I learned these integrals last week in my complex analysis course! My professor did the general case of the integrals from 0 to ∞ of cos ax² dx (call it I₁) and of sin ax² dx (call it I₂), with a > 0. This means I₁ + iI₂ = integral from 0 to ∞ of exp(iax²) dx (call it I₃). We then set up the same contour and path parameterizations that you did and obtained the expression for I₃, whose real part and imaginary part are equal = I₁ = I₂ = √2/4 * √(π/a), and the case of a = 1 gives the solution as in the video. A slightly different approach for the definition of f and still very elegant!
12:18 Doesn't that require the use of DCT or MCT? To be able to switch the order of the limit operation and integration? The function doesn't seem to be monotone, taking MCT out of the picture, so how would DCT work here? Or is there perhaps a different theorem that comes into play?
How do you know when to sue a specific contour because some have semi-circle or box or pizzaslice or keyhole, etc. and I don't understand when to use which
@@maths_505 wow !! awesome, didn't expect that. Thought you would do pure math. I'm doing a very similar study, seeing your grasp you will do fantastic I'm sure.
@@sciencelover-c2j No, they can be evaluated only as definite integrals (unless we use special functions). The classical methods of solving indefinite integrals (integration by parts, substitutions, etc.) do not work. Essentially this means that there are no elementary functions that we know, whose derivative is cos(x²) or sin(x²).
Hi,
"ok, cool" : 2:42 , 4:15 , 6:26 , 7:52 , 11:27 , 12:18 ,
"terribly sorry about that" : 8:53 , 9:16 , 11:09 , 12:52 .
This is one of the comments of all time
Very helpful, thank you
I learned these integrals last week in my complex analysis course! My professor did the general case of the integrals from 0 to ∞ of cos ax² dx (call it I₁) and of sin ax² dx (call it I₂), with a > 0. This means I₁ + iI₂ = integral from 0 to ∞ of exp(iax²) dx (call it I₃). We then set up the same contour and path parameterizations that you did and obtained the expression for I₃, whose real part and imaginary part are equal = I₁ = I₂ = √2/4 * √(π/a), and the case of a = 1 gives the solution as in the video. A slightly different approach for the definition of f and still very elegant!
This was how I first learnt how to solve them! 😊
wonderful teaching sir
Thank you my friend
Beautiful beyond words 😍
This video made contour integrals click for me!
@@bennettkinder5654 glad it helped. There are lots more in the contour integration playlist.
12:18 Doesn't that require the use of DCT or MCT? To be able to switch the order of the limit operation and integration? The function doesn't seem to be monotone, taking MCT out of the picture, so how would DCT work here? Or is there perhaps a different theorem that comes into play?
What app are you using? It looks super smooth.
I think its the samsung notes app
How do you know when to sue a specific contour because some have semi-circle or box or pizzaslice or keyhole, etc. and I don't understand when to use which
Experience is a great teacher
i loved it
what kind of master/specialization are you doing bro?
Astrophysics bro
@@maths_505 wow !! awesome, didn't expect that. Thought you would do pure math. I'm doing a very similar study, seeing your grasp you will do fantastic I'm sure.
could you do it using Imaginary and Real parts of (e^i(x^2))
Nice ❤
Which whiteboard application you are using
Samsung notes
@@maths_505 thanks for your help. Kindly rate the tablet good for teaching Wacom One
Remarkable 2
Apple ipad pro
Etc
What is the same Integral but without limits ?? Can I use the UV method to solve it?
Both integrals have no elementary solution when they are indefinite.
@marioangelov113 so ,can I solve it by udv ??
@@sciencelover-c2j No, they can be evaluated only as definite integrals (unless we use special functions). The classical methods of solving indefinite integrals (integration by parts, substitutions, etc.) do not work. Essentially this means that there are no elementary functions that we know, whose derivative is cos(x²) or sin(x²).
@marioangelov113 That's mean when they come as indefinite (open integral ,no limits) ,we can't make integration for them?
@@sciencelover-c2j yes
first
Second
Third
Fourth😂😂
just had to solve them in my homework, solved using double angle formula🙃
How would that work? Im a bit rusty but isnt the double angle formula for sin(A+B), not sin(A²)?
@@nott_applicable cos2A = 1 - 2sin^2(A)
@@phylIsin²(x) is something different than sin(x²)
@@nott_applicable oh youre right, saw it wrong😅
I had sin^2(A)