Another way: First set x = e^t. This transforms the integral to \int_0^\infty t^{2011} e^{-2010t} dt. Then set t = s/2010 to get 1/(2010^2012) \int_0^\infty s^{2011} e^{-s} ds. The integral \int_0^\infty s^{n} e^{-s} ds should be known to be n!; thus we end up with 2011!/(2010^2012).
Wow I love the way you notice your mistakes as you are solving. When I observe a mistake while you are solving, I know in my mind that you will figure it out
This’s just the Laplace transform of t^n, where t=ln(x) and n=2011. The answer is 2011!/(2010)^2012. Without referring to the Laplace transform table, we can use the integration by part, along with mathematical induction to get the general formula for n, which is n!/(n-1)^(n+1), thereby the given answer.
I think a more general way of expressing this technique is to define I_{a,b} as the integral of (ln x)^a x^-b over the interval. This way, you only need to integrate by parts once to see that I_{a,b} = (a/b) I_{a-1,b}.
An easy approach to follow is using Laplace transforms, substitute x=exp(t) so that integral transforms to integral 0 to ♾️, exp(-2010)t.t^2011dt which yields (2011)!/(2010)^2012 using Laplace transform of t^n as gamma(n+1)/s^n+1
The exposition of the idea is a bit chaotic. I would propose to consider Iₙ = ∫(lnx)ⁿdx/xᵐ⁺¹. Integration by parts (IBP stands for Initial-Boundary Problem rather than this) implies Iₙ= (n/m) Iₙ₋₁ with I₀=1/m. By induction, Iₙ = n!/mⁿ⁺¹. Finally, choose n=2011, m=2010.
13:45 for anyone confused by this, the x actually comes from the denominator of the ln(x) integration part (for du as shown on the left board), since x^-2010 = 1/x^2010, leaving the parts to be multiplied (to be multiplied by the denominator factor of x) is always nulls out the +1 to the exponent from the original anti-derivative to stay the same at 1/x^2011 (or x^-2011) no matter how many integrations you perform. i get what he meant though.
I solved it in the head in under 30 seconds. Here is my solution. 1. Put ln x = t. 2. Multiply and divide by x so that you get the term 1/x outside the expression in the power seperately to make your dx. 3. Since ln x was t, x = e^t. Also, the limts of integral would change from 1 to inf to 0 to inf. And we have our new integral - e^t(t/e^t)^2011. 4. Further simplification makes it e^-2010t * t^2011. With the limits from 0 to inf. 5. This integral looks familiar isnt it? Only if I had s inplace of 2010 and n inplace of 2011... so let me just replace it with those values. 6. So it's now integral from 0 to inf of e^-st * t^n dt. Bingo!!! That's the Laplace Transform of t^n and we know its formula is n!/s^(n+1). 7. What now? Just replace s with 2010 and n with 2011 and it's done! Done in the head without solving any integral or doing any calculations except for some substitutions! And that's how you solve a 19 min long solution in your head in less than a minute (in about 10-15 seconds to be precise if you are precise if you are in practice of solving mentally)
Could you please do the same interlgral with smaller powers, say from 1 to 10 as boundaries, and ((ln (x)/x))10. We need to see if the answer will also be the factorial. Thank you.
I tried to generalize this for the indefinite integral of any (ln(x)/x)^k and got something like this: - k!/(x^(k-1)) ( sum(n=0, k-1)[(ln(x))^(k-n)/((k-n)! (k-1)^(n+1))] + 1/((k-1)^(k+1)) ) +C Can someone verify this? Also I've put some values into this on a calculator and noticed that it only works for positive real values of k and becomes undefined or completely off in case of negative k, I'd like to know why this is the case.
Substitution gives Gamma function but we can derive reduction by parts and use it Do you want proposition for video Calculate d^n/dt^n (1/sqrt(1-2xt+t^2)) My observations d^n/dt^n (1/sqrt(1-2xt+t^2)) = \sum\limits_{k=0}^{\lfloor\frac{n}{2} floor}\frac{a_{n-2k}(x - t)^{n-2k}}{(1 - 2xt + t^2)^{\frac{1}{2} + n - k}} Coefficients a_{n-2k} are integers To calculate them calculate d^{n+2}/dt^{n+2} and rearrange terms in the sum
Here is a fun way. Let's just start by the function I(a) = Int[xᵃ, 1, inf] dx (a < -1) First step is to simply integrate this in terms of x. Everyone knows I(a) = xᵃ ⁺ ¹ / (a+1) | 1 -> inf, which is just -1 / (a+1) (please note that we are working under the assumption that a < -1, if a is not in this range then the integral does not exist) Here comes the fun part. We will use Leibniz Integral Rule by simply differentiating under the integral sign, and don't forget to differentiate the RHS as well: I'(a) = Int[ ∂/∂a(xᵃ), 1, inf ] dx = d/da [-1 / (a+1)] = Int[ xᵃ * ln(x), 1, inf] dx = 1 / (a+1)² Keep differentiating both sides! I"(a) = Int[ ∂/∂a [xᵃ * ln(x)], 1, inf ] dx = d/da [1 / (a+1)²] = Int[ xᵃ * ln²(x), 1, inf] dx = -2 / (a+1)³ I'''(a) = Int[ ∂/∂a [xᵃ * ln²(x)], 1, inf ] dx = d/da [1 / (a+1)²] = Int[ xᵃ * ln³(x), 1, inf] dx = 3! / (a+1)⁴ From now, I think you already see a pattern here, so let's just differentiate until the n-th degree. I'ⁿ(a) = Int[ xᵃ * lnⁿ(x), 1, inf] dx = (-1)ⁿ ⁺ ¹.n! / (a+1)ⁿ ⁺ ¹ We want to calculate the integral when a = -2011, and n = 2011, so substitute: Int[ x ⁻ ² ⁰ ¹ ¹ * ln² ⁰ ¹ ¹(x), 1, inf] dx = (-1)² ⁰ ¹ ².2011! / (-2011 + 1)² ⁰ ¹ ² = 2011! / 2010² ⁰ ¹ ².
The substitution u=lnx works here perfectly. It transforms the integral to ∫₀^∞ u²⁰¹¹exp(-2010u)du. Now another substitution 2010u=s transforms it to 2010⁻²⁰¹²× ∫₀^∞ s²⁰¹¹exp(-s)ds. The latter integral is Γ(2012)=2011!. Hence the answer: 2011!2010⁻²⁰¹². It might be interesting to estimate this value, which is but a huge number divided by another huge number. For instance, by the Stirling formula: n! ≈ √(2πn )(n/e)ⁿ. Or nⁿ≈ n! eⁿ/√(2πn). Its application with n=2010 gives us an estimate of the integral as ( √(4020π )2011! ) / (2010! 2010²e²⁰¹⁰) ≈ √(2π )/(√(2010) e²⁰¹⁰). Thus, the value of the integral is very small. This is because ln(x)/x is a small number for x>1. It doesn't exceed 1/e (achieved at x=e). So (ln(x)/x)ⁿ vanishes as n→∞, and quite rapidly.
@@sobolzeev sir, I like that and You. May I know whether You use any specific application or extension on your mobile phone to write mathematical symbols or You work on computer?
@@alphazero339 It was written from a Pixel phone, with a use of Math Keyboard. It is not 100% good for writing Maths. However, it contains some symbols and a wide range of subscript and superscript symbols in addition to Google keyboard, also rich in symbols.
Hi prime, I'm kind of confusing from the 2nd board - third line int lnx²⁰⁰⁹ is not where to be found, I need to inquire, can you pls explain this because it looks more challenging to me. Thx
Because it is an improper definite integral. Look on the internet the definition of definite integral, indefinite integral and improper integral. You will totally understand!😊
Can you make some videos on integrals (sostitution and integration by part) or recommend me some other videos to watch pls. Because i have some pretty basic knoledges of this two methods but i did not understand much.
I used u sub u = ln(x) then noticed the integral was in the form of a 'laplace transform' I = integral from 1 to infty of (lnx/x)^2011 dx u = lnx => x = e^u => x^2010 = (e^u)^2010 = e^(2010u) du/dx = 1/x at x = 1, u = 0 as x approaches infty, u approaches infty too I = integral from 1 to infty of (lnx)^2011 / x^2011 dx I = integral from 1 to infty of ((lnx)^2011 / x^2010)(1/x) dx I = integral from 0 to infty of (u^2011 / e^(2010u) du since u is a dummy variable and we are evaluating a definite integral, let u = t I = integral from 0 to infty of (u^2011)e^(-2010u) du I = integral from 0 to infty of (t^2011)e^((-2010)t) dt Note L{t^n} laplace transform of t^n = integral from 0 to infty of (t^n)e^(-st) dt = n!/s^(n+1) where s > 0 So our original integral I = L{t^2011} | _ s = 2010 Therefore I = 2011!/2010^(2011+1) = 2011!/2010^2012 (Scrolling down I can see someone did a similar method)
Another way: First set x = e^t. This transforms the integral to \int_0^\infty t^{2011} e^{-2010t} dt. Then set t = s/2010 to get 1/(2010^2012) \int_0^\infty s^{2011} e^{-s} ds. The integral \int_0^\infty s^{n} e^{-s} ds should be known to be n!; thus we end up with 2011!/(2010^2012).
Great!
Right, I remember that back in the college, this is one of 2 forms of Gamma function
Решение намного короче авторского. Видно,что обладатель бейсболки далеко не всегда находит лучшие варианты.
Great content. Please go on with this kind of math contest content, that's interesting.
Wow I love the way you notice your mistakes as you are solving. When I observe a mistake while you are solving, I know in my mind that you will figure it out
So nice of you
We are usually advised to choose U as a function which can be differentiated repeatedly until we get a constant interger
with you math is not bored anymore!
This’s just the Laplace transform of t^n, where t=ln(x) and n=2011. The answer is 2011!/(2010)^2012. Without referring to the Laplace transform table, we can use the integration by part, along with mathematical induction to get the general formula for n, which is n!/(n-1)^(n+1), thereby the given answer.
I think a more general way of expressing this technique is to define I_{a,b} as the integral of (ln x)^a x^-b over the interval. This way, you only need to integrate by parts once to see that I_{a,b} = (a/b) I_{a-1,b}.
An easy approach to follow is using Laplace transforms, substitute x=exp(t) so that integral transforms to integral 0 to ♾️, exp(-2010)t.t^2011dt which yields (2011)!/(2010)^2012 using Laplace transform of t^n as gamma(n+1)/s^n+1
How do you deal with the lower bound using this approach? Change of variables?
The exposition of the idea is a bit chaotic. I would propose to consider
Iₙ = ∫(lnx)ⁿdx/xᵐ⁺¹. Integration by parts (IBP stands for Initial-Boundary Problem rather than this) implies
Iₙ= (n/m) Iₙ₋₁
with I₀=1/m. By induction,
Iₙ = n!/mⁿ⁺¹.
Finally, choose n=2011, m=2010.
13:45 for anyone confused by this, the x actually comes from the denominator of the ln(x) integration part (for du as shown on the left board), since x^-2010 = 1/x^2010, leaving the parts to be multiplied (to be multiplied by the denominator factor of x) is always nulls out the +1 to the exponent from the original anti-derivative to stay the same at 1/x^2011 (or x^-2011) no matter how many integrations you perform. i get what he meant though.
That's what he said...
@@JavedAlam24 huh don't remember this video. i guess at the time i maybe found it slightly confusing; the way it was originally explained.
1. I love your cap 2. Your writing and your explanation are perfect 3. You convinced me...your gran(gran(gran(.....(.granpa is NEWTON.
🤣🤣🤣🤣
Nice problem. Thanks for the video.
This would be a good integral to practice proof by induction.
I solved it in the head in under 30 seconds. Here is my solution.
1. Put ln x = t.
2. Multiply and divide by x so that you get the term 1/x outside the expression in the power seperately to make your dx.
3. Since ln x was t, x = e^t. Also, the limts of integral would change from 1 to inf to 0 to inf. And we have our new integral - e^t(t/e^t)^2011.
4. Further simplification makes it e^-2010t * t^2011. With the limits from 0 to inf.
5. This integral looks familiar isnt it? Only if I had s inplace of 2010 and n inplace of 2011... so let me just replace it with those values.
6. So it's now integral from 0 to inf of e^-st * t^n dt. Bingo!!! That's the Laplace Transform of t^n and we know its formula is n!/s^(n+1).
7. What now? Just replace s with 2010 and n with 2011 and it's done! Done in the head without solving any integral or doing any calculations except for some substitutions! And that's how you solve a 19 min long solution in your head in less than a minute (in about 10-15 seconds to be precise if you are precise if you are in practice of solving mentally)
Could you please do the same interlgral with smaller powers, say from 1 to 10 as boundaries, and ((ln (x)/x))10. We need to see if the answer will also be the factorial. Thank you.
I tried to generalize this for the indefinite integral of any (ln(x)/x)^k and got something like this:
- k!/(x^(k-1)) ( sum(n=0, k-1)[(ln(x))^(k-n)/((k-n)! (k-1)^(n+1))] + 1/((k-1)^(k+1)) ) +C
Can someone verify this? Also I've put some values into this on a calculator and noticed that it only works for positive real values of k and becomes undefined or completely off in case of negative k, I'd like to know why this is the case.
Substitution gives Gamma function
but we can derive reduction by parts and use it
Do you want proposition for video
Calculate d^n/dt^n (1/sqrt(1-2xt+t^2))
My observations
d^n/dt^n (1/sqrt(1-2xt+t^2)) = \sum\limits_{k=0}^{\lfloor\frac{n}{2}
floor}\frac{a_{n-2k}(x - t)^{n-2k}}{(1 - 2xt + t^2)^{\frac{1}{2} + n - k}}
Coefficients a_{n-2k} are integers
To calculate them calculate
d^{n+2}/dt^{n+2} and rearrange terms in the sum
This problem unlike other problems require creative mind to be used nice problem
Here is a fun way. Let's just start by the function I(a) = Int[xᵃ, 1, inf] dx (a < -1)
First step is to simply integrate this in terms of x. Everyone knows I(a) = xᵃ ⁺ ¹ / (a+1) | 1 -> inf, which is just -1 / (a+1) (please note that we are working under the assumption that a < -1, if a is not in this range then the integral does not exist)
Here comes the fun part. We will use Leibniz Integral Rule by simply differentiating under the integral sign, and don't forget to differentiate the RHS as well:
I'(a) = Int[ ∂/∂a(xᵃ), 1, inf ] dx = d/da [-1 / (a+1)]
= Int[ xᵃ * ln(x), 1, inf] dx = 1 / (a+1)²
Keep differentiating both sides!
I"(a) = Int[ ∂/∂a [xᵃ * ln(x)], 1, inf ] dx = d/da [1 / (a+1)²]
= Int[ xᵃ * ln²(x), 1, inf] dx = -2 / (a+1)³
I'''(a) = Int[ ∂/∂a [xᵃ * ln²(x)], 1, inf ] dx = d/da [1 / (a+1)²]
= Int[ xᵃ * ln³(x), 1, inf] dx = 3! / (a+1)⁴
From now, I think you already see a pattern here, so let's just differentiate until the n-th degree.
I'ⁿ(a) = Int[ xᵃ * lnⁿ(x), 1, inf] dx = (-1)ⁿ ⁺ ¹.n! / (a+1)ⁿ ⁺ ¹
We want to calculate the integral when a = -2011, and n = 2011, so substitute:
Int[ x ⁻ ² ⁰ ¹ ¹ * ln² ⁰ ¹ ¹(x), 1, inf] dx = (-1)² ⁰ ¹ ².2011! / (-2011 + 1)² ⁰ ¹ ² = 2011! / 2010² ⁰ ¹ ².
The substitution u=lnx works here perfectly. It transforms the integral to
∫₀^∞ u²⁰¹¹exp(-2010u)du. Now another substitution 2010u=s transforms it to
2010⁻²⁰¹²×
∫₀^∞ s²⁰¹¹exp(-s)ds.
The latter integral is Γ(2012)=2011!. Hence the answer: 2011!2010⁻²⁰¹².
It might be interesting to estimate this value, which is but a huge number divided by another huge number. For instance, by the Stirling formula:
n! ≈ √(2πn )(n/e)ⁿ.
Or nⁿ≈ n! eⁿ/√(2πn).
Its application with n=2010 gives us an estimate of the integral as
( √(4020π )2011! ) /
(2010! 2010²e²⁰¹⁰)
≈
√(2π )/(√(2010) e²⁰¹⁰).
Thus, the value of the integral is very small. This is because ln(x)/x is a small number for x>1. It doesn't exceed 1/e (achieved at x=e). So (ln(x)/x)ⁿ vanishes as n→∞, and quite rapidly.
@@sobolzeev sir, I like that and You. May I know whether You use any specific application or extension on your mobile phone to write mathematical symbols or You work on computer?
@@alphazero339 It was written from a Pixel phone, with a use of Math Keyboard. It is not 100% good for writing Maths. However, it contains some symbols and a wide range of subscript and superscript symbols in addition to Google keyboard, also rich in symbols.
∫ x²
Hi prime, I'm kind of confusing from the 2nd board - third line int lnx²⁰⁰⁹ is not where to be found, I need to inquire, can you pls explain this because it looks more challenging to me. Thx
Woooooow
You forgot to change the limits of integration while taking the minus out after IBP
wouls this integral be a good candidate for Feinman's method?
Yes but you have to take the derivative 2011 times, take the derivative 3 or 4 times and the pattern will emerge
That chalk is so nice
Sir can you explain why there is no +c at the end?
Because it is an improper definite integral. Look on the internet the definition of definite integral, indefinite integral and improper integral. You will totally understand!😊
It describes the area of a specific function - the function on the inside for 1≤x≤∞. There's only one area, so we can only expect one answer.
Best approach is using gamma function
An even better one is using Laplace Transform
Your investigation leads to a proper solution but I feel that you miss an inductive proof.
Can you make some videos on integrals (sostitution and integration by part) or recommend me some other videos to watch pls. Because i have some pretty basic knoledges of this two methods but i did not understand much.
+ i agree, would be nice to have some videos explaining concepts, not just using them and assuming the person watching knows them
@@tai0fps If you need to brush up on basic integration techniques Im sure there are hundreds of example problems on youtube, just do a quick search
I have made more videos on ALL integration techniques than any other topic on this channel. Just search. I also assume you are new to my channel.
@@PrimeNewtons thank u
Sir you are great 😊
I tried it first by own self and came to watch the vdo my ans matches you
2010-2011-2012
thank u sir
Wow!
Con Laplace si fa prima
Plzz Give me a hi sir ❤
u=ln(x)
du=dx/x
dx=e^u•du
I=int[0,♾️](u^2011•e^-2010u)du
p=2010u
dp=2010du
I=2010^-2012•int[0,♾️](p^2011•e^-p)dp
I=2011!/(2010^2012)
=2011!/((2010)²⁰¹²)
I just used gamma function
he forgot to write down dx
First pls pin
I used u sub u = ln(x)
then noticed the integral was in the form of a 'laplace transform'
I = integral from 1 to infty of (lnx/x)^2011 dx
u = lnx => x = e^u => x^2010 = (e^u)^2010 = e^(2010u)
du/dx = 1/x
at x = 1, u = 0
as x approaches infty, u approaches infty too
I = integral from 1 to infty of (lnx)^2011 / x^2011 dx
I = integral from 1 to infty of ((lnx)^2011 / x^2010)(1/x) dx
I = integral from 0 to infty of (u^2011 / e^(2010u) du
since u is a dummy variable and we are evaluating a definite integral, let u = t
I = integral from 0 to infty of (u^2011)e^(-2010u) du
I = integral from 0 to infty of (t^2011)e^((-2010)t) dt
Note L{t^n} laplace transform of t^n = integral from 0 to infty of (t^n)e^(-st) dt = n!/s^(n+1) where s > 0
So our original integral I = L{t^2011} | _ s = 2010
Therefore I = 2011!/2010^(2011+1) = 2011!/2010^2012
(Scrolling down I can see someone did a similar method)