I like the Feynman technique a lot because of how elegant and simple it is. However, because the u-sub method is more straightforward, and when attempting to tackle an unknown integral, you should always try all the other integration tools in your arsenal before the Feynman technique. Therefor, from a practicality standpoint, I prefer the u-sub method because it’s one someone may be more likely to try and be successful with when dealing with unknown integrals

Another way to tackle this integral is to write the integrand as the integral from 0 to 2 of x^ada and exchange the order of integration with Fubini's Theorem. This is basically just using Feynman's trick but honestly I prefer using other methods if I can help it since they feel a bit more intuitive

@ChaitanyaTappu

2 жыл бұрын

I'll get behind this because I always found it hard to verify the hypotheses of the Lebesgue DCT. Fubini/Tonelli is usually easier

@aneeshsrinivas9088

2 жыл бұрын

itsa me fubini

3:50 was actually pointed out to me by an old professor, Dr. Lambers, at University of Southern Mississippi. I'm nowhere near that clever myself! 😝

@drpeyam

2 жыл бұрын

Omg, hi Floyd!!! Thanks again for the suggestion 😁

I like both methods, Peyam! Great video!

Cool! Just watched one of your feynman integral videos yesterday, and now found this one that is a minute old :)

I like Feynman method because it kind of swoops in out of nowhere and solves all sorts of weird integrals. Your videos are boss and very dope! thanks!

Great video, that's so cool! By the way in Italy the u letter is not used to do substitution integration, we rather use t

amazing!

I love the way you talk and represent things ☺️

@drpeyam

2 жыл бұрын

Thank you!!!

These 2 method I've seen from blackpenredpen and Michael Penn. However your presentation let me remind and appreciate yours. Thank you for your share.😊

Good job

A different approach would be to introduce x= Exp[-t].This leads to the Integral of (Exp[-t]-1)*Exp[-t]/t. If you now set J[s]=Integral ((Exp[-t]-1)*Exp[-s*t]/t) from 0 to Inf. Then it is a standard Feynman-case ,which is easily solved.

You are going to the dark place of Reddit. What next Wikipedia? We are all doomed I tell you.

using Schwinger parametrization for log[x] will make evaluating it easier.

@drpeyam

2 жыл бұрын

What’s that?

@AlamKhan-ly5qn

2 жыл бұрын

@@drpeyam the first equation of link en.m.wikipedia.org/wiki/Schwinger_parametrization

They both are very interesting

Sheeeesh early !! Love you Dr.

Both cool 😎

Hello, I'm from Indonesia, I like to watch explanations about mathematics, and I only suggest giving Indonesian subtitles so that I understand the explanation that is explained.

Nice video ! When we had f'(t)=1/(t+1) integrate: f(t)=ln(t+1)+C We know f(0)=0=C Then f(t)=ln(t+1) Then: f(2)=ln(3) In fact it is the same way to do but in an other point of view.

@MurshidIslam

2 жыл бұрын

How did you get f(0) = 0?

@ericdenadai1556

2 жыл бұрын

@@MurshidIslam We saw that in the video and it is logique. f(t)=int(x^(t)-1)/ln(x)) Then f(0)=int(x^(0)-1/(ln(x)) f(0)=int(0) f(0)=0 (Int(0)=0 because int(0*dx)=[C]=C-C=0

@MurshidIslam

2 жыл бұрын

@@ericdenadai1556 Thank you.

I love Feynman way♥!

Intéressant

Hey, I am a freshman right now and exploring mathematics as my minor. I have not yet lerned about feynman Integral but I have a question @2:20 on where does f(2) come from?

woa that's cool

Use lnx=-u write and use some Laplace transform

@drpeyam

2 жыл бұрын

Care to elaborate?

@breadsneakypeaky1104

2 жыл бұрын

@@drpeyam he uses the Laplace transform for (e^(3t)-e^t)/t, which is ln[(s-3)/(s-1)], plug in s = 0 and we get the result.

Everything Feynman is good

To do the first technique rigorously, would you first need to check f us well-defined, or is that check contained in the calculation?

@Keithfert490

2 жыл бұрын

Why wouldn't it be well defined?

@deadfish3789

2 жыл бұрын

@@Keithfert490 In this case, I agree it's pretty obvious, but it may not be if the function went to infinity on the interval, or if the interval had infinite length.

@WareCzech

2 жыл бұрын

@@deadfish3789 To argue this formally you can use the Leibniz integral rule for Lebesgue integration (see en.wikipedia.org/wiki/Leibniz_integral_rule#Measure_theory_statement) which has very weak conditions that are easily satisfied in this case.

@soufianenajari8900

2 жыл бұрын

@@deadfish3789 For the feynman thechnique ? In fact you need to prove that f is well defined AND that f is differentiable, which is a theorem but i don't find it in english on the web so i can't link it to you

Feynman's technique is far better than u-sub in this case.

Do you think your friend bprp could do these integrals? (I couldn´t do them.)

@drpeyam

2 жыл бұрын

Of course!!

From the problem, it looks to be valid, but could you perhaps explain why Fubini's condition is satisfied so that we can swap the order of the integrals?

@fakechuck7659

2 жыл бұрын

The intuitive explanation is that you're summing up the slices of a volume and it doesn't matter which axis you slice along as you are still counting all of the parts. It's similar to the simple commutative property of multiplication.

@jeff13379001

2 жыл бұрын

@@fakechuck7659 I know it works in the finite case, but in this case, one of the bounds is infinite. I have seen instances where we can't do this, so I want to know why this one works.

Can u try to change the integrals in 06:26 to Y axis in another one ?

@drpeyam

2 жыл бұрын

?

Like 2nd mathod because it had 2 integrals 🤣

Bro i thought it was reddit feed lol😁😫😫😁😫😁

Ellen Degeneres!!! I knew it