Kind of weird how this implies that part of the function flips around the x-axis for n of different parity. Though, since log(t) is negative between 0 and 1 it doesn't matter: the additional minus simply cancels out the negativity of the logarithm for odd n.
HA! I knew you'd leave that limit as homework. Rewrite t(ln 1/t)^(k+1) as (ln 1/t)^(k+1) / (1/t) Substitute u = 1/t. Now the limit goes to infinity lim as u->infinity of (ln u)^(k+1) / u Do L'Hopital's Rule lim as u-> infinity of (k+1)(ln u)^k * 1/u in the numerator and 1 in the denominator = (k+1)(ln u)^k / u. Very similar to the original limit, just with an extra factor and a reduced exponent. Keep applying this formula until the exponent is reduced to 1. You'll have the limit as u->inf of (k+1)! (ln u) / u. Apply L'Hopital one more time to get 0.
The integral is equivalent to the traditional gamma(z+1) definition since it was just a substitution that transforms one into the other (x = ln(1/t) as in the 2nd part of the video).
Just let I(t) = ∫₀¹ (1/x)^t dx = 1/(1-t). Differentiating n times under the integral sign gives Dⁿ I(t) = ∫₀¹ (ln(1/x))^n (1/x)^t dx. Normal differentiation gives us Dⁿ I(t) = Dⁿ (1/(1-t)) = n! * 1/(1-t)^(n+1). Setting t=0 gives us the result in the video: Dⁿ I(0) = ∫₀¹ (ln(1/x))^n dx = n!
I find this identity very interesting. Using indirect differentiation under the integral, Evaluating \int_{0}^1(lnt)^n dt We know that \int_{0}^1x^m=1/(n+1) Now Differentiating the above results n times and setting m=0 We get (-1)^n n! Therefore A=\int_{0}^1(lnt)^n dt=(-1)^n n! Now back to the question: \int_{0}^{1}(-lnt)^ndt=(-1)^n A =n!
Just asking: Since 0! = 1 based on convention or definition, might it not be better to use n = 1 as the base case for the induction, since 1! really does equal 1, without invoking any special convention or definition?
In your base case, you run into problems at the endpoints of the interval. At the left endpoint, the expression inside the exponent goes to infinity and, at the right endpoint, it goes to 0: both (infinity)^0 and 0^0 are indeterminate forms. It's probably not hard to show that it approaches a limit of 1 at both ends. But there is something to show there.
@John Martin The limit of 1/t as t goes to zero from the right is infinity so I am not sure how that helps. And the notation for the integral says 0 to 1 not anything about a limit.
a bit of nitpicking: the first argument of the base case technically implies that the integrand converges, which it doesn't, thus the integral may or may not converge which would be a required step, as infinity ^ 0 is indeterminant form which may or may not have a solution
The integral should be taken in the improper sense, that is as the integral from eps>0 to 1 with epsilon going to zero from above. For those values the integrand is well defined. So ln(1/t)^0 = 1
I just created a recurrence relation and used integration by parts. Iₙ = ∫ ln(1/t)^n dt Iₙ = ∫ 1 * ln(1/t)^n dt Iₙ = [t*ln(1/t)] (evaluated from 0 to 1) - ∫ -n(t)(1/t)(ln(1/t))^(n-1) dt Iₙ = ∫ n(t)(1/t)(ln(1/t))^(n-1) dt Iₙ = ∫ n(ln(1/t))^(n-1) dt Iₙ = n ∫ (ln(1/t))^(n-1) dt Iₙ = n * Iₙ₋₁ So, by observation, it's clear that this pattern continues down to I₀. I₀ = ∫ 1 dt = 1 (evaluated from 0 to 1) So, any Iₙ = n! For the record, any instance of an integral in this comment is designated to be a definite integral between 0 and 1. Unicode doesn't really allow for inline super and subscripts, so I had to approximate a well-written proof as best I could :)
@@rodrigomarinho1807 Yes, I know if (ln (1/t)) is anything other than infinity (in other words, when t is anything other than 0), then you're fine. But the integral is from 0 to 1. When t=0, (ln(1/t)) = infinity.
Michael, there were some audio issues in this video. The volume appears to go up and down significantly as you move closer to and farther from the microphone, or when you turn your head. There also appear to be deep reverberations transmitted to the microphone when you write on the board, particularly noticeable when you’re writing on the uppermost part of the board (although, personally, I find these deep reverberations to be somewhat pleasant, although momentarily distracting).
Such a great problem. I'm going to show my students when we return from holidays
We can Use fact ln(1/t) = -ln(t)
For simplicity
If you don't mind the (-1)^k factor that is. That is a mosquito in an integration problem.
Kind of weird how this implies that part of the function flips around the x-axis for n of different parity.
Though, since log(t) is negative between 0 and 1 it doesn't matter: the additional minus simply cancels out the negativity of the logarithm for odd n.
HA! I knew you'd leave that limit as homework.
Rewrite t(ln 1/t)^(k+1) as (ln 1/t)^(k+1) / (1/t)
Substitute u = 1/t. Now the limit goes to infinity
lim as u->infinity of (ln u)^(k+1) / u
Do L'Hopital's Rule
lim as u-> infinity of (k+1)(ln u)^k * 1/u in the numerator and 1 in the denominator = (k+1)(ln u)^k / u. Very similar to the original limit, just with an extra factor and a reduced exponent.
Keep applying this formula until the exponent is reduced to 1. You'll have the limit as u->inf of (k+1)! (ln u) / u. Apply L'Hopital one more time to get 0.
You should have a video on the Laplace transform. Often parts of the solutions to integral problems that you do reduce to Laplace transforms.
Can this serve as an alternative definition of gamma(z+1) ? Or is there something in complex analysis that would make the two integrals different?
The integral is equivalent to the traditional gamma(z+1) definition since it was just a substitution that transforms one into the other (x = ln(1/t) as in the 2nd part of the video).
I thought of that too . The Traditional Gamma function
I believe neither integral is defined for Re(z)
Amazing as always, thanks Michael!
What a great and clear problem!
Thank you, professor!
We can have Gamma function after substitution
or reduction after integration by parts
9:16 Re(a+1) > 0
Hi Michael. With this video you remained about one fascinating topic - fractional integration. Maybe some video/videos on this? Just a suggestion.
Just let I(t) = ∫₀¹ (1/x)^t dx = 1/(1-t).
Differentiating n times under the integral sign gives Dⁿ I(t) = ∫₀¹ (ln(1/x))^n (1/x)^t dx. Normal differentiation gives us Dⁿ I(t) = Dⁿ (1/(1-t)) = n! * 1/(1-t)^(n+1).
Setting t=0 gives us the result in the video:
Dⁿ I(0) = ∫₀¹ (ln(1/x))^n dx = n!
9:26 No homework today because I’m resting to celebrate my birthday! 🎂
Happy birthday
Happy Bday!!
happy birthday!
Happy birthday!!!!
happy birthday!
compi can verify too:
*Assuming[n \[Element] Integers \[And] n > 0, Integrate[Log[1/t]^n, {t, 0, 1}]] == n! // FullSimplify*
I like induction problems. They always seems "magic" !
Thank you
A very pleasant video to watch ! Thanks Michael !
Wow...I do love your sharing!
Great problem !!! Thank you !!!
I find this identity very interesting.
Using indirect differentiation under the integral,
Evaluating \int_{0}^1(lnt)^n dt
We know that \int_{0}^1x^m=1/(n+1)
Now
Differentiating the above results n times and setting m=0
We get
(-1)^n n!
Therefore
A=\int_{0}^1(lnt)^n dt=(-1)^n n!
Now back to the question:
\int_{0}^{1}(-lnt)^ndt=(-1)^n A
=n!
This identity's useful to me at the moment - thanks!
Just asking: Since 0! = 1 based on convention or definition, might it not be better to use n = 1 as the base case for the induction, since 1! really does equal 1, without invoking any special convention or definition?
In your base case, you run into problems at the endpoints of the interval. At the left endpoint, the expression inside the exponent goes to infinity and, at the right endpoint, it goes to 0: both (infinity)^0 and 0^0 are indeterminate forms.
It's probably not hard to show that it approaches a limit of 1 at both ends. But there is something to show there.
Outstanding!
All in all, it was just using Induction to prove or finding through the substitution method.
THAT'S REALLY EXCITING, LET'S PUT THAT IN A SHIRT!!!!!
Hi,
For fun:
0:05 : "our goal is to",
5:49 : "let's get rid of this".
Turns out it was very exciting!
On a scale from alpha, beta etc. I'd rate it as gamma.
Wow when I saw that substitution, the gamma/pi function really jumped out at me
yes so it is true for n € R. Interesting, if this alternative gamma function can be used for advanced stuff or extend that on Z, or futher stuff...
1/t at t=0 is undefined. So how can you evaluate at the 0 limit of the integral?
@John Martin The limit of 1/t as t goes to zero from the right is infinity so I am not sure how that helps. And the notation for the integral says 0 to 1 not anything about a limit.
@@charlesglidden557 noob
@@ojasdeshpande7296 Thank you for your informative replay
U-sub:
ln(1/t) = u => I = Integral [0..inf] (u^(n+1-1) * exp(-u) * du) = Г(n+1) = n!
Oh, its a second part of the video, sorry.
a bit of nitpicking: the first argument of the base case technically implies that the integrand converges, which it doesn't, thus the integral may or may not converge which would be a required step, as infinity ^ 0 is indeterminant form which may or may not have a solution
The integral should be taken in the improper sense, that is as the integral from eps>0 to 1 with epsilon going to zero from above. For those values the integrand is well defined. So ln(1/t)^0 = 1
@@firemaniac100 yes, you have to show that limit exists, before you can just replace it with 1.
@@Ensivion For every interval [a, 1] with 0
I just created a recurrence relation and used integration by parts.
Iₙ = ∫ ln(1/t)^n dt
Iₙ = ∫ 1 * ln(1/t)^n dt
Iₙ = [t*ln(1/t)] (evaluated from 0 to 1) - ∫ -n(t)(1/t)(ln(1/t))^(n-1) dt
Iₙ = ∫ n(t)(1/t)(ln(1/t))^(n-1) dt
Iₙ = ∫ n(ln(1/t))^(n-1) dt
Iₙ = n ∫ (ln(1/t))^(n-1) dt
Iₙ = n * Iₙ₋₁
So, by observation, it's clear that this pattern continues down to I₀.
I₀ = ∫ 1 dt = 1 (evaluated from 0 to 1)
So, any Iₙ = n!
For the record, any instance of an integral in this comment is designated to be a definite integral between 0 and 1. Unicode doesn't really allow for inline super and subscripts, so I had to approximate a well-written proof as best I could :)
Cool integral
this is gamma function!!!
Can we prove whether this is true for n not being intergers?
The RHS is gamma of n+1 instead of n!
Absolutely, but that would mean that you couldn't use induction :)
Of course
This is just another awesome representation of the gamma function.
To prove that, you will have to use
x=-lnt
@@jserleakabay7614 ok, thanks you.
Can you prove it why 0!=1 ?!
TIL This formula was found in 1730 by David Bernoulli.
I am sorry but 0! is not equal to 1
good
Differentiation under the integral sign makes proving this identity pretty simple too!
Hey, Michael, how can you casually claim that (ln (1/t))^0 is 1, when ln (1/0) is infinity?
N=0, not t
@@rodrigomarinho1807 Yes, I know if (ln (1/t)) is anything other than infinity (in other words, when t is anything other than 0), then you're fine. But the integral is from 0 to 1. When t=0, (ln(1/t)) = infinity.
Sweet
Beautiful
I solved it correctly in my imagination without watching the video in 2min
Michael, there were some audio issues in this video. The volume appears to go up and down significantly as you move closer to and farther from the microphone, or when you turn your head.
There also appear to be deep reverberations transmitted to the microphone when you write on the board, particularly noticeable when you’re writing on the uppermost part of the board (although, personally, I find these deep reverberations to be somewhat pleasant, although momentarily distracting).
i smell some gamma funtion hiding somewhere
Yeah ..... taking x=-lnt or
e^(-x)=t
You will get the integral representation of the Gamma function
gammmmmmmma my favorit
trivially.
very
evident gamma function for any Russian child )))
It's too easy
One of the basic formulae of gamma function
Yeet
Have a Great Day Everyone.
God Loves You and so do I.
Peace!
\o/