Another way to solve it: let F(a) to be the integral of (1-e^(-ax)/x)^2 (on ]0,+infinity[) so that the integral we want to evaluate is F(1). F is differentiable on ]0,+infinity[ and F'(a) = 2 ln(2) (it is a Frullani integral), so that F(a)=2ln(2) a + C. Take the limit when a goes to 0+ to get F(a)=2ln(2)a : F is in fact linear ! And we get 2ln(2) for F(1).
Integration by parts two times leads to int 2ln(x)(exp(-x) - 2exp(-2x))dx, in second term we make replacement t = 2x, so int 2ln(x)exp(-x)dx - 2 int ln(t/2) exp(-t))dt. The log part cancels, and we get elementary integral 2ln(2)int exp(-x)dx = 2ln(2)
The integral is the limit as epsilon goes to 0+ of the integrand between epsilon to infinity. Then, developing the square we have the integrals of 1/x^2, minus twice the integral of e^(-x)/x^2 plus the integral of e^(-2x)/x^2. Taking in the third integral the change y=x/2, the last two integrals almost cancel each other exceot fior the integral from epsilon to 2(epsilon) of e^(-x)/x^2. Finally using a Taylor expansionj with three terms for e^(-x) we obtain that the integral is 2Ln2 plus an error term of order epsilon and we are done.
I would start by rewriting ((1-exp(-x))/x)^2 as (1-exp(-x))^2/x^2 and then by integration by parts Then substitution u = exp(-x) leads us to the integral 2*Int((u-1)/ln(u),u=0..1) From this moment it is easy to use Leibniz rule for integration or to use double integral Int((u-1)/ln(u),u=0..1) via Leibniz rule for integration Let I(t) = Int((u^t - 1)/ln(u),u=0..1) I'(t) = Int(u^{t},u=0..1) I'(t) = 1/(t+1) I(t) = ln(t+1)+C I(0) = 0 I(t) = ln(t+1) I(1) = ln(2) so result is 2ln(2)
Integral of ln(z+a) = (z+a)ln(z+a) - z, NOT (z+a)ln(z+a) - (z+a), check the by parts calculations Afterwards the z cancel out and you're left with just the logs, good thing the 1 cancels out by itself with the limits
I really want to use a vanishing contour integral at infinity but I can't find a contour that would work. It's hard because f(z) only has a removable singularity and is not even.
Bit worried about your application of Fubini's Thm 3 mins in. Looks better if you work backwards. Start with: double int [1, 1] exp (x(y+z)) dy dz Now write the above as an iterated integral in which we're first int wrt y and then z = int [0, 1] { int [0, 1] exp (x(y+z)) dy } dz Factor the exponential, and take the factor exp xz outside the y int (treating it as a const.) = int [0, 1] { int [0, 1] exp xy dy } exp xz dz Now factor the whole y-int out of the z integral, treating _that_ as a const. = { int [0, 1] exp xy dy }. { int [0, 1] exp xz dz } which is what we want
Beautiful! I love this kind of problems. I learned some unexpected tricks! Thank you so much!
either "adding zero" or "multiplying by one"
Found the logician
every time lol
Another way to solve it: let F(a) to be the integral of (1-e^(-ax)/x)^2 (on ]0,+infinity[) so that the integral we want to evaluate is F(1).
F is differentiable on ]0,+infinity[ and F'(a) = 2 ln(2) (it is a Frullani integral), so that F(a)=2ln(2) a + C.
Take the limit when a goes to 0+ to get F(a)=2ln(2)a : F is in fact linear ! And we get 2ln(2) for F(1).
integration by parts + Frullani, so no need to introduce parameter.
I think that in order to prove Frullani, you have to do something somewhat similar to what Michael does in the video.
@@DanielPoupko Right, one way to compute Frullani integrals is to use Fubini, which is what Michael did.
Integration by parts two times leads to int 2ln(x)(exp(-x) - 2exp(-2x))dx, in second term we make replacement t = 2x, so int 2ln(x)exp(-x)dx - 2 int ln(t/2) exp(-t))dt. The log part cancels, and we get elementary integral 2ln(2)int exp(-x)dx = 2ln(2)
2*ln(2) = ln(4)
You can do integration by parts and obtain integral of 2 (exp(-x) - exp(-2x))/x which is Frullani integral equal to 2*exp(-0)*ln(2/1) = 2ln2
What was stopping us from just doing the power series of e^-x and integrating through that?
The integral is the limit as epsilon goes to 0+ of the integrand between epsilon to infinity. Then, developing the square we have the integrals of 1/x^2, minus twice the integral of e^(-x)/x^2 plus the integral of e^(-2x)/x^2. Taking in the third integral the change y=x/2, the last two integrals almost cancel each other exceot fior the integral from epsilon to 2(epsilon) of e^(-x)/x^2. Finally using a Taylor expansionj with three terms for e^(-x) we obtain that the integral is 2Ln2 plus an error term of order epsilon and we are done.
++, I've also used this method)
Maybe combine the logs before applying integration by parts
I would start by rewriting ((1-exp(-x))/x)^2 as (1-exp(-x))^2/x^2 and then by integration by parts
Then substitution u = exp(-x) leads us to the integral
2*Int((u-1)/ln(u),u=0..1)
From this moment it is easy to use Leibniz rule for integration
or to use double integral
Int((u-1)/ln(u),u=0..1)
via Leibniz rule for integration
Let I(t) = Int((u^t - 1)/ln(u),u=0..1)
I'(t) = Int(u^{t},u=0..1)
I'(t) = 1/(t+1)
I(t) = ln(t+1)+C
I(0) = 0
I(t) = ln(t+1)
I(1) = ln(2)
so result is 2ln(2)
Alernatively after integration by parts we can calculate Laplace transform of (1-exp(-t))/t and plug in s=1
Frullani was here.
Taking the natural log a lot
Ima guess a series expansion
Never get tired of this: " ... and that's a good place to stop!! " 😄
woot woot \o/\o/
Integral of ln(z+a) = (z+a)ln(z+a) - z, NOT (z+a)ln(z+a) - (z+a), check the by parts calculations
Afterwards the z cancel out and you're left with just the logs, good thing the 1 cancels out by itself with the limits
you can use Feinman's technique, actually, I was suprised you didn't
This is what I was thinking
my dumbass used ramanujan's master theorem lol (with taking a limit because Γ(z) has a pole in z = -1)
It went from dx to dz dy dx then to 2 ln2 too too incredible too? 🙂
I thought maybe Feynman's technique or Laplace transforms
Whyb use x y and z as triple integrsl..why not just x and y and then convert to polar coordinated maybe? I dont see any advantage..
I really want to use a vanishing contour integral at infinity but I can't find a contour that would work. It's hard because f(z) only has a removable singularity and is not even.
Fubini's approach should have had been presented.
I was waiting to add zero or multiply by one in a particularly clever way. Oh well... 🤷♀
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in the integration in x, what happens if y+z=0?
it only happens in one point y = z = 0, so it will not affect the answer. look at it as an improper integral
@@lagnugg Sure. but it was worth mentioning. At least.
@@harrymattah418 fair, but i understand why the author omitted the explanation
Bit worried about your application of Fubini's Thm 3 mins in.
Looks better if you work backwards. Start with:
double int [1, 1] exp (x(y+z)) dy dz
Now write the above as an iterated integral in which we're first int wrt y and then z
= int [0, 1] { int [0, 1] exp (x(y+z)) dy } dz
Factor the exponential, and take the factor exp xz outside the y int (treating it as a const.)
= int [0, 1] { int [0, 1] exp xy dy } exp xz dz
Now factor the whole y-int out of the z integral, treating _that_ as a const.
= { int [0, 1] exp xy dy }. { int [0, 1] exp xz dz }
which is what we want
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asnwer= 1xd
The trick is to skip to the end of the video, perhaps? That trick works for most integrals that are presented like this...
Guess he is gonna use x = 1/x and then use the fact that expressed via trig functions that’s pretty advanced tho