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
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?
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.
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.
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.
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
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.
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 With the Lebesgue integral, what would the result be? I don't know anything about measure theory mixed with complex analysis and especially multivalued functions with branches.
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
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.
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
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)
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
There's some real trickeration going on here! The Riemann surface of (-1)^z is a helicoid: a sequence of sheets connected together along branch cuts on the negative real axis, because -1 = e^(ln(-1)) and the log is multivalued on the negative reals. However, the integral itself is along some contour from z=0 to z=1 on a particular sheet. (It doesn't matter which contour is used because the exponential function is entire). The "multivaluation" of this integral comes by specifying the sheet, i.e. the particular value of ln(-1) as mentioned at 4:35, which appears in the antiderivative. The principle value of the integral, of course, uses ln(-1) = i pi, which results in the original answer 2i/pi. If you want to be really clever, you can specify that the "z=0" point is on one sheet and the "z=1" point is on another sheet, by taking a path that loops around z=0 some number of times (i.e. traveling up or down the helicoid) before landing on z=1. The particular contour doesn't matter, except for the winding number around z=0. The value of ln(-1) differs between the start and end points of the contour because they are on different sheets, so you can get even more values out of this integral, i.e. i/(2n+1)pi + i/(2m+1)pi, where m and n are the numbers of the sheets (e.g. the principal values are on sheet number 0). If you form a simple, closed contour (e.g. z=0 on sheet m to z=1 on sheet n; z=1 to z=0 on sheet n; z=0 on sheet n to z=1 on sheet m; and z=1 to z=0 on sheet m) the formulas above for integrating along each leg of the contour will all add to zero. This is expected, since as noted above the exponential function is entire so this contour does not enclose any poles. Yay for making complex integration... complex.
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.
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
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.
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.
For those of you unfamiliar to the maddening delights of Complex Analysis. Don't think in terms of area to extend your intuition of integrals into complex numbers, but rather to the fundamental theorem of calculus. Then add the real variable calculus fact that the primitive or anti-derivative of the exponential of any non-negative base involves the logarithm and that complex logarithms are multi valued.
I solved this in a really weird informal/visual way: First off, -1 represents a rotation of 180 degrees (and no scaling), so (-1)^x is a rotation of 180*x degrees. These rotations correspond to points on the unit circle on the complex plane. For x ranging from 0 to 1, these go from 0 degrees to 180 degrees, making up the upper half of the circle. Since the x is picked uniformly between 0 and 1, all the points of this half-circle are weighted equally and their weights add up to one. That means we're just looking for the center of mass of the upper half of the unit circle. Since the half-circle is symmetric across the real axis, the real component of the center must be 0. To get the imaginary component, we look at how the vertical positions are distributed. For any angle theta, the vertical component is sin(theta). So we integrate sin(theta) from theta=0 to pi, giving us 2, but then we've given the sines a total weight of pi, so to get the average, we divide by pi. That gives us the final answer of 2/pi. As you can tell, my informal approach got me the right answer (possibly faster than if I had tried it by computing the integral rigorously). However, there's plenty of reasons to do a rigorous proof. A rigorous proof is much less likely to be considered "insufficient" by other people. Someone else could complain that switching between angles and complex numbers, and between averages and integrals, requires that I speak more precisely, mention the theorems I'm implicitly using, etc.
The polar form of (-1) = cosΠ+ i sinΠ By De-Moiver's Thm, (-1)^x = [cosΠ + i sinΠ]^x = CosΠx + i SinΠx. So its integral is 1/Π [ SinΠx - i CosΠx] . Applying the limits from 0 to 1 , we get the answer 2i/Π. I think this is simple
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.
First you have to define (- 1) ^x , If x is real. Then you can try to integrate the (complex) function....there are many ways to define a function f(x) = (-1)^ x.
hey blackpenredpen. I have a problem for you! the antiderivative of log(arctan(x)). This function is fun since in the real world it's an entire function on its domain, and in the complex world it's the composition of two multi valued functions!
This has no antiderivative in terms of any standard mathematical functions according to Wolfram at least. So there would be no way to express the result :(
mike4ty4 no but the power series of it is very nice. xlog(arctan(x)) + Σc(n)*arctan(x)^(2n+1) /(2n+1) where c(n) is the nth coefficient of the maclaurin series of tan(x)
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.
The solution has a problem. Let's calculate ∫ 1dx (0-->1) = ∫ 1^x dx (0-->1) = ∫ e ^ (2i*pi*x) dx (0-->1) = e ^ (2i*pi*x) (0-->1) /(2i*pi) = 0, but actually the result should be 1. The major reason the describe by @Yann Le Dréau, the formula e^ (a*b) = (e^a)^b only sustained when a is real number or b is integer
@@moeidmehri2153 well, the anti derivative is -ln(cos(x)-sin(x)) but plugging in 0 and pi/2 won't work (gives you a complex result) since the integral is improper because you're dividing by 0 half way through. So either the function your teacher was dealing with was different or he did some funky stuff.
This is a very bizarre question, in my opinion. I'm not sure if the answer really makes sense. One thing to keep in mind is that roots like (-1)^(1/2) and (-1)^(1/3) etc have multiple solutions in the complex plane. So, maybe there is some deeper meaning whereby the choice of n defines which of the solutions you choose? In a sense, there is infinite solutions to (-1)^(1/2). Perhaps these infinite solutions correspond to n in some way? For example: (-1)^(1/2) = (e^(i * pi *(2 * n + 1)))^(1/2) = e^(i * pi * (n + 1/2))
This one is ambiguous until you choose which particular branch of the complex logarithm you wish to use. And you have to consistently use that branch for all steps in a calculation or else you will get a nonsense answer. This is incredibly important to keep in mind in certain areas of physics and engineering.
Eek. What is a negative number to a “power”? The polar decomposition establishes a association of -1 with an equivalence class of polar representation, pairs of real numbers, magnitude and phase, -1,0 has the polar decomposition 1, 2 pi n with n an integer. -1 to the x has the polar decomposition exp(I pi x + 2 I pi n x). = 1, i pi x + 2 I pi n x that for irrational x can result in an equivalence with any unit modulis complex number. You can define the integral with all n=0, a “principal value”, but why? Doesn’t that need to be justified, stated up front, something? What precludes using alternate n, even n(x)?
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?
Sorry I didn't follow the bit where you put the 2 n π in the integral. I think the derivative of i π p + 2 n π wrt x is i π, not i π + 2 π, so the integral between 0 and 1 is 1 / i π( e^ (i π(1) + 2 n π) - e^ i π (0) + 2 n π)) so I dint see why the 2 n π terms don't just go, because e^iπ(1) + 2 n π) = -1 for all n.
No this trick doesn't works, you have to explain why you can multiplicate the powers ! The formule (e^a)^b = e^ab works when b is an interger but it is not always true. For example if we assume that is always true, then -1 = e^ipi = e^(2 i pi/2) = (e^(2 i pi))^1/2 = 1^(1/2) = sqrt(1) = 1
You can eliminate the i and get a real number result: 2i/pi = 2/p
XD
Best comment
yes, this result is used in string theory
p is imaginary though (it is about -3.1415i )
angery react
This is one of the times where the calculations make sense, it’s just I have no idea what I’m actually calculating
YES, Please more on complex logarithms!
ln -x = i.pi.ln x
This is integrals
@@johnny_eth This was a year ago, but ln(-x) = ln(x) + ln(-1) = ln(x) *+* i(pi)
@@johnny_eth but it's illegal to put negative number in log then how you put negative in log..........
4:01 "i don't like to be on the bottom, i like to be on the top."
Lol
hmm ok
Lol i
i noticed, i should have not
( ͡° ͜ʖ ͡°)
bluechalkredchalk
No bluechalkredchalkwhitechalk
Whitechalkredchalkbleuchalk
Redchalkwhitechalkbluechalk. AMERICA, F*CK YEAH!
Thank you very much blackpenredpen!! Very cool answer
Luca Zara bellissima domanda! :D
Cia
It's a false answer
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
Ey are you still watching
OK, now integrate 1^x using the substitution 1 = e^42πi
attyfarbuckle Then the antiderivative is 1/(42πi) e^(42πix) in which case the integral from 0 to 1 is 0.
Integrating 1^x dx is just integrating 1 dx
Is just x :)
From zero to one we just get 1 :)
@@skylardeslypere9909 whooooosh
afterfarbuckle, y u gadda complicate things? keep it simple farbucke!
Why does this happen? What's the mistake in using euler form ?
Using the rule that Antiderivative(b^x) = b^x/Ln(b), this work just fine too.
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?
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.
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.
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.
Because Complex Functions have 4D space but integral is just 2D. So 4D spaces have infinitely many 2D spaces.
this is the black magic
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
you are doing a great job through this channel. Keep going. It makes me happier
Fun with imaginary numbers, I love it! Thank you bprp!
This has so far been the most satisfying video I've seen in 2018
"ISN'T IT?"
It is!
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!
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.
Keep the complex math coming!!! I love it
I had no clue how to solve this one, as soon as I saw euler, i was like :that's a realllyyyyy smart way :)
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 :)
If you want to go really far down that route you will need to dump Riemann integration in favor of Lebesgue integration (measure integral) :)
@@mike4ty4 With the Lebesgue integral, what would the result be? I don't know anything about measure theory mixed with complex analysis and especially multivalued functions with branches.
Love your videos. I already know these things, but I like your presentation and explanations.
Thanks!!!!
Did Dr. Payam steal your markerboard?
ZipplyZane, OMG that's what I wanted to comment!
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
Gorgeous! Well done mate!
Really cool! Hope you'll make videos about other complex integrals in the future :)
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.
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
0:02 "Hello darkness my old friend..."
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.
Oh sorry, I had not watched the end of the video yet lol
Best video so far !!! :)
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)
*Beautifully explained!*
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
The function is discontinuous over the interval, the integral doesn't exist. It doesn't meet the existance condition of definite integral, U=L.
U need some love!
We can also use i^2 in place of -1
There's some real trickeration going on here! The Riemann surface of (-1)^z is a helicoid: a sequence of sheets connected together along branch cuts on the negative real axis, because -1 = e^(ln(-1)) and the log is multivalued on the negative reals. However, the integral itself is along some contour from z=0 to z=1 on a particular sheet. (It doesn't matter which contour is used because the exponential function is entire). The "multivaluation" of this integral comes by specifying the sheet, i.e. the particular value of ln(-1) as mentioned at 4:35, which appears in the antiderivative. The principle value of the integral, of course, uses ln(-1) = i pi, which results in the original answer 2i/pi.
If you want to be really clever, you can specify that the "z=0" point is on one sheet and the "z=1" point is on another sheet, by taking a path that loops around z=0 some number of times (i.e. traveling up or down the helicoid) before landing on z=1. The particular contour doesn't matter, except for the winding number around z=0. The value of ln(-1) differs between the start and end points of the contour because they are on different sheets, so you can get even more values out of this integral, i.e. i/(2n+1)pi + i/(2m+1)pi, where m and n are the numbers of the sheets (e.g. the principal values are on sheet number 0).
If you form a simple, closed contour (e.g. z=0 on sheet m to z=1 on sheet n; z=1 to z=0 on sheet n; z=0 on sheet n to z=1 on sheet m; and z=1 to z=0 on sheet m) the formulas above for integrating along each leg of the contour will all add to zero. This is expected, since as noted above the exponential function is entire so this contour does not enclose any poles.
Yay for making complex integration... complex.
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.
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.
wonderful as always
We can easily convert the expression into e to the power something and then the integral become more easy
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.
427 likes 0 dislikes. love it. This might be the first UA-cam video I have seen with 0 dislikes. Congrats!
Such simple, but so cool answer !!
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
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.
an imaginary integer is an integer that lies perpendicular to the real axis. that's the geometric interpretation^^
Sorry I meant the geometrical interpretation of a complex integral
maybe there is none?
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.
If you use i²=-1 instead, you end up with a result of -1/ln(i) which is a more compact way for the solution.
For those of you unfamiliar to the maddening delights of Complex Analysis. Don't think in terms of area to extend your intuition of integrals into complex numbers, but rather to the fundamental theorem of calculus. Then add the real variable calculus fact that the primitive or anti-derivative of the exponential of any non-negative base involves the logarithm and that complex logarithms are multi valued.
I like how random your problems are, but that you always find out the answers
i was glad & reliefed at the same time when you eventually "brought in" the "blue chalk" ... !!!
Yes sir, i agree to your hyphysical solution. Thanks
Fantastic video!
That was a satisfying integral to solve!
I solved this in a really weird informal/visual way:
First off, -1 represents a rotation of 180 degrees (and no scaling), so (-1)^x is a rotation of 180*x degrees. These rotations correspond to points on the unit circle on the complex plane. For x ranging from 0 to 1, these go from 0 degrees to 180 degrees, making up the upper half of the circle. Since the x is picked uniformly between 0 and 1, all the points of this half-circle are weighted equally and their weights add up to one. That means we're just looking for the center of mass of the upper half of the unit circle. Since the half-circle is symmetric across the real axis, the real component of the center must be 0. To get the imaginary component, we look at how the vertical positions are distributed. For any angle theta, the vertical component is sin(theta). So we integrate sin(theta) from theta=0 to pi, giving us 2, but then we've given the sines a total weight of pi, so to get the average, we divide by pi. That gives us the final answer of 2/pi.
As you can tell, my informal approach got me the right answer (possibly faster than if I had tried it by computing the integral rigorously). However, there's plenty of reasons to do a rigorous proof. A rigorous proof is much less likely to be considered "insufficient" by other people. Someone else could complain that switching between angles and complex numbers, and between averages and integrals, requires that I speak more precisely, mention the theorems I'm implicitly using, etc.
I cannot describe beauty and splendor
Super cool - thank you!
Nice solution !
The polar form of
(-1) = cosΠ+ i sinΠ
By De-Moiver's Thm,
(-1)^x = [cosΠ + i sinΠ]^x =
CosΠx + i SinΠx.
So its integral is 1/Π [ SinΠx - i CosΠx] . Applying the limits from 0 to 1 , we get the answer 2i/Π.
I think this is simple
Hola, quería saludarte y felicitarte por tu trabajo. Te sigo desde los 1000 subcriptores :)
I liked before the video started
Thank you!
That's cuz he knows it's You and You always produce such great content.
Cool! Euler's equation is very useful.
He is asserting his dominance on us
Looks at thumb nail, that equals infinity. Was not expecting the Euler identity substitution. Very cool.
Love the change to the chalkboard :)
Hello from Norway! A problem for you: prove a correlation between integral of ((lnx)^ndx) from 0 to 1 and (n!)
The thing that comes to my mind is Euler's identity!
Limit of x to 0 (3x-Sin3x)/(2x-Sin2x)
Amazing, this is so cool!
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.
Wow !!cool 😎. i never thought that intergral gonna get to a complex value 😮
直接用-1=(i)^2代入,再用a^x積分=a^x/lna可得這結果為 i^x/(2lni)(上下限為1,0]
You should try to integrate from -1 to 0 a similar function using residues and see what happens.
First you have to define
(- 1) ^x , If x is real.
Then you can try to integrate the (complex) function....there are many ways to define a function
f(x) = (-1)^ x.
You should use white chalk and blue chalk so that you can invert the colors in editing and get blackchalkredchalk.
Great stuff as always dude. Isn't it!
Parabéns meu amigo! Gostei muito do vídeo!
hey blackpenredpen. I have a problem for you! the antiderivative of log(arctan(x)). This function is fun since in the real world it's an entire function on its domain, and in the complex world it's the composition of two multi valued functions!
This has no antiderivative in terms of any standard mathematical functions according to Wolfram at least. So there would be no way to express the result :(
mike4ty4 no but the power series of it is very nice. xlog(arctan(x)) + Σc(n)*arctan(x)^(2n+1) /(2n+1)
where c(n) is the nth coefficient of the maclaurin series of tan(x)
Nicholas Jenkins try e^arctanx which can actually be calculated, its a really fun integral!
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.
The solution has a problem. Let's calculate ∫ 1dx (0-->1) = ∫ 1^x dx (0-->1) = ∫ e ^ (2i*pi*x) dx (0-->1) = e ^ (2i*pi*x) (0-->1) /(2i*pi) = 0, but actually the result should be 1.
The major reason the describe by @Yann Le Dréau, the formula e^ (a*b) = (e^a)^b only sustained when a is real number or b is integer
you are best mathologer
if you can solve this question:
integrate:(sinx+cosx)/(cosx-sinx) with down limit:0 up limit:pi/2
@@__ocram__ it looks like but my teacher solve that and I'm confusing
@@moeidmehri2153 well, the anti derivative is -ln(cos(x)-sin(x)) but plugging in 0 and pi/2 won't work (gives you a complex result) since the integral is improper because you're dividing by 0 half way through. So either the function your teacher was dealing with was different or he did some funky stuff.
The result of this integral is -2 /(ln(-1)) if you use i^2 = -1.
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😙
This is SO cool!
How come there are many answers for finding the area under that function? What's the geometric? interpretation?
The geometric interpretation is that you can get from the point (-1)^0 to (-1)^1 more than one way in the complex plane
Bombelus also (-1)^x isn’t a continuous function as any irrational or transcendental number you plug in would lead to a complex number.
It is continuous (in fact analytic) as a complex function
This is a very bizarre question, in my opinion. I'm not sure if the answer really makes sense.
One thing to keep in mind is that roots like (-1)^(1/2) and (-1)^(1/3) etc have multiple solutions in the complex plane.
So, maybe there is some deeper meaning whereby the choice of n defines which of the solutions you choose?
In a sense, there is infinite solutions to (-1)^(1/2). Perhaps these infinite solutions correspond to n in some way? For example: (-1)^(1/2) = (e^(i * pi *(2 * n + 1)))^(1/2) = e^(i * pi * (n + 1/2))
Sebastian Schweigert square root -1 is just i and -i, there aren’t infinite solutions
Why i put -1=e^iπ
Why we not put
-1=i^2 Then solve
Int i^2x
Take i^2=a and solve
int of a^x=(a^x/loga)
Final answer is -1/logi
I think its because the video is wrong saying that -1=i^2
This one is ambiguous until you choose which particular branch of the complex logarithm you wish to use. And you have to consistently use that branch for all steps in a calculation or else you will get a nonsense answer. This is incredibly important to keep in mind in certain areas of physics and engineering.
amazing, hanks a lot
Eek. What is a negative number to a “power”? The polar decomposition establishes a association of -1 with an equivalence class of polar representation, pairs of real numbers, magnitude and phase, -1,0 has the polar decomposition 1, 2 pi n with n an integer. -1 to the x has the polar decomposition exp(I pi x + 2 I pi n x). = 1, i pi x + 2 I pi n x that for irrational x can result in an equivalence with any unit modulis complex number. You can define the integral with all n=0, a “principal value”, but why? Doesn’t that need to be justified, stated up front, something? What precludes using alternate n, even n(x)?
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?
Makes you realise that any integral of an exponential will have multiple solutions, including e^x
Its 4 am where i live and one more time, im thirsty for your knowledge
white chalk red chalk
Now I see why we only studied line integrals in complex analysis lol
Wow, really good video
I was expecting the general integral f(x)=int[(-1)^t,dt,0,x] to be in open form. Surprised that it wasn't!
Ty so mush doctor
But why are there multiple answers? WE MUST GO DEEPER
Please, more complex integration
Hey WhiteChalkPinkChalk, can we learn more about summations?
2:20 De Moivre's formula:
exp(z) ^ n = exp(nz)
Only works when n is an integer (Z)
So I don't think you can do this manipulation :)
It doesn't really work for all kind of numbers?
It's kind of tricky. Here he is assuming that sqrt(1) = -1. Not sure if it is well defined
You can use de moivre theorem in step 3 too.
Sorry I didn't follow the bit where you put the 2 n π in the integral. I think the derivative of i π p + 2 n π wrt x is i π, not i π + 2 π, so the integral between 0 and 1 is 1 / i π( e^ (i π(1) + 2 n π) - e^ i π (0) + 2 n π)) so I dint see why the 2 n π terms don't just go, because e^iπ(1) + 2 n π) = -1 for all n.
No this trick doesn't works, you have to explain why you can multiplicate the powers ! The formule (e^a)^b = e^ab works when b is an interger but it is not always true. For example if we assume that is always true, then -1 = e^ipi = e^(2 i pi/2) = (e^(2 i pi))^1/2 = 1^(1/2) = sqrt(1) = 1