I'm glad you show that because it's exactly the same approach I had the first time I tried to solve the Fresnel's integral. I was surprised to find the good result, yet knowing it was illegal 😀
in this case it works, because the integrand is holomorphic everywhere and the contour integral of exp(-x^2) at infinity vanishes for Re(x^2)>0. What you did was rotating the integration path by 45°, just staying in the region where the function does not blow up at infinity.
I have an example where it does NOT work: f(x) = 1/(x^2 - 4x + 5) int[-inf to inf] f(a x) = pi/a. for real "a" that works. And as long as complex "a" has a low enough argument, it still works. But for a=i, the integral is just 0 😊
But you assume a>0, basically it's not legit. But if you assume as a complex number, I think it might lead to some complex analysis. Answer is correct but method doesn't seem legit.
For arg(a)< pi/4, the substitution turns the integral into a line that extends to points where |e^(-z^2)| decays exponentially. Hence you can make a contour that connects to the original real bounds with an outer integral that approaches zero. (This also works for arg(a) = pi/4, our convergent integral). Since e^(-z^2) is entire, we have int(complex bounds) - int(real bounds) = 0. Hence you can say the bounds are unchanged..
If you assume at the start that 'a' is a complex number, all you need for the integral to converge is Real(a) >0. Let 'a' lie on the upper right quadrant of the unit circle, 'a' = cos(phi) + i*sin(phi). As phi goes from 0 to pi/2, the real part of 'a' is always > 0 and the integral exists. So in the limit phi --> pi/2 and a --> i, you get the desired result.
This sounds correct, but I'd have to think about it. I think the important part in this is that we need to take the limit to make this proof rigorous. Good observation.
Correct, but you have to show three things 1. The function of A and the function of i becomes uniformly arbitrary close to each other, over finite length, as A approaches i. 2. The integral of the tail with i becomes arbitrairy close to 0 if the tail is chosen small enough (hard part) 3. The integral of the tail of the integral with A becomes arbitrary close to 0, if the tail becomes small enough (basically the same as step 2) You get a neat triple bound, which you want to show approaches 0, then you are done.
Fresnel’s integrals were always one of those topics that felt impossible to grasp fully. I remember how much easier it became once I started using tools that broke down the steps clearly. SolutionInn’s AI study tool does exactly that-it simplifies even the trickiest integrals and helps you understand the logic behind them. If anyone else is struggling with topics like this, it’s definitely worth exploring
The original integral only requires Re(a)>0. The result in terms of the square root is analytical in the complex plane except for the origin and the negative real axis. So what you do, is that you define the integral for imaginary real as the analytical continuation (which is unique).
I remember when I tried this in the past and it seems to work. But with complex numbers the integration bounds will change, so you need to show using contour integration that it is equal to the same integral over the real line
Similarly, I once used the fractional derivative for a “fractional order pole” to evaluate an integral using residue theorem and it gave the right result, but only bc the branch cut coincidentally didn’t interfere this time
The bounds might not change. If nothing else, exp(-ix²) being entire at least means it's path independent. Although I guess it would preferably need to vanish at ∞, and this wouldn't necessarily.
When I first encountered the feesnel integrals, I also used the method of converting to an exponential with power I and the making a function I(t) where our target integrals was the real value of I(-i). Then I took the lapplace transform and after further simplification and inverse lapplace transform, you get the famous result sqrt(pi/2).
At the beginning we say that 'a' must be positive for the integral to converge, because we are in the world of real numbers. The question that must therefore be asked is what condition should a complex 'a' fulfill for the integral to converge? Personally I cannot evaluate the convergence of an integral of a complex function. Moreover, what does mean negative and positive inifity in a complex world ? However, mathematics is known to be consistent, so ...
My method might be equivalent, but I was less worried about it being okay because I've seen this sort of trick in physics before. I did it by just considering the integral of e^(-x^2), and then letting u=x*sqrt(i). The RHS is still sqrt(pi), but on the left you get an e^(-i*pi/4) from the dx and you have an integral over e^(-i*u^2). From there I divided out e^(-i*pi/4) and matched real and imaginary parts like your solution. I think it's legit since the integral is still with a>0, it's just we transformed it into a form that gave us the sines and cosines before meddling with the RHS by dividing out the phase.
I think your proof as given doesn't let you conclude that the integrals converge, but does let you conclude that, if the integrals do converge, they converge to the values you found.
I think you need to prove that the integral of e^(-ax^2) from -inf to inf equals to the same integral, but from -inf*a to inf*a. Since you can put complex numbers into a, we cannot say that, for example, inf*(1+i)=inf. So we would need to do some complex analysis, but I think the approach is correct, but need some rigorous things
i wonder: does using i for a mean that when doing u sub, considering the bounds of the integral should go from -i infinity to i infinity instead? wondering if that changes anything.
I have an example where it does NOT work: f(x) = 1/(x^2 - 4x + 5) int[-inf to inf] f(a x) = pi/a. for real "a" that works. And as long as complex "a" has a low enough argument, it still works. But for a=i, the integral is just 0 😊
If something is valid for Gauss it does not mean it is valid for others. The Gaussfunction neither has poles or zeros in the complex plane, but your example has.
@@grafrotz5286 that's how i quickly found my example 😀 but the Gauß integral is only nice up to Re(a)=0. So be careful with complex integrals and use contour integrals to be sure - taking care of poles in the plane and at infinity.
@@deinauge7894 I am not sure if we discuss the same problem. I understand that we have an Integral of a real function with a complex parameter. That is slightly different from an integral of a complex function in the complex plane?
@@grafrotz5286 in these cases (the integral in the video and my example) a change in the parameter is equivalent to stretching and rotating the integration path in the complex plane. And everything is nice until you hit a singularity. As you know, if I understand your first reply correctly 😀
@@deinauge7894 the path is -oo to +oo along the real axis. And this path remains on the real axis even if the function exp(-a x²) has a complex parameter a. There is no change from the real x to a complex z, the path of integration remains real.
Legit or not it is pretty. The way 'i' splits the surface between the two fields, and you can see it "going" to the original and real sqrt(pi) as you decrease the portion of 'i' to 0 (within unity).
It works by Cauchy's integral formula. You can prove that the integral on the imaginary axis is the same as on the real axis, by using a quarter circle contour whose arc converges to 0 by Jordan's lemma (Which requires only that the integrand*z goes to zero **pointwise** on the arc, which it does in infinite radius, except on the imaginary axis but it's only a single point) Sorry, this was meant for the integral from 0 to infinity. Pretty obvious how you get the result.
i has to be greater than zero, because otherwise we're forced to say that -i isn't less than zero, which is absurd. i is the same as 1 * i, and -i is the same as -1 * i. If i weren't positive, then the first wouldn't true, and if -i weren't negative, the second wouldn't be true. So why is this the case? It's because the choice of axes is arbitrary. We could just as readily label the horizonal axis imaginary and the vertical real. Now what we called i takes on the value of 1, and 1 is always positive. And that's why the magnitude of the square root of i ... 1/sqrt(2) + i*1/sqrt(2) ... is also 1.
The only thing we ask of i is that i²=-1. If we surreptitiously switched it for (-i)²=-1, nobody would know the difference. That's why there's no ordering on C. We can also just prove it by contradiction. If i>0, then i×i>0×i, then -1>0 If i0, then i×(-i)
We require a > 0 in the real world so that the integral is finite. In the complex world, I think the requirement is Re(a) >= 0 and a != 0 but the margins of this comment are too small to give a proof. In that case, your approach is valid.
You write the complex number in polar form, with the argument chosen so that it is bigger than -pi and at most +pi. Then you take the square root of the modulus and half of the argument. The resulting number (in polar form) is the principal square root of the given number.
It's not valid because in order to get a valid result, the limit of the function at infinity must be 0. That is the case when a > 0, but not when a = i. The situation is similar for the infinite geometric sum where the formula is only valid for an argument with absolute value less than 1, but not on the edge case.
Definitely something seems not right, as you choose a root of i, so there is no reason to think that the result of the integral is just this root and not the other one.
A false assumption means the outcome is not guaranteed to be valid; however, validity is also not forbidden. You got the right answer; however, you didn't prove that it is the right answer.
When calculatung the Gauss Integral with i, you didn't change the bounds of the integral (with a>0 you didn't have to because inf stays inf and -inf stays -inf anyway) so therefore I think it's not legit 😅
Yes, but he has defined a>0 which the imaginary unit “i” does not satisfy. The answer still comes out correct but the correct method needs contour integration.
This would not work, as (ax)^2 is not what is in the exponent, as it it only the x being squared. To make the whole exponent squared, he need to have (sqrt(a)x)^2, so when he does u sub, the whole exponent becomes u, which yields the Gaussian integral.
Are you sure you were listening correctly? Isn't the first way you'd get taught contour integration through a parametrisation γ that gives you your contour? You'd then do z=γ=γ(t), dz=dγ=γ'(t) dt
@@xinpingdonohoe3978 Yeah, but what you mentioned is not a substitution, but a parametrization. The way he re-arranges u= sqrt(a) x looks like a substitution to me. Maybe it works because the path is only on the real axis. I am honestly not sure.
u = sqrt(a) times x. You differentiate both sides and as a derivative of "sqrt(a) times x" you get "sqrt(a) times x"? But how? Sqrt(a) seems to be a constant, so it's derivative is zero, yet you just write it there. This doesn't seem right to me. I may be wrong though.
he wrote sqrt(a) dx, which is definitely correct. The derivative of a constant on its own would be 0, but when it is attached to an x with degree 1, when you differentiate by power rule you just end up with sqrt(a)
I think over ℂ, the square root of i can have two values, so the square root doesn't define a function on ℂ, and you can obtain two results without knowing which one is the correct one. 4o
It does define a function, just not a single valued one, so without further specification the integral is left in slight shambles of ambiguity. What's 4o though?
What if we changed the Fresnel’s integral What if we changed a Fresnel's integral?
ruclips.net/video/FwwJ1LgJp_w/видео.html
Please sir you again take a live class of 100 question integration
In calculus, when referring to region of convergence in complex numbers, we Might use Re(i) >=0. So maybe this formula works for 2+i
I'm glad you show that because it's exactly the same approach I had the first time I tried to solve the Fresnel's integral. I was surprised to find the good result, yet knowing it was illegal 😀
in this case it works, because the integrand is holomorphic everywhere and the contour integral of exp(-x^2) at infinity vanishes for Re(x^2)>0.
What you did was rotating the integration path by 45°, just staying in the region where the function does not blow up at infinity.
I have an example where it does NOT work:
f(x) = 1/(x^2 - 4x + 5)
int[-inf to inf] f(a x) = pi/a.
for real "a" that works. And as long as complex "a" has a low enough argument, it still works.
But for a=i, the integral is just 0 😊
But you assume a>0, basically it's not legit. But if you assume as a complex number, I think it might lead to some complex analysis. Answer is correct but method doesn't seem legit.
it is legit, but it needs few more steps to show. When a is a complex number with Re(a) > 0 then the limit Re(a) --> 0 exist as long as Im(a) != 0
For arg(a)< pi/4, the substitution turns the integral into a line that extends to points where |e^(-z^2)| decays exponentially. Hence you can make a contour that connects to the original real bounds with an outer integral that approaches zero. (This also works for arg(a) = pi/4, our convergent integral). Since e^(-z^2) is entire, we have int(complex bounds) - int(real bounds) = 0. Hence you can say the bounds are unchanged..
@@taterpun6211Yeah, Contour Integration works
If you assume at the start that 'a' is a complex number, all you need for the integral to converge is Real(a) >0. Let 'a' lie on the upper right quadrant of the unit circle, 'a' = cos(phi) + i*sin(phi). As phi goes from 0 to pi/2, the real part of 'a' is always > 0 and the integral exists. So in the limit phi --> pi/2 and a --> i, you get the desired result.
but Re(i)=0
@@Metaverse-d9fSurely Re (i) d.n.e?
This sounds correct, but I'd have to think about it. I think the important part in this is that we need to take the limit to make this proof rigorous. Good observation.
7:16 I am gonna use this next itme >_
Correct, but you have to show three things
1. The function of A and the function of i becomes uniformly arbitrary close to each other, over finite length, as A approaches i.
2. The integral of the tail with i becomes arbitrairy close to 0 if the tail is chosen small enough (hard part)
3. The integral of the tail of the integral with A becomes arbitrary close to 0, if the tail becomes small enough (basically the same as step 2)
You get a neat triple bound, which you want to show approaches 0, then you are done.
Fresnel’s integrals were always one of those topics that felt impossible to grasp fully. I remember how much easier it became once I started using tools that broke down the steps clearly. SolutionInn’s AI study tool does exactly that-it simplifies even the trickiest integrals and helps you understand the logic behind them. If anyone else is struggling with topics like this, it’s definitely worth exploring
The original integral only requires Re(a)>0. The result in terms of the square root is analytical in the complex plane except for the origin and the negative real axis. So what you do, is that you define the integral for imaginary real as the analytical continuation (which is unique).
but Re(i)=0
Exactly, you extend the result to Re(a)=0 using the analytical continuation, which again, is unique.
I remember when I tried this in the past and it seems to work. But with complex numbers the integration bounds will change, so you need to show using contour integration that it is equal to the same integral over the real line
Similarly, I once used the fractional derivative for a “fractional order pole” to evaluate an integral using residue theorem and it gave the right result, but only bc the branch cut coincidentally didn’t interfere this time
The bounds might not change. If nothing else, exp(-ix²) being entire at least means it's path independent. Although I guess it would preferably need to vanish at ∞, and this wouldn't necessarily.
man I love higher lever math bprp
This isn't higher leveled. At all.
@@tfg601 oh ok my bad
@@tfg601Still a bit more than typical undergrad curriculum
When I first encountered the feesnel integrals, I also used the method of converting to an exponential with power I and the making a function I(t) where our target integrals was the real value of I(-i). Then I took the lapplace transform and after further simplification and inverse lapplace transform, you get the famous result sqrt(pi/2).
At the beginning we say that 'a' must be positive for the integral to converge, because we are in the world of real numbers. The question that must therefore be asked is what condition should a complex 'a' fulfill for the integral to converge? Personally I cannot evaluate the convergence of an integral of a complex function. Moreover, what does mean negative and positive inifity in a complex world ? However, mathematics is known to be consistent, so ...
-∞ is the left end of the real axis. ∞ (or +∞) is the right end of the real axis.
My method might be equivalent, but I was less worried about it being okay because I've seen this sort of trick in physics before. I did it by just considering the integral of e^(-x^2), and then letting u=x*sqrt(i). The RHS is still sqrt(pi), but on the left you get an e^(-i*pi/4) from the dx and you have an integral over e^(-i*u^2). From there I divided out e^(-i*pi/4) and matched real and imaginary parts like your solution. I think it's legit since the integral is still with a>0, it's just we transformed it into a form that gave us the sines and cosines before meddling with the RHS by dividing out the phase.
I am not sure of the steps, but the final answer looks okay.
I think your proof as given doesn't let you conclude that the integrals converge, but does let you conclude that, if the integrals do converge, they converge to the values you found.
This sure felt surreal
You can use contour integration to find integral exp(-ax^2) from -inf to inf for complex a. Then the rest is chill
I think you need to prove that the integral of e^(-ax^2) from -inf to inf equals to the same integral, but from -inf*a to inf*a. Since you can put complex numbers into a, we cannot say that, for example, inf*(1+i)=inf. So we would need to do some complex analysis, but I think the approach is correct, but need some rigorous things
i wonder: does using i for a mean that when doing u sub, considering the bounds of the integral should go from -i infinity to i infinity instead? wondering if that changes anything.
Would you ever consider sitting an entire STEP 3 paper like you have with Further maths papers? I think it would make quite an enjoyable watch
I have an example where it does NOT work:
f(x) = 1/(x^2 - 4x + 5)
int[-inf to inf] f(a x) = pi/a.
for real "a" that works. And as long as complex "a" has a low enough argument, it still works.
But for a=i, the integral is just 0 😊
If something is valid for Gauss it does not mean it is valid for others. The Gaussfunction neither has poles or zeros in the complex plane, but your example has.
@@grafrotz5286 that's how i quickly found my example 😀 but the Gauß integral is only nice up to Re(a)=0. So be careful with complex integrals and use contour integrals to be sure - taking care of poles in the plane and at infinity.
@@deinauge7894 I am not sure if we discuss the same problem. I understand that we have an Integral of a real function with a complex parameter. That is slightly different from an integral of a complex function in the complex plane?
@@grafrotz5286 in these cases (the integral in the video and my example) a change in the parameter is equivalent to stretching and rotating the integration path in the complex plane. And everything is nice until you hit a singularity. As you know, if I understand your first reply correctly 😀
@@deinauge7894 the path is -oo to +oo along the real axis. And this path remains on the real axis even if the function exp(-a x²) has a complex parameter a. There is no change from the real x to a complex z, the path of integration remains real.
Legit or not it is pretty. The way 'i' splits the surface between the two fields, and you can see it "going" to the original and real sqrt(pi) as you decrease the portion of 'i' to 0 (within unity).
It works by Cauchy's integral formula. You can prove that the integral on the imaginary axis is the same as on the real axis, by using a quarter circle contour whose arc converges to 0 by Jordan's lemma (Which requires only that the integrand*z goes to zero **pointwise** on the arc, which it does in infinite radius, except on the imaginary axis but it's only a single point)
Sorry, this was meant for the integral from 0 to infinity. Pretty obvious how you get the result.
can everyone from different countries participate BMT?
Please sir you make again the 100 question series of integration
Sir can you make a vedio don short tricks of quadratic equations
i has to be greater than zero, because otherwise we're forced to say that -i isn't less than zero, which is absurd. i is the same as 1 * i, and -i is the same as -1 * i. If i weren't positive, then the first wouldn't true, and if -i weren't negative, the second wouldn't be true. So why is this the case? It's because the choice of axes is arbitrary. We could just as readily label the horizonal axis imaginary and the vertical real. Now what we called i takes on the value of 1, and 1 is always positive. And that's why the magnitude of the square root of i ... 1/sqrt(2) + i*1/sqrt(2) ... is also 1.
The only thing we ask of i is that i²=-1. If we surreptitiously switched it for (-i)²=-1, nobody would know the difference. That's why there's no ordering on C. We can also just prove it by contradiction.
If i>0, then i×i>0×i, then -1>0
If i0, then i×(-i)
As a physics major, as long as the answer is right, the method doesn't matter to me! 🤣🤣🤣
We require a > 0 in the real world so that the integral is finite. In the complex world, I think the requirement is Re(a) >= 0 and a != 0 but the margins of this comment are too small to give a proof. In that case, your approach is valid.
a!≠0 (never)
@@Metaverse-d9fIt means not equal
@@Metaverse-d9f I didn't have the not equals symbol so I used a common computing equivalent. I was not implying a factorial.
Hey Steve! I came across a problem… e^z = ~e. Can you solve that? Please show us how. :)
How do you define the square root in C ?
You write the complex number in polar form, with the argument chosen so that it is bigger than -pi and at most +pi. Then you take the square root of the modulus and half of the argument. The resulting number (in polar form) is the principal square root of the given number.
√z is that which squares to z. There are two branches, cut along the negative reals. One has Re(√z)>0 and the other has Re(√z)
It's not valid because in order to get a valid result, the limit of the function at infinity must be 0. That is the case when a > 0, but not when a = i.
The situation is similar for the infinite geometric sum where the formula is only valid for an argument with absolute value less than 1, but not on the edge case.
So good!
Definitely something seems not right, as you choose a root of i, so there is no reason to think that the result of the integral is just this root and not the other one.
A good point.
A false assumption means the outcome is not guaranteed to be valid; however, validity is also not forbidden. You got the right answer; however, you didn't prove that it is the right answer.
When calculatung the Gauss Integral with i, you didn't change the bounds of the integral (with a>0 you didn't have to because inf stays inf and -inf stays -inf anyway) so therefore I think it's not legit 😅
Brilliant I liked it very much and I think that it's legal because someone does a proof that calculus working on immiginary number's too.
Yes, but he has defined a>0 which the imaginary unit “i” does not satisfy. The answer still comes out correct but the correct method needs contour integration.
Why sqrt a times x tho, why Not just set u =ax?
This would not work, as (ax)^2 is not what is in the exponent, as it it only the x being squared. To make the whole exponent squared, he need to have (sqrt(a)x)^2, so when he does u sub, the whole exponent becomes u, which yields the Gaussian integral.
are you sure i > 0?
It is not.
@@xinpingdonohoe3978 i know
Interesting!
What is 3u? What does u means?
That confused me too...the best I got is "i
@jeffstryker2419 Ha! It makes sense !
I don't know if i is positive, but is the square root of -1 positive? It's a square root afterall.
i>0 → i×i>0×i → -1>0
i0 → i×(-i)
Yes, i>0 and i0 as well.
i 0, but could it be negative? In which case, bprp would need to show the equality holds for -i
Indeed, sometimes i is negative.
@@brian554xxI is never negative, -I is (* by -1, pos*neg=neg). I is just 1i, so that’s like saying 1 is negative. It’s not, but -1*1 is
Now look who's a negative one. 🤭
If i>0 then i²>0i which means -1>0
The complex plane can't be ordered. i is not positive or negative.
speak about Fourier series pls
Brazil 🇧🇷🇧🇷
I thought there is no substitution rule for complex integrals? Thats at least what my tutors told me.
Are you sure you were listening correctly? Isn't the first way you'd get taught contour integration through a parametrisation γ that gives you your contour? You'd then do z=γ=γ(t), dz=dγ=γ'(t) dt
@@xinpingdonohoe3978 Yeah, but what you mentioned is not a substitution, but a parametrization. The way he re-arranges u= sqrt(a) x looks like a substitution to me. Maybe it works because the path is only on the real axis. I am honestly not sure.
@@friedrichfreigeist3292 same thing.
u = sqrt(a) times x.
You differentiate both sides and as a derivative of "sqrt(a) times x" you get "sqrt(a) times x"? But how? Sqrt(a) seems to be a constant, so it's derivative is zero, yet you just write it there. This doesn't seem right to me. I may be wrong though.
he wrote sqrt(a) dx, which is definitely correct. The derivative of a constant on its own would be 0, but when it is attached to an x with degree 1, when you differentiate by power rule you just end up with sqrt(a)
You’re not sure? Then I won’t try it at home.
LGTM
Merge & let QA sort it out
i > 0 for sure because i is 0 + 1i and 0 is 0 + 0i in complex world, so this is legit
the result "legit" is correct but your argumentation is wrong. Complex numbers have no '' relation, you cannot order them this way.
That makes zero sense (and is wrong). i is neither positive or negative.
If it's just the imaginary line, i is greater than 0
i is not greater than 0.
Illegal
It will be legit if you allow 'a' be complex and put this into limit to 'i' from 'Re(a) > 0'.
In what way does a=i satisfy Re(a)>0?
@@xinpingdonohoe3978 it doesn't. But doing right limit thing is fine.
I think you do a bitcoin math
I think over ℂ, the square root of i can have two values, so the square root doesn't define a function on ℂ, and you can obtain two results without knowing which one is the correct one.
4o
It does define a function, just not a single valued one, so without further specification the integral is left in slight shambles of ambiguity.
What's 4o though?
@@xinpingdonohoe3978 the 4o is from ChatGpt, my English is not so good, so I used it to enhance the comment. 🤣
@@xinpingdonohoe3978 I was trying to enhance my comment using Ai, so it has written the 4o. Sorry 😅.