Beyond the Gaussian integral
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- Опубліковано 2 чер 2024
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Here's a challenging integral for Calculus 2 students! We will evaluate the improper integral 1-e^(-1/x^2) from 0 to infinity. This integral actually converges thanks to the "1-". This integral is closely related to the most famous integral in the world, the Gaussian integral. In fact, we can use integration by parts, u-substitution, and L'Hosptial's rule to make the connection with the Gaussian integral. This can be given to any calculus 2 or calculus 3 students if they would like to have a challenging integral. #calculus #mathforfun #blackpenredpen
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0:00 My pick for the most famous integral in the world
0:14 integral of 1-e^(-1/x^2) from 0 to inf
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_Beyond the Gaussian Integral_ would make for a smashing title of a TV-series, starring BlackPenRedPen. Doubt if it would be any more intriguing than his YT videos, though.
😆 thanks!
I solved it using Feynman’s method, i.e. by inserting alpha into the exponent and differentiating with respect to it, then performing a simple change of variables u = 1/x. You will then get a regular gaussian, then integrate the result with the respect to alpha. The integration constant is zero since I(0) = 0.
I knew someone would try that lol
Great explanation as always - thanks a lot👌
Can you make more toughest integrals seems to know some interesting ones....😎😎🗿
Cease existing
Great!
Here's how I did it: letting 1/x^2=t gives for the integral I = 1/2 int_0^oo t^(-3/2) (1-e^-t)dt which can easily done using integration by parts and the integral representation of the gamma function.
Thank you Blackpenredpen for being an integral part of my life. I am in class 9 but your videos have helped me to learn and ace differentiation and integration. HATS OFF MAN FOR SUCH A WONDERFUL EXPERIENCE.
An *integral* part of your life?
no pun intended?
THANKS PROFESOR !!!!, VERY INTERESTING !!!!!!
The way he holds the pen is legendary
8:42 finally finally finally!
Very nice
really interesting❤❤
Glad you think so!
Marvelous.
It cured my tummy ache which I've had all morning!
I love calculus
Sir! Can u make a course on integration from basics to top highest level pls
Waiting for it...
I like this
Question: When you do integration by parts, why do you sometimes take the plus or minus integral of the last row and sometimes you don’t
I think it's usually when that line contains a 0
At the very end lim x-> infinity e^(-1/x^2) should be e^0 = 1 (so 0*(1-1) also equals 0)
find integral sqrt(1+x+sqrt(1+x))dx
Here's the solution:
∫ √(x+√(x+1)+1) dx
//
// Preform a U-sub:
// Let u = √(x+1)
// x+1 = u²
// x = u²-1
// dx = 2u du
//
= ∫ 2u √(u²+u) du
= ∫ (2u+1-1) √(u²+u) du
= ∫ (2u+1)√(u²+u) du - ∫ √(u²+u) du
= ⅔√(u²+u)³ - ∫ √(u²+u) du
//
// Complete the square:
//
= ⅔√(u²+u)³ - ∫ √((u+½)²-¼) du
= ⅔√(u²+u)³ - ½∫ √((2u+1)²-1) du
//
// Perform a Hyp-sub:
// Let cosh(θ) = 2u+1
// sinh(θ) dθ = 2 du
//
= ⅔√(u²+u)³ - ¼∫ √(cosh²(θ)-1) sinh(θ) dθ
= ⅔√(u²+u)³ - ¼∫ sinh²(θ) dθ
//
// This is a common integral:
// ∫ sinh²(x) dx = ½(sinh(x)cosh(x)-x) + C
//
= ⅔√(u²+u)³ - ⅛(sinh(θ)cosh(θ)-θ) + C
//
// Back substitute to u:
// cosh(θ) = 2u+1
// sinh(θ) = √(cosh²(θ)-1) = √((2u+1)²-1)
// θ = cosh⁻¹(2u+1)
//
= ⅔√(u²+u)³ - ⅛(2u+1)√((2u-1)²-1) + ⅛cosh⁻¹(2u+1) + C
//
// Expand the perfect square:
//
= ⅔(u²+u)√(u²+u) - ¼(2u+1)√(u²+u) + ⅛cosh⁻¹(2u+1) + C
//
// Factor out ⅟₁₂ √(u²+u)
//
= ⅟₁₂ √(u²+u) (8(u²+u) - 3(2u+1)) + ⅛cosh⁻¹(2u+1) + C
= ⅟₁₂ √(u²+u) (8u²+2u-3) + ⅛cosh⁻¹(2u+1) + C
//
// Back substitute to x:
// u = √(x+1)
// u² = x+1
//
⅟₁₂ √(x+√(x+1)+1) (8x+2√(x+1)+5) + ⅛cosh⁻¹(2√(x+1)+1) + C.
Is there a simpler geometric demonstration?
Bro speak about anything and i would listen.
Your are ln(legend)^legend mathematician
Could you differentiate the prime counting function?
It’s 0 everywhere except at prime numbers where it’s undefined
So. What if there’s a half iterate Gaussian function ?
Bro inventing more stuff 🗿
Meanwhile college students students: 💀☠️
老師好，我想問羅畢達定理趨近於無限的版本要怎麼證明
Ah yes, the famous mathematician Carlian Friedrichian Gaussian lol.
Sir, is an “Improper Integral” the same thing as a “Nonelementary Integral”?
Can you please do a video explaining the difference and shed some light on this topic, thanks a lot 🙏
I believe an integral is improper when there is infinity involved and you gotta take limits, not the value itself.
No, that's totally different notions. For checking, is it an *improper integral* or not, you need to plug in bounds of integration to the function you integrate. If you have something indefinite, such like 1/0, 0/0, or smth, this is an *improper integral* (because you need to take a limit). Also, if one or both of the bounds are infinity, this is exactly an *improper integral* (because you can't just "plug in" infinity without taking a limit).
A *nonelementary* integral is more about indefinite integrals. That means, that the antiderivative of the integrating function cannot be "explained" in terms of numbers, variables, additions, multiplications, powers and so on. Such an antiderivative is called "nonelementary". For example, sine is an *elementary function* ('cause in complex analysis we express sine as difference of exponents). But error function isn't, so an (indefinite) integral of the function e^(x^2) is called *nonelementary* .
Proving the integral is nonelementary is veeeeery hard to do on your own. You need to use some really massive techniques to do that. So here you need to rely on your feelings. For example, you can safely assume that e^(x^3) can't be nicely integrated so integral is *nonelementary*, because of its similarities to error function.
@@andreyfom-zv3gpOk, but what if you have an integral that you cannot obtain an antiderivative for, neither by regular (elementary) functions nor by special (nonelementary) functions, but you can still do a definite integration on it (on a specific region) or express its antiderivative using series (like Taylor series), what do you call this type of integral?
@@BassemFanari Elementary functions are those who we consider to be "simple", hence the name. Anything else that is not elementary is a non elementary function. Special functions are a "subclass" of non-elementary function which can be seen regularly in practical applications. It doesn't matter the integrand, as long as it is continuous, it will always be the case that the function has an antiderivative. Writing it down elegantly is a different story but it does exists.
An improper integral is a definite integral that involves limits, commonly involving infinity in the bounds, while a nonelementary integral is an indefinite integral that has no solution in the terms of standard mathematical function. If you want to evaluate a nonelementary definite integral, you have to use advanced techniques such as complex analysis.
What would Chen Lu do?
can you do something easy? like 4x+6=2
His name was Gauss.
cool
Yesss 🏋️🏋️🏋️💯💫
I solved with Feynman’s method, isn’t it.
how are you always smiling
Am I the only one using Taylor expansion instead of l'Hôpital rule ? to get limits ?
C’est juste que les ricains ils sont louches
Where's your pokeball?
😢
Is calculas can understand by a 7 th grader
No Johnny please go do your homework
Yes
hey im the 10kth viewer 🥸
Minus was neglected twice, I guess then minus wasn't neglected at all.