Clever choice of techniques used to arrive at the result of these two integrals. Thanks for sharing. Excellent explanation and methodology.

Nice. Dr Peyam used a very similar method to evaluate the related Gaussian integral on his channel. But idk if I would say this uses Feynman's trick or the Leibniz integral rule because the integrand of I(t) isn't a function of the variable x and parameter t, rather the parameter t is the upper bound of the integral. I mean you could say it uses it but it's essentially just the FTC in this case.

@njiawuatuaValentine-ug2jn

5 ай бұрын

But does this integral converges

After making the substitution y=x/t, doesn't the bounds of integral change from 0 to 1, to 0 to t??

@maths_505

Жыл бұрын

Yeah I figured that out when reviewing the video. The upper limit doesn't bother the integral because the complex exponential is an oscillatory function in t so it thankfully doesn't blow up. So in essence we're still integrating zero with zero and infinity as the limits of integration which still checks out to zero.

@illumexhisoka6181

Жыл бұрын

@@maths_505 but if we are integrateing 1/t fro 0 to t the answer will be 1 no matter what t is

@maths_505

Жыл бұрын

@@illumexhisoka6181 the integration was being carried out w.r.t x with t held constant. So it turns out we're evaluating the integral of zero from zero to infinity....which is again zero.

@illumexhisoka6181

Жыл бұрын

@@maths_505 yes The integral 1/t w.r.t x from 0 to t when t->∞ equal to 1

@xtremeplayz3606

10 ай бұрын

​@@maths_505 I still don't get this, if the exponential part was linear in x, then it might have been an easy conclusion, but here it is quadratic right? So how can we be sure?

Awesome video....coincides with a Putnam question so happy days.

At 17:21, you write down the bounds of integration as 0 to 1 before you substitute y = x/t. The correct bounds should be from 0 to t. This change means that it is no longer sufficient to show that the integrand tends to 0 for large t. This also means that x/t does not tend towards 0 for large x, which could have also affected the result.

@maths_505

Жыл бұрын

You're integrating zero from zero to infinity They key here is that the limit is being taken w.r.t to t while the integration is being carried out w.r.t x.....that's what keeps the math from falling apart here....although you could separate the real and imaginary parts and go through the same steps and that would give the same result albeit with a better explanation now that I think of it

@akossomogyi3489

Жыл бұрын

@@maths_505 if you apply the same logic to lim_{t -> inf} int_0 ^t ( 1/(1+x^2/t^2) * 1/t) dx it says that the value is 0 while it is actually pi/4 (regardless of t), so this definitely needs more care to sort out. Saying that a function goes to 0 [even uniformly] while the interval you integrate on goes to infinity can pull the result anywhere (i.e. basically inf x 0 = 0 type logical error)

@ach3456

8 ай бұрын

@@maths_505 I don't believe that explanation is sufficient, as there are several functions where the integrand goes to 0 but the integral does not because of a t on that integration bound. 1/(t+x)^(1/2) for example. You need to show the integral apart from that 1/t is bounded, which you could by considering the weird quasi-periodicity and boundedness of e^(ix^2).

3 ай бұрын

the substitution actually transform the bound of integration from [0,1] to [0,1/t] (if x = 0, then y = 1/t) so the integral actually disappear even faster.

@audreychambers3155

3 ай бұрын

@ If y = 1 and y = x/t, x = t The bounds are from 0 to t, not 0 to 1/t.

Beautiful. I personally thought Flammy's video felt a little stiff and counterintuitive, but you explained it methodically :) Now the Fresnel integrals ain't got nothing on me 😈

@daddy_myers

Жыл бұрын

Oh, and speaking of integrating under the integral sign to use the Fresnel integral, don't you mean using the Laplace Transform to solve it?

@maths_505

Жыл бұрын

@@daddy_myers you know me so well....

@daddy_myers

Жыл бұрын

@@maths_505 But wait, you already have a video on that. Unless... *Are you going to Laplace Transform the complex exponential?*

@maths_505

Жыл бұрын

@@daddy_myers suiiiiiiiiiiiiiiii

@daddy_myers

Жыл бұрын

@@maths_505 😲😲😲 Hype.

Thank u

Wouldn't it be easier to just set the integral as a Gaussian integral then evaluate the integral as a=-i. It gave the same answer and it didn't take me more than a minute

My eyes: *sees cosine and sine in the same sentence* My brain: "EULER EULER EULER EULER!"

What's the program used as a board?

Now I understood why tables are made byhearted from reverse in childhood

The final value of both original integrals should be sqrt(pi/2). The number sqrt(pi/8) is only half of that value.

@andre-z

Жыл бұрын

Definitely

@violintegral

Жыл бұрын

@@andre-z k

@OP or anyone who knows the answer....can you prove the existence of these improper integrals before evaluating them? Otherwise, it doesn't make much sense. I mean, the integral of sin(x) over any unbounded interval does not exist, why would sin(x^2) be any different?

@sin2Pi

Жыл бұрын

sin(x^2) is highly oscillatory (freq goes to infinity as x goes to +/- infinity), see attached plot, here: drive.google.com/file/d/141UX0Dj7AqpIf7ffOkVDoev_x3fmFt3V/view?usp=share_link

@daddy_myers

Жыл бұрын

@@sin2Pi This can actually be proved quite easily. ∫sin(x^2)dx from -inf to inf can be reduced through a neat substitution, namely: t = x^2 First, since this is an even function, I'll double the integral and have it go from 0 to inf. So now, the integral reduces under the substitution to: ∫sin(t)/sqrt(t) dt from 0 to inf Through Dirichlet's test for improper integrals, we can tell this integral is convergent, as it has the form of a bounded function multiplied by a monotonically decreasing function. Most importantly, it has no anomalies along the interval of integration (i.e: continuous), and the limit exists at both its endpoints (being zero for 0, and zero for inf. The second fact is crucial for this test to work). Since through a substitution we showed that the original integral reduced to this problem, we can safely say that they're both equal and both convergent.

@maths_505

Жыл бұрын

@@daddy_myers stole the words right from the keyboard! Great explanation bro I'm proud of ya!!!

@daddy_myers

Жыл бұрын

@@maths_505 Thanks man :) I'm actually super proud of myself on this one, since I had just learned it quite recently.

@sin2Pi

Жыл бұрын

@@daddy_myers Awesome! Forgot about Dirichlet test, which proves the conditional (non absolute) convergence. Thanks.

@ 19:15 the LHS should be sqrt(I(t)), not just I(t).

@maths_505

Жыл бұрын

Yeah my bad...

17:40 it's unclear why you are allowed to let y=x/t, thus changing y from what appears to be an independent constant in t, to suddenly bring dependent on t and affected by taking the limit. I assume it's just the reversal of your much earlier t-substitution, but it would be clearer of you made this explicit. With all the manipulation in between you lose sight of the fact that "y" is actually "y(t)" i.e. a function of t that should be included when considering the limit.

@maths_505

Жыл бұрын

Basically I was moving from the y world to the x world in both of which t is a constant so I was just manipulating that constant nature of t along with the oscillatory of the complex exponential term It could equivalently have been y=z/t. However, now that I think of it, all that algebra in between can prove a bit distracting. I'll take care of it next time. Cool insights as always.

@zunaidparker

Жыл бұрын

@@maths_505 no worries. Yet another great integral technique I must add. Yeah the main thing to deconfuse is that y was always a function of t since the first substitution. I think a simple fix would be a slight change in language. Instead of saying "let's let y=x/t" which makes it seem like we are choosing to make y suddenly dependent on t, you should rather call back to the original substitution, something like "remember from our earlier substitution that y=x/t, now when we reverse that substitution you'll see how everything works out neatly...".

@zunaidparker

Жыл бұрын

Actually, I think I spotted a mistake. When you do the substitution y=x/t, you need to change your limits of integration when moving to the dx domain. Instead of 0 to 1 it becomes 0 to t. So when you take the limit you have an indeterminate form: your integrand is going to zero but your limit of integration is going to infinity. I knew something felt off to me about this step. Can you please double check if my observation is correct?

@maths_505

Жыл бұрын

@@zunaidparker yeah I figured that out when reviewing the video. The upper limit doesn't bother the integral because the complex exponential is an oscillatory function in t so it thankfully doesn't blow up. So in essence we're still integrating zero with zero and infinity as the limits which still checks out to zero.

@maths_505

Жыл бұрын

@@zunaidparker after all, the antiderivative of 0 is C so taking limits of C as t approaches infinity and zero are both C....so yeah thankfully it's still zero

The only way you could have improved this is if you had some way of weaving in the Gamma functions

18:21 with the last substitution [x = y/t] also the bounds of integration have to change [0, 1] -> [0 to 1/t] and that makes the integral contribution to the result disappearing even faster. So the final result is correct.

>says it’s the “coolest way” >uses feynman’s technique instead of contour integration smh my head

@maths_505

Жыл бұрын

Well honestly even contour integration pales in comparison to my video on the fresnel integrals using basic complex analysis. That's definitely the most elegant. I've provided a link to that in the description