integration (measure integral) :)

[@marcinbednara3825]

And what's the geometric interpretation of this result? When x move from 0 to pi, we get complex unit semi-circle. This line has the center of gravity in the point (0,2i/pi)

[@mathematicadeestremo6396]

We can easily convert the expression into e to the power something and then the integral become more easy

[@friedkeenan]

The best part is that you also found the average value of (-1)^x on [0,1], which makes it even crazier that it has multiple answers. I love complex numbers

[@Supware]

Great stuff as always dude. Isn't it!

[@jamesstewart2524]

427 likes 0 dislikes. love it. This might be the first KZbin video I have seen with 0 dislikes. Congrats!

[@michelkhoury1470]

Nice solution !

[@AnuragKumar-io2sb]

Wow !!cool 😎. i never thought that intergral gonna get to a complex value 😮

[@Quwertyn007]

Jeez these vids are cool! :D ...i still have problems wrapping my head around stuff like an integral having infinitely many solutions though x.x

[@TehCaprone]

So the area of this integral can be from complex to almost 0 for very big n values?

[@adamkangoroo8475]

This is SO cool!

[@hatsadythongin6100]

You should also point out where this integration is applied, especially in Physics or Electronics or any others. That will be more interesting to know.

[@dr.husseinalgusab4638]

amazing, hanks a lot

[@rybaplcaki7267]

Can you make video about analytic continuation of zeta function? How we get it?

[@user-fp6pi6wi5l]

Thanks for the video. And by the way, what is music in the end ?

[@ratnarajwora2674]

Hello blackpenred, can you do your next video on how to fix the area under the curve x^4 + y^4 = 2xy

[@tahajoedelhoum5508]

Ty so mush doctor

[@finalbossd]

Perfect! Now, if I ever get into an integral fight, I will be well prepared.

[@MsPataso]

Wow, really good video

[@tannercypret3171]

Looks at thumb nail, that equals infinity. Was not expecting the Euler identity substitution. Very cool.

[@srpenguinbr]

Actually, e^n*pi*i is -1, where n is an odd integer. So we would have 2/n*pi as our answer. there are infinitly many answers.

[@srpenguinbr]

Oh sorry, I had not watched the end of the video yet lol

[@qdav5]

Cool! Euler's equation is very useful.

[@PeterBarnes2]

You should use white chalk and blue chalk so that you can invert the colors in editing and get blackchalkredchalk.

[@davisouza7762]

Holy this was nice

[@soumyajotiroy9185]

As i am 11 th standard student Often this comes in my mind whatis integration or differentiation complexe no. Really your technique impresse me . Thanks for your help.

[@restcure]

As you move across the screen, the camera gets lighter and darker. Maybe turn of the auto exposure?

[@Dlmlai7255]

We can also use i^2 in place of -1

[@workforyouraims]

Man you are clear and whenerver I have free time I enjoy some math from you,but can you explain how is it possible that the area under (-1)^x has an imaginary value.I understand the calculations but I can not grasp the implication.

[@cesarmendoza8959]

Esta integral se resuelve más fácil y rápido usando integrate[(i^2)^x] = integrate[(i)^2x]=(1/2)(i^x/ln(i) de 0 a 1; de esta forma la integral es inmediata.

[@RomanNumural9]

You should try to integrate from -1 to 0 a similar function using residues and see what happens.

[@blue_blue-1]

Kudos!

[@evanmorrison3232]

Oh boy, just add a unit step and you can have some oscilitory response functions. Extra fun.

[@jemcel0397]

0:03 - 0:08 - Let us all press "F" to pay respect for the fallen red chalk. And blackpenredpen might've probably went super saiyan off the video.

[@sebmata135]

I don't understand the result. What is the geometrical interpretation of having multiple answers for an integral? What is the geometrical interpretation of a complex integral for that matter? *edit: integer->integral

[@lkjkhfggd]

sebmata don't know the answer for the second question, but the function e^itheta spins endlessly around the origin, so there's infinite ways you can go from 0 to 1 because you can changed the number of times you spin around.

[@AndDiracisHisProphet]

an imaginary integer is an integer that lies perpendicular to the real axis. that's the geometric interpretation^^

[@sebmata135]

Sorry I meant the geometrical interpretation of a complex integral

[@AndDiracisHisProphet]

maybe there is none?

[@angelmendez-rivera351]

There really is no interpretation. It’s just a generalization of the concept of an integral. Much like how there is no interpretation of the concept of the generalization of Harmonic numbers to complex arguments, but it is still a nice concept to have.

[@imperialrecker7111]

He is asserting his dominance on us

[@flamingpaper7751]

What if we did an integeral from -1i to 1 of x^i for instance? Would it work if you travel from one axis to another axis?

[@purim_sakamoto]

最初こういう動画見たら、すっげ！マジック！って思ったけど 見慣れてくると「こう定義しましたので」と言ってる動画なんだなーって思うようになってきました😙

[@12lfc321]

Hello from Norway! A problem for you: prove a correlation between integral of ((lnx)^ndx) from 0 to 1 and (n!)

[@dwijbhatt3822]

Hi blackpenredpen. Hope you are doing good in this pandemic. Can you explain application part of such complex function integrals? Also, here ans is function of n. (i.e. ans=f(n).) What does it look like area under the curve.

[@ahmedhamed4773]

I cannot describe beauty and splendor

[@vbcool83]

The thing that comes to my mind is Euler's identity!

[@garzoness]

Very Good!

[@zamkove1278]

I solved this using a quicker method First you rewrite (-1)^x as i^2x Now you just have to divide by the derivative of the power and the natural log of i and you can get the anti derivative of i^2x As you might know, i can be written as e^(pi/2)i and when you take the natural log of that you just get (pi/2)i The 2s cancel out on the denominator and you are left with (i^2x)/(pi*i) You can fix this to be (i^(2x-1)/pi) Plug in 1 and 0 and subtract them and you get (2/pi)i

[@arjunbhardwaj2883]

Thank you so much.....

[@Greasyhair]

Set x = 2y. Integral changes from (-1)^x to (-1)^2y from 0 to 1/2. Then the integration answer is simply 2(1/2-0) = 1.

[@ajaymishra7212]

You can use de moivre theorem in step 3 too.

[@papsanlysenko5232]

Hello, i just came up with this quation, what is int from 0 to 1 of ln(x^2+e^x)? Im not quite sure if its really possible, may be some complex analysis i dont know...

[@lorenzobarbolla7988]

Please, more complex integration

[@usmanismail4130]

This is really interesting

[@laoskdjfasoldkfj3116]

Whats would be a possible way, when you have got e.g. the integral from 0 to 1 of (-2)^x?

[@naragames789]

Limit of x to 0 (3x-Sin3x)/(2x-Sin2x)

[@_deleted_12472]

Does this mean that the value of the integral can be arbitrarily small? (Since 2*i/(pi+2*n*pi) -> 0 as n -> infinity?)

[@lkjkhfggd]

Can you explain what the result of 2i/pi means? Isn't the integral the area between the curve and the x axis? Then what does an imaginary area mean? Or does the integral not represent that in the complex plane

[@qqqquito]

If this is an integral on the complex plane, shouldn't we be concerned with what the path is from 0 to 1? Or is the result independent of the path?

[@yausimon9549]

The function is discontinuous over the interval, the integral doesn't exist. It doesn't meet the existance condition of definite integral, U=L.

[@blackpenredpen]

U need some love!

[@leodip97]

I've been wondering, are there any relationships a=b where both a and b are real, but the relationship can only be proved using complex numbers? The question is not well posed, so I'll need you to understand what I mean (and perhaps formulate better the question), but when I say that a and b are real I mean that both of those don't need complex numbers to be defined. For example, -1=exp(iπ) is indeed a=b, and both a and b are real, but exp(iπ) has i in it, and that's a problem. Any ideas?

[@jeffreyluciana8711]

Excellent

[@user-gd6er8qo1h]

I hope you define the integral complex

[@fernandobolanos7748]

Hello Mr. BlackPenRedPen. I think there's a mistake in the procedure. Integrating ⌠e^iπx·dx has to be done by variable substitution: for example u = iπ·x, and du = iπ·dx; which means that the boundaries of the integral must also be changed: x=0 → u=0 and x=1 → u=iπ !!! I think that when changing from ⌠e^iπx·dx to (1/(iπ))·⌠e^u·du the upper integrating boundary had to be changed from 1 to iπ. Am I wrong?

[@christoskettenis880]

Very nice and cool indeed!

[@kushagragupta7051]

Are all the answers equal? If they are, then can I cancel all the 2i, reciprocate the terms and get π=3π=5π=7π and so on?