One crazy looking integral
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- Опубліковано 6 лют 2025
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In a field burdened by so much technical vernacular, refreshing to see adjectives like crazy and weird getting used all over the place!
"Evaluate this wack-ass integral, yo."
There is a progression. I wonder, what will come after "crazy and weird?"
@@mrng1724 wild has gotten a little air time
Doesnt everyone agree what he does at 4:30 with y is not something anyone will think of ever unless they saw it before..and what he does is a bit similar to Feynman's technique..
You can directly apply Leibniz rule after placing a parameter in front of tan^2. With a Pythagorean identity, you split the result in two. One is solved with u-sub to the Gaussian integral. The other is the original parameterized integral. You then have a first order linear ODE. Either 0 or infinity can be used for initial values.
12:54 Have a good day everyone ! ✨
Thanks! You too.
General form is: integral from 0 to pi/2 exp(-u (tan x)^2) dx = pi/2 exp(u) erfc(sqrt(u))
Don't the u→z and z→t substitutions invert each other?
Then we could have simply upgraded from e^(-u^2)/(1+u^2) to e * exp(-y(1+u^2))/(1+u^2)
Yea or just solve by regular substitution right?
Yay! Another integral that results in something with both pi and e in it.
Lovely technique!
This was easy to understand, but to actually come up with all these steps? Maybe a final bonus exam question at Michael Penn St. U.
Nice video and cool integral. I did it using Feynman technique by posing I(t) = int[0 to inf] e^(-t²x²)/(1+x²) dx which gave the same result
If I could make a suggestion, I always feel mathematics relies on "yet another trick" that makes a solution possible. Where does this trick come from? It would be nice to maybe see an exploratory video where you don't know the trick upfront, but try different things. I'm not even a mathematician, but I imagine that to be what a research mathematician does.
If you're known for using tricks in mathematics, you could be called a "mathemagician."
Agree, it feels like: "But wait! There is more!..."
In the case of integrals, there is a single unspoken (and untaught) rule for tackling hard integrals: turning a single integral into multiple integrals whenever possible can likely simplify it. Doing this has potential to turn a single integral of a crazy function into a multiple integral of a rational function (or some other basic form), and rational functions are ALWAYS possible to integrate in terms of elementary functions. In this video, introducing a double integral did not simplify it into a rational function, but it still simplified into a familiar form. In general, people tend to think that the only way to really break down integrals into multiple parts is through addition. But turning a single integral into a multiple integral gives us a second (usually overlooked) method for breaking an integral into multiple parts: through multiplication of integrals (because a multiple integral can be turned into a multiplication of integrals, as long as the variables can be separated. Even if the variables cannot be separated, computing a multiple integral of a rational function can be much simpler than the original integral). Also, sometimes changing the integral into a multiple integral can make other tricks apparent. For example, the well-known method of change of coordinates for computing the Gaussian integral, etc. This is the trick of tricks when it comes to integration. There are a bunch of other smaller tricks, but this single trick can make someone much better at integration.
Another powerful trick is parametrization. Basically, make the integral you're trying to compute just a specific case of a more general form, by introducing a new variable into the integral. It turns out that sometimes a general integral can be easier to compute. This is basically how the "Feynman's" method works. A variation on the Feynman's method is taking a Fourier Transform or the Laplace Transform (or some other transform that puts you in a frequency domain) of the parametrized integral. This method has potential to work because of the amazing convolution theorem that makes certain integrals easier to compute in the frequency domain.
Once you got to the integral of e^{-u^2}/(1 + u^2) you could have written 1/(1 + u^2) = integral from 0 to infinity of e^{-t(1 + u^2)} dt. Then switch the order of the integration. I've seen this trick for proving gamma function identities.
Evaluated the integral prior to watching the video. At first I made the same substitution that Michael did. After that, in honor of the differential equation videos on Math Major, the technique I used to evaluate this integral is Feynman's technique and evaluate it using differential equations. It's just a first order linear differential equation.
I've just done the same. It seems much simpler! You can use the standard integrating factor technique for linear first order differential equations.
The zeroth integral method outlined really IS Feynmann's method, just with the order of differentiation wrt y and definite integration wrt y switched around, and with y used instead of the usual a.
Do you like variable substitution? I added a bit of variable substitution into your variable substitution.
He is probably the best math teacher on utube.
So, the evaluation of the integral ends up being another integral after endless substitutions. Nice.
8:46 I would use "u sub 1" or "u sub 0" or something like that.
Tbh i get inspiration of what i should use based on the video's length 😂😂
Hi Dr. Penn!
Try the integral x*ln(1+x)/(1+x^2) from x=0 to x=1 and do a video on it
You can also solve this with a Laplace transform if you make one more substitution after the first one.
You can make the integral into
1/2*Int(exp(-t)/(sqrt(t)*(t+1))dt from (0, inf)
Let s be a parameter and our Integral becomes
I(s) = 1/2*Int(exp(-s*t)/(sqrt(t)*(t+1))dt from (0, inf)
This is just the Laplace transform of
t^(-1/2)/(t+1)
and we can use a unique result for the Laplace transform of the product of functions, provided certain integrals exhibit the appropriate convergence (which for these functions they do).
Laplace Transform (f(t)*G(t)) =
int(g(y-s)*F(s))dy from (s,inf)
Here just recognize that G is the Laplace transform of g, and F is the Laplace transform of f.
thank you alot.🌷
I miss the time when math used to have numbers :')
This function is in fact the geometric derivative of (sin(x))^2
0.5 e π erfc(1) ≈ 0.671647
Dude a lot of substitution to end with a special function? 😅
How lovely
Please wear a magician’s hat next time as you pull another one of your awesomely inspired zeroth integrals out of seemingly nowhere!😄As Paul Daniels used to say, “Now that’s magic!”
You’ll like it. Not a lot, but you’ll like it.
wait, how did y appear in the equation all of a sudden? :)
Why not solve the integral at 2:05 using the Residue Theorem?
The arc integral are not bounded, due to essential singularity of exp(-R^2) for R=Infinity*i
@@vascomanteigas9433 Ahhhh you are correct. Thank you!
Either way the residue theorem would given e*pi/2, and fail the erfc(1) part.
It may exists a crazy trick for integrals with f(z)=exp(-z^2)/(P(z)), where P(z) are a polynomial.
@@vascomanteigas9433 thank you! Was wondering about this:)
Now thats one crazy wacky looking intergral 💯💯
I solved it by expanding e^(-tan^2(x)) and using the beta and gamma function and the euler reflection formula and got pie*e/2
Im so confused
Edit: i now realized that the real part of z1 and z2 must be bigger than zero in the beta function
You need the Analytical Continuation of beta function, which are Simple, but could cause problems.
Ah, new math tricks. Fun, thanks.
Hey can anyone please help me find a prime number whose digits add up to give 3.....
One example is 3 ( obviously)
If digits add up to 3, then the 3 divides that number. So 3 is only such number that is prime
I'm afraid 3 is the only one. The numbers of the forms 2...1 and 1...1...1 are divisible by 3
@@alre9766 Thanks a lot man...
U know...I was playing around with prime numbers the other day..and I found that every prime number ( I tested for prime numbers between 1 and 100 only ) has, what i call , a *reversible twin* ....
For example
For 17 the digits add up to give 8
Similarly the digits of 53, another prime number, ads up to give 8
So I called 17 and 53 reversible twins....
I found the same with so many primes such as 5 & 41 and 7&43 and 19&37...and so on....
But I was struggling to find the same for 3........
That is the reason I came for help here.....
Also, do u know any other prime number which doesn't have a reversible twin...
Or is 3 the only one??
If yes, then please let me know....
🙏🙏
@@nikhilsoni2403 67 does not follow the rule, I guess
The 0 integral is the feynman technique backwards
What does it do?
Drink each time he says "substitution"
Nice
good
I'm surprised you didn't use the Feynman technique (differentiation under the integral sign). I guess your "zeroth" integral is essentially the same thing.
It's the same
How would you apply Feynman technique? I tried , and got I(t)=c/t . We would like to use I(0) = π/2 to get c but we cant as far as I can see. However please let me know if I missed something very curious
@@domc3743 i used feyman trick, what i did was ; I(a)=integral from 0 to inf of exp(-u²) * exp(a*(u2+1))/u²+1
Integrating it back was quite difficult, however I(1) is easy to calculate so we get c easily, then we use I(0) and its done
Set I(t) = integral of exp(-t*x²) /(1+x²) and apply Feynmans technique . You get I(t) = I'(t) + sqrt(pi/t)/2. Integrate and it should do the trick. Integral of exp(-t)/sqrt(t) is error function in diguise
@@diniaadil6154 yes , I was hoping we could get it directly from the thumbnail integral . Substitution is a bit messy but straightforward
too many substitutions
U know...I was playing around with prime numbers the other day..and I found that a prime number ( I tested for prime numbers between 1 and 100 only ) has, what i call , a *reversible twin* ....
For example
For 17 the digits add up to give 8
Similarly the digits of 53, another prime number, ads up to give 8
So I called 17 and 53 reversible twins....
I found the same with so many primes such as 5 & 41 and 7&43 and 19&37...and so on....
But I was struggling find the same for 3........
But then I realized that any number whose digits add up to give 3 is not a prime at all...because it will be divisible by 3
Does anyone know any other prime number which doesn't have a reversible twin ?
Or is 3 the only one??
If yes, then please let me know....
🙏🙏
Edit: I am just in high-school so I don't know much about number theory..
Thats a really dumb question to ask... are you a 7 year old??..lol...
Seems strange to define z = u^2 + 1 - and hence u = sqrt(z - 1) - only to later define a separate variable t = sqrt(z - 1), meaning that t = u. And then we go on to reuse "u" for a different defintion! If we had just recognized that t = u, there wouldn't have been a need to reuse "u" in the first place!
But moreover, undoing a substitution implies that the u --> z substitution was never necessary in the first place. If we were only going to subsequently undo it, we could have just applied the subsequent substitution to the function defined in terms of u instead.
In bengali.
Pai na dashamik.
So many substitutions 🤯 what exactly is the final answer???
about 0.67
hectic
This is the scariest video on the internet. Calculus is terrifying.
Calculus is love. I swear calculus, linear algebra and differential equations made me into what i am.
@@kalujny I can understand algebra and trigonometry very well, and I can even resolve integrals and differential equations by following the tried-and-true steps, but what I'm referring to is the fact that I don't understand the intuition behind Calculus and why the aforementioned steps work. I've watched so many people on UA-cam explaining it and I still don't get it:
MY BIG MENTAL BLOCK: How can dx or dy be infinitely small and added an infinite number of times (i.e. rectangles) to calculate the area underneath the curve so precisely. Isn't infinity infinite? How do you add up an infinite number of rectangles between two coordinates? Why does the integration method work?
@@dannybodros5180 I'm by no means a math teacher. But I think the way to understanding integrals are limit then derivative then the integral. The limit is the foundation upon which the calculus builds. Sorry if this sounds obvious.
@@kalujny Limits are fine. It's the notion of infinitesimals that I'm struggling to comprehend. The idea that something can be infinitely small but not zero. Like the number 0.000001 for example. Why not just say dx/dy = 0.000001 instead? The result would still be accurate enough anyway. Why does it have to be infinitely small and how does that even work? I just have trouble visualizing it. That's why I don't understand the intuition of Calculus, and I think it's the core reason why so many people find it so difficult. If someone would explain it in a way that makes sense, I think Calculus wouldn't have the bad reputation it has.
Hey...This is not related to the video..but is every natural no. a multiple of a prime no???
Yes, because every number is either prime or a multiple of primes.
Yes, every natural number except 1 is a multiple of a prime number. Follows from the fundamental theorem of arithmetic which states every natural number > 1 has a prime factorization
@@MyOneFiftiethOfADollar a unique prime factorization, i might add
Pin plz
no?...
asnwer=1dz x isit
Damn I arrived at the desired result but spent days trying to figure out the value of the complementary error function at 1. I got so disappointed I couldn't work it out I thought I had failed but it turns out michael did the exact same thing.
Hey can anyone help me find a prime number whose digits add up to give 3.....
One example is 3 ( obviously)
Hey can anyone help me find a prime number whose digits add up to give 3.....
One example is 3 ( obviously)
Such prime does not exist, because if the sum of digits is 3 then the number itself is divisible by 3 ;)
@@JakubMatuszczyk Thanks a lot man...
U know...I was playing around with prime numbers the other day..and I found that every prime number (but I tested for prime numbers between 1 and 100 only ) has, what i call , a *reversible twin* ....
For example
For 17 the digits add up to give 8
Similarly the digits of 53, another prime number, ads up to give 8
So I called 17 and 53 reversible twins....
I found the same with so many primes such as 5 & 41 and 7&43 and 19&37...and so on....
But I wad struggling find the same for 3........
That is the reason I came for help here.....
Thanks a lot man...
Also, do u know any other prime number which doesn't have a reversible twin...
If yes, then please let me know....
🙏🙏
@@nikhilsoni2403 number theory is not my area of interest, I have never heard of such property (it's interesting, though!). Ask on facebook groups, maybe someone will help you.