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! 😊

Beautiful beyond words 😍

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 .

@Dedicate25

2 сағат бұрын

This is one of the comments of all time

wonderful teaching sir

@maths_505

Ай бұрын

Thank you my friend

i loved it

What app are you using? It looks super smooth.

@axeitor

Ай бұрын

I think its the samsung notes app

could you do it using Imaginary and Real parts of (e^i(x^2))

Nice ❤ Which whiteboard application you are using

@maths_505

Ай бұрын

Samsung notes

@usmansaleem3173

29 күн бұрын

@@maths_505 thanks for your help. Kindly rate the tablet good for teaching Wacom One Remarkable 2 Apple ipad pro Etc

what kind of master/specialization are you doing bro?

@maths_505

Ай бұрын

Astrophysics bro

@Unidentifying

Ай бұрын

@@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.

just had to solve them in my homework, solved using double angle formula🙃

@alex_bor

Ай бұрын

How would that work? Im a bit rusty but isnt the double angle formula for sin(A+B), not sin(A²)?

@phylI

29 күн бұрын

@@alex_bor cos2A = 1 - 2sin^2(A)

@alex_bor

29 күн бұрын

​@@phylIsin²(x) is something different than sin(x²)

@phylI

29 күн бұрын

@@alex_bor oh youre right, saw it wrong😅 I had sin^2(A)

∫[0,∞] sin(x²)dx = Im ( ∫(0,∞ )e^(ix²) dx ∫[0,∞] cos(x²)dx = Re ( ∫(0,∞ )e^(ix²) dx

first

@prudhvi634

Ай бұрын

Second

@maths_505

Ай бұрын

Third

@7yamkr

Ай бұрын

Fourth😂😂

What is the same Integral but without limits ?? Can I use the UV method to solve it?

@marioangelov113

Ай бұрын

Both integrals have no elementary solution when they are indefinite.

@user-zg8ny5tp4g

Ай бұрын

@marioangelov113 so ,can I solve it by udv ??

@marioangelov113

Ай бұрын

@@user-zg8ny5tp4g 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).

@user-zg8ny5tp4g

Ай бұрын

@marioangelov113 That's mean when they come as indefinite (open integral ,no limits) ,we can't make integration for them?

@marioangelov113

Ай бұрын

@@user-zg8ny5tp4g yes