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.
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
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).
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.
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).
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 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
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 ...
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.
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).
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.
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.
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.
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 😅
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.
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.
About "legitimacy" - it's easy. You needed a > 0 in the real case because otherwise the thing diverges. And now you "replaced" it with something non-real at all. So you definitely would need to show the new "thing" you constructed will converge.
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)
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.
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.
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.
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
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
man I love higher lever math bprp
This isn't higher leveled. At all.
@@tfg601 oh ok my bad
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 >_
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
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).
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.
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.
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 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
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 ...
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
So good!
You can use contour integration to find integral exp(-ax^2) from -inf to inf for complex a. Then the rest is chill
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.
As a physics major, as long as the answer is right, the method doesn't matter to me! 🤣🤣🤣
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).
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.
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. 🤭
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.
are you sure i > 0?
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.
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 😅
Please sir you make again the 100 question series of integration
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.
Sir can you make a vedio don short tricks of quadratic equations
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.
Interesting!
About "legitimacy" - it's easy. You needed a > 0 in the real case because otherwise the thing diverges. And now you "replaced" it with something non-real at all. So you definitely would need to show the new "thing" you constructed will converge.
Hey Steve! I came across a problem… e^z = ~e. Can you solve that? Please show us how. :)
can everyone from different countries participate BMT?
I thought there is no substitution rule for complex integrals? Thats at least what my tutors told me.
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)
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.
It will be legit if you allow 'a' be complex and put this into limit to 'i' from 'Re(a) > 0'.
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.
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.
If it's just the imaginary line, i is greater than 0
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.
You’re not sure? Then I won’t try it at home.
LGTM
Merge & let QA sort it out
Illegal
What is 3u? What does u means?
That confused me too...the best I got is "i
@jeffstryker2419 Ha! It makes sense !
I think you do a bitcoin math
Sorry to be "that guy"... You don't pronounce the "s" in Fresnel, its fruhh-NELL. 😬