I Computed An Integral That Breaks Math

🙏Support me by becoming a channel member!

Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.

#math #brithemathguy #integral

[BriTheMathGuy]

I'd like to address some of the comments on this video. Many have (correctly) pointed out that the method in the video is not exactly mathematically rigorous. I like to think of it in the same way first semester calculus students think of continuity: "a function is continuous if I can draw it without picking up my pencil". Is this approach rigorous? - No, but it's intuitive and leads to the correct answer. There are other ways to get the same answer in the video. (For example: using x^dx = e^(lnx*dx) and expanding the Taylor Series)

I'd probably need the opinon of an expert mathematical analysis (which I am certainly not), but I think the most dubious part of the solution is turning part of the integrand into a limit of h. I believe that it only works in this situation (and it might be coincidence) because we preserve the traditional dx at the end of the integrand. We certainly cannot do this is every circumstance since it would make most every integral equal 0. (for example: ∫xdx ≠ ∫ lim h->0 (x*h) = 0. )

I think, despite the lack of rigor, the method is interesting and worth thinking about. Regardless of how you feel about the video's methods, I hope you enjoyed it and I appreciate all of the comments! (especially the ones requiring us to to think critically)

[Ciencillamente]

Mathematicians: "Physicist misuse differentials by using them as fractions"
Also BriTheMathGuy:

[tomasgerasimidis]

3:01 "This is a good candidate to use Lobey Towel's Rule"

[melchiortod]

I love that face he makes while he's concentrated. He looks like he's disgusted by the math😂😂 love it

[mcxzx]

Though I am not convinced by "dx" is "Iim Δx to 0" thing approach, 'cause in differential geometry, we know dx is a 1-form on the manifold(here is ℝ). But x^dx can be thought of as an element of C*(ℝ, ℝ) or Λ*(ℝ), the exterior algebra on ℝ. Like how we did to some characteristic class, x^dx can be expanded into Taylor series, with the multiplication of differential forms replaced by wedge product. Luckily, dxΛdx=0, lot of terms are 0, so x^dx-1 is equal to ln(x)dx. Furthermore, ℝ is one dimensional and contractable, so every 1-form is an exact form, we can find a 0-form(scalar field) F so that dF=(x^dx-1), and F is exactly the integral of ln(x), which gives out the answer in the video

[rody]

I do not know if it is just me, but I have always hated math, until I took Calculus. I am now fascinated. Everything just comes together and makes a lot more sense.

[gotbread]

It is an interesting question, but it feels like an abuse of notation.

[Cannongabang]

It's "Hospital rule" since that's where you end up everytime after using it and forgetting about Taylor.

[Pabloable]

I *knew* there had to be a reason for that -1 in the original problem. The funny thing is, each individual step was something I had done before, but I'd never have considered stringing them together in this way to solve this problem.

[jehmarxx]

What the heck? I made my profile pic as a joke, but with very little awareness that it may make sense.

[hugojoly]

Very nice problem ! I actually got it in a different manner : x^dx = exp(dx*lnx) and then I used the mcLaurin series exp(x)= 1+x+xx/2... et said any power of dx was negligible compared to dx.

[aioia]

I tried computing the indefinite form of the integral numerically and I it behaves like the indefinite integral of lnx, so you are right

[lucascostavalenca]

No way, this is amazing! Never expected such an elegant answer!!

[BriTheMathGuy]

🎓Become a Math Master With My Intro To Proofs Course! (FREE ON YOUTUBE)

[LukePalmer]

Another way, same answer:

x^dx = e^(ln x dx)

Taylor expand around dx = 0 (taking dx as just a variable, unrelated to x)

= 1 + ln x dx + 1/2 (ln x)^2 dx^2 + ...

Since we are integrating, we can take dx to be a first-order infinitesimal, i.e. assume dx^2 = 0.

= 1 + ln x dx

And the rest flows like warm butter.

[faisalkarimi]

Just something interesting that gives the same answer:
Re-write x^(dx) in terms of the taylor series of exp(t) around t=0 (where we substitute t=ln(x)dx). this transforms our integral into: n=1 to ∞ ∫∑(ln(x)dx)^n . the sum goes from n=1 rather than n=0 because the -1 in the original integrand cancels this term out. we can consider every term of order greater than or equal to 2 as equal to zero, which leaves us with just the first term in the expansion ∫ln(x)dx = xlnx - x + C

[___Truth___]

How do you make sense of what you're doing, you make it look intuitive

[drewpocernich]

Length of the video: 4:20
I applaud thee big brain sir.

[christoffer]

I tested this in Python, and indeed it works. Here is the code if anyone is interested:

import time
import matplotlib.pyplot as plt
import numpy as np

a = 1
b = 2


def integrand(x):
return (x ** dx) - 1

def antiderivative(x):
return x * np.log(x) - x

dx_range = np.logspace(-6, -1, 1000)

S_list = list()
exact_list = list()

start = time.process_time()
for dx in dx_range:
x = np.arange(a, b, dx)
S = np.sum(integrand(x))
exact = antiderivative(b) - antiderivative(a)

S_list.append(S)
exact_list.append(exact)

plt.loglog(dx_range, S_list, label='Riemann sum')
plt.loglog(dx_range, exact_list, label='antiderivative xlog(x) - x')

end = time.process_time()

plt.xlabel('$dx$')
plt.legend()

print(f'ran in {end - start} seconds')
plt.show()

[NaniuTherapist]

Another way for this unambigous expression is:

x^{dx} = e^{ln(x) dx}

and then expand the exponential as a series:

1+ ln(x) dx + ln^2(x) d^2(x)/2 + ...

of course one has to be very careful at several steps in order for this to be well-defined. I personally find this series more interesting than the standard log(x) integral.