What if we changed a Fresnel's integral?

[@blackpenredpen]

I modified the Gaussian integral: kzbin.info/www/bejne/oJWvkKFtoayeq68si=ldFQ04-IlI6tf6r9

[@fireballman31]

simplifying the root 2 only to undo it is so real

[@damyankorena]

it's not just real. Its algebraic xd

[@appybane8481]

@@damyankorena I don't think pi is algebraic.

[@thenightjackal]

​@@appybane8481 yeah it's more trigonometric if anything /j

[@Sci-Fi-Mike]

It's so real, but for some students, it's also "complex." My math humor isn't very good, as you can see...

[@stephenbeck7222]

@@appybane8481transcendental comment

[@FacultyofKhan]

Nice. Now do the integral that Cleo solved on the math stack exchange without any explanation!

[@Rando2101]

The one with the inverse tangent of the golden ratio?

[@rafaelpascualortega3956]

Just yesterday I had a vector calculus exam where we solved the Fresnel integral using a conservative field and the Gaussian integral, and today you upload this video, what a lovely coincidence.

[@robdev02]

I'd really like to see the complex analysis approach.

[@RandyKing314]

sometimes it’s good to see some trial and error in a problem solve instead of a clean pre-scripted solution…thanks BPRP!

[@d-hat-vr2002]

Such a nifty question! My solution is similar, but somewhat cleaner in my opinion. First step: I use a substitution of u=1/x (rather than u=1/x²). This reveals that ∫0..∞ sin(1/x²) dx = ∫0..∞ sin(x²)/x² dx Second step: to integrate ∫0..∞ sin(x²)/x² dx use integration by parts (DI method) where D=sin(x²) and I=1/x² This very cleanly produces ∫0..∞ sin(x²)/x² dx = -sin(x²)/x + 2 * ∫0..∞ cos(x²) dx Third step: to evaluate the -sin(x²)/x term at 0, multiply by x/x and then let u=x² so lim u→0 to evaluate the -sin(x²)/x term at ∞, use squeeze theorem.

[@markerguy]

First time reaching in 38 secs. Gonna watch the full vid now!!

[@spoopy1322]

Really enjoyed the video ❤

[@cameronspalding9792]

@ 0:18 One can verify it converges in the Lebesgue sense using the comparison test: 1. For x in (0,1) |sin(1/x^2)|

[@General12th]

So good!

[@blackpenredpen]

@@General12th thank you!

[@Dreamprism]

2:56 I saw here roughly why it converges. If we split the integral at u=1, then the first part where u approaches 0 we have sin(u) basically the same as u and therefore sin(u) * u^(-3/2) is basically 1/u^(1/2) as we near the vertical asymptote, which we know converges by a p-Test with p1. So abs(our inntegrand) will have an integral that converges to something smaller. Then abs conv implies regular converges of our integral. This is just a rough way of looking at it, but it does make me confident it will converge.

[@abvd92840]

This was on the MIT integration bee 2024 or 2023 I don't remember. I was just solving them for fun and got stuck at this part.

[@mathiastoala7777]

That thumbnail is just so eviiiil lmaoo 🗣️🔥

[@nightytime]

7:57 "due diligence"

[@blackpenredpen]

Thanks!

[@scottleung9587]

Very nice!

[@saharhaimyaccov4977]

Great way

[@Ben-wv7ht]

Maybe I'm overdoing it , but I used Ramanujan's master theorem to solve this. The IBP was smarter tho.

[@Happy_Abe]

How do we solve the cos(x^2) integral?

[@mismis3153]

cos(x^2) = Real(e^(-ix^2)).

[@Samir-zb3xk]

4 methods I know of: 1: Use Euler's formula to transform cos(x²) into an exponential function (Real part of e^(-ix²)). Then you can use the Gaussian integral to solve it (after a substitution) 2: Define a function f(t) = (0 to ∞) ∫ cos(tx²) dx Then take the Laplace transform of both sides. The resulting integral is nasty but solvable. (I would factor the the denominator using complex numbers then partial fraction decomposition). Once you have L{f(t)} simplified, you can solve for f(t) and plug in f(1). 3: Sub u = x², then the resulting can be solved using the Mellin transform of cos(x) 4: Sub u = x², then convert cos(u) = Real part of e^(-iu), then the resulting can be solved using the Laplace transform of t^(-1/2) Btw the last 3 methods also requires you to know how to calculate (-1/2)! = √π using the gamma function, which you also need to the know the Gaussian integral for.

[@CKNGAI-r8x]

​@@Samir-zb3xk For 4 a easier way is realise that it not only it's the tranform of t^-1/2 but also s=i so instead of gamma(1/2)/(s^1/2) which is still a headache and take simply the Re(gamma(1/2)/(i^1/2))

[@hcgreier6037]

07:18 l'Hospitals rule: for u => 0 the sin(u)/u gives a 0/0 expression again! So there is a small lapsus. Of course, doing l'Hospitals rule again, one gets cos(u)/1, and for u => 0 this is 1/1 = 1.

[@stephenbeck7222]

It’s a standard limit, learned before derivatives, that sin(x)/x goes to 0 as x goes to 0.

[@hcgreier6037]

@@stephenbeck7222 Sorry, but sin(x)/x goes to *1* as x goes to 0.

[@bain8renn]

hes made videos about this, but this is a really bad idea, as this exact limit is used to solve the derivative of sinx, meaning its circular reasoning. its neat it works out with l'hopitals rule, but in order to prove that the derivative of sine is cosine, you need the limit, and if you try to use L'H to solve the limit, you need the derivative of sine. there is a way to evaluate the limit with squeeze theorum and geometry that gets the result without causing problems

[@ciorchinos]

the problem is when you introduced the imaginary component, since it is imaginary it does not respect teh initial condition of a>0 since it is an emaginary value, also you cannot subtiture the value with pi because of the same reason. (we should have a rule in math about this substitutions meaning that we need to chack the nature of parameters introduced )

[@Nain115]

Can I get one of tour t-shirts? I live in Nicaragua tho :c

[@cdkw2]

this kind of reminds me of old bprp

[@felixmellgren8116]

Why do you always use the letters u & t when doing a substitution

[@KaiserBob99]

Convention.

[@stetzelspretzels930]

Can you explain the graph of x^x=y^y please? I've been trying for weeks and can't figure it out.

[@centristterrorist]

Xln(x)=Yln(y) Y/X=ln(x)/ln(y), ifk if this helps but thats that

[@nyctophile445]

Hello sir.. i have a very beautiful integral.. please make a video on it's solution... integral: ∫√ (e^(√ x))dx

[@Metaverse-d9f]

use gamma function

[@aitordominguez6482]

how

[@RezerdPrime]

Lets try now \int_{-infty}^{+infty} 1 - cos(tan(1/x)) dx

[@EHMM]

nice

[@Metaverse-d9f]

You can just solve for x, and then find dx.

[@ethohalfslab]

No! Don't change my integral >_

[@heinrich.hitzinger]

Fresnel: 😔

[@VideoFusco]

No offense, but this is really a barbaric way to apply the substitution method in an integral. If you then use the relation u=x^(-2) to rewrite x^3 in terms of u, why not immediately invert the relation by writing x=u^(-1/2) and then derive dx from there? It immediately comes dx=-(1/2)u^(-3/2)du without having to do all those intermediate steps mixing two different variables in an integral that has only one variable! In Italy, anyone who did something like this in a university exam would fail immediately. And even in high school this would determine a significant lowering of the evaluation.

[@d-hat-vr2002]

Math is difficult. You cannot force your mind to see the most elegant derivation. So we cannot blame others for failing to see the most elegant derivation. If you know a better derivation, then share it in a spirit of generosity rather than criticism and competitiveness. Calling someone a "barbarian" by implication is not polite. La matematica è difficile. Non puoi forzare la tua mente a vedere la derivazione più elegante. Quindi non possiamo biasimare gli altri per non aver visto la derivazione più elegante. Se conosci una derivazione migliore, allora condividila con uno spirito di generosità piuttosto che di critica e competitività. Chiamare qualcuno "barbaro" per implicazione non è educato.

[@VideoFusco]

@@d-hat-vr2002 It's not a matter of elegance, but of formal correctness (which is fundamental in mathematics): if you change variables you cannot write an expression that contains both variables at the same time. It's incorrect.

[@rachanamathssolutions]

Scientific Calculator Gives Ans. No Solution

[@abuabdullaahiwaaaishatah8235]

Since When Did u bald urself Sir Did Fresnel Do it as a punishment for changing his integral

[@PromeaurixhenXhivel]

i have this queation. Limit[Divide[1,h]\(40)Integrate[f\(40)y\(44)x\(44)t+h\(41),{y,p\(40)x\(44)t\(41),p\(40)x\(44)t+h\(41)},{x,a\(40)t\(41),a\(40)t+h\(41)}],h->0])

[@jamelbenahmed4788]

First ?

[@AndDiracisHisProphet]

nice

[@blackpenredpen]

Thanks!!