x^s/(e^x - 1) = x^s * e^-x / (1 - e^-x) to get a/1-r form where r=e^-x and a=x^s*e^-x so is sum_{n=1}^{inf} x^s * e ^{-nx} and the integral becomes the sum {n=1}^{inf} of the integral between 0 and inf of x^s * e ^{-nx} dx . Now these integrals are obviously very close to the gamma function. That's where I cheated and used Wolfram to get an expression in terms of the gamma function. I won't write the answer then because it may be a spoiler for anyone who actually wants to work it out.
A video just popped up on my feed which by coincidence is pretty much the identical problem (just starting from the other side) .. Black pen red pen : Zeta function in terms of Gamma function. I prefer starting from the integral though it's really pretty how everything drops out in the summation.
Substitution u = cos(x)^2 Geometric seris expansion Exchanging order of summation and integration Integration by parts Can be reduced to Basel problem in elementary way, without Zeta function
That's a really cool integral ! It reminds me of a Michael Penn video from late April 2023, where the title said something about pi, but there wasn't any discussion about where the pi came from, but the first few steps you introduced are brilliant!
Okay! All of these are great for STEP preparation but I'd recommend you take a look at my most recent video because recursive integrals of that sort come up all the time.
Very nice trick with that geometric series there. Funnily enough I actually have a video which answers the challenge called "Bose Integral" back when I had my small whiteboard and my parents were still filming my videos lmao. Also 1k subs incoming soon?!
Challenge problem : let's take e^x common and try to use the same gp idea integral(x^5 e^-x (1 + e^-x + e^-2x + e^-3x + ...) dx) from 0 to infinity = integral(x^5 sum(e^-nx) dx) n from 1 to infinity, integral from 0 to infinity = sum (integral(x^5 e^-nx) dx) from 0 to infinity) n from 1 to infinity When we integrate this by parts, we'll get both power of x varying and power so n showing up in the denominator, so to avoid that creating t^5 e^-t dt will be a better bet nx = t dx = dt/n x^5 = t^5/n^5 sum(integral(t^5 e^-t /n^6 dt) = sum(1/n^6 integral(t^5 e^-t dt)) integral(t^5 e^-t dt) : D I + t^5 e^-t - 5t^4 -e^-t + 20t^3 e^-t - 60t^2 -e^-t + 120t e^-t - 120 -e^-t + 0 +e^-t Because the limits are 0 and infinity, any term having both t and e^-t will invariably give 0. So the integral will simply collapse down to -120 e^-t (limits from 0 to infinity). Plugging in infinity this will give 0, so the answer is 120 (from plugging in 0). back to the summation : sum(120/n^6) [from n = 1 to infinity] = 120 zeta(6) btw I absolutely loved the video, the summation step is pure genius!
My proof for the challenge problem will use a bit more rigor than some may be used to, namely a bit of introductory special function theory and some measure theory (I'm sure that there are other methods, this is just the first that came to my mind). This is mainly for didactic reasons, but I also wish to lay a firm foundation for the bound on the main result. Another commenter multiplied both the numerator and denominator by e⁻ˣ and used the geometric series to turn the target integral into an integral of sums; however, they stopped short of justifying the interchange of summation and integration and did not give the answer. I will take a similar approach, using a substitution to make the crux of my rigorous argument for interchange more readily apparent. Our goal is to evaluate I(s)=∫₀ ᪲xˢ/(eˣ-1)dx. (1) We start by making the substitution x=ln(1/u), so we have dx=-1/u du, e⁻ˣ=u, and 1/(eˣ-1)dx=-1/(1-u)du. For the bounds of integration, we obtain u=0 as x approaches infinity and u=1 as x approaches 0. Therefore, we have ∫₀ ᪲xˢ/(eˣ-1)dx=∫₀¹lnˢ(1/u)/(1-u)du. (2) Since |u|0 (k=0 coincides with an alternate definition for the gamma function, Γ(s)=∫₀¹lnˢ(1/t)dt for Re(s)>-1, and thus still causes no issue); however, it is trivial to show that lim_(u→1⁻) (uᵏlnˢ(1/u))=0 for all k∈ℂ and Re(s)>0. For s along the imaginary axis, we have ∫₀¹Σ_(k≥0) |uᵏ(ln(1/u))ᵇⁱ|du=∫₀¹Σ_(k≥0) uᵏ|(ln(1/u))ⁱ|ᵇdu =∫₀¹Σ_(k≥0) uᵏ·1ᵇdu=∫₀¹1/(1-u)du. (3.1) This diverges, so (3) is not Lebesgue integrable for s with Re(s)=0 (furthermore, (2) diverges). Next, for Re(s)cu for all u∈(0,1). With this, we can write ∫₀¹Σ_(k≥0) |uᵏlnˢ(1/u)|du=∫₀¹Σ_(k≥0) uᵏ(ln(1/u))ᴿᵉ⁽ˢ⁾du >c∫₀¹Σ_(k≥0) uᵏ⁺¹du=c∫₀¹u/(1-u)du. (3.2) This also diverges; combining this with our results from (3.1), (3) is not Lebesgue integrable for any s with Re(s)≤0, and (2) diverges for the same s. Since our integral is well-behaved for all other s and has no other singularities within our domain of (0,1), it follows that it is bounded for any s such that Re(s)>0. Therefore, I(s) is Lebesgue integrable on the product measure of the counting measure on ℕ₀ and the Lebesgue measure on (0,1) if and only if Re(s)>0. By Fubini's theorem, we can freely swap integration and summation, and we have the following: I(s)=Σ_(k≥0) ∫₀¹uᵏ(ln(1/u))ˢdu. (4) To proceed, we can introduce the upper incomplete gamma function (a generalization of the usual gamma function Γ(s):=∫₀ ᪲tˢ⁻¹e⁻ᵗdt for Re(s)>0), defined as Γ(s,x):=∫ₓ ᪲tˢ⁻¹e⁻ᵗdt for complex s and x with positive real parts. By the fundamental theorem of calculus, ∂/∂x Γ(s,x)=-xˢ⁻¹e⁻ˣ. Examining this derivative and the new form of I(s) in (4), we can deduce I(s)=Σ_(k≥0) (k+1)⁻ˢ⁻¹[Γ(s+1,(k+1)ln(1/u))]₀¹ =Σ_(k≥1) k⁻ˢ⁻¹[Γ(s+1,0)-lim_(v→∞) Γ(s+1,v)], (5) where v=k ln(1/u). By the definitions given above, we have Γ(s+1,0)=Γ(s+1) and lim_(v→∞) Γ(s+1,v)=0. Therefore, we can conclude ∫₀ ᪲xˢ/(eˣ-1)dx=Γ(s+1)Σ_(k≥1) k⁻ˢ⁻¹=Γ(s+1)ζ(s+1). □ This result holds for any complex s satisfying Re(s)>0, as shown above.
@@OscgrMaths Thank you! Loved working on this one. We take a lot of mathematical results for granted, but it's super instructive to look at them and figure out exactly how and why they work.
Yes! For the challenge you might recognise from the bose einstein equation that it's equal to zeta (s+1) gamma(s+1). Which one did you get an answer of 8pi^6/63 for?
@@OscgrMaths Yes! For the challenge, I used the specific case of the Bose-Einstein integral where s = 5, the integral becomes gamma(6)zeta(6) and I estimated zeta(6) by applying the formula involving Bernoulli numbers, which combined with gamma(6) = 5! = 120, leads to the result of 8pi^6/63! Also, thanks a lot for your high-quality videos! Keep up the good work:)
Don't forget... the great thing about maths is that if you can prove it then you've proven something that is always true!! Unlike art or science which are changing and developing, maths may become more and more complex - but something true won't ever be untrue. So I guess you're discovering universal truths... But if that's not enough it's also extremely useful in everything (physics, engineering, economics etc...)
“The scientist does not study nature because it is useful to do so. He studies it because he takes pleasure in it, and he takes pleasure in it because it is beautiful." ― Henri Poincaré, Science and Method
Don't forget to try the challenge problem at the end and comment your answers!
x^s/(e^x - 1) = x^s * e^-x / (1 - e^-x) to get a/1-r form where r=e^-x and a=x^s*e^-x so is sum_{n=1}^{inf} x^s * e ^{-nx} and the integral becomes the sum {n=1}^{inf} of the integral between 0 and inf of x^s * e ^{-nx} dx . Now these integrals are obviously very close to the gamma function. That's where I cheated and used Wolfram to get an expression in terms of the gamma function. I won't write the answer then because it may be a spoiler for anyone who actually wants to work it out.
You should mention why you're allowed to swap the integral / summation. It would help people learn.
Gamma (s+1)×zeta (s+1) is the answer for your last integral
@@DihinAmarasigha-up5hf Perfect!! How did you do it?
A video just popped up on my feed which by coincidence is pretty much the identical problem (just starting from the other side) .. Black pen red pen : Zeta function in terms of Gamma function. I prefer starting from the integral though it's really pretty how everything drops out in the summation.
The summation step is genius. I am loving your videos!
Thanks so much for the comment!! I loved that step too, it's things like that that make the integral worth doing for me.
Substitution u = cos(x)^2
Geometric seris expansion
Exchanging order of summation and integration
Integration by parts
Can be reduced to Basel problem in elementary way, without Zeta function
That's a really cool integral ! It reminds me of a Michael Penn video from late April 2023, where the title said something about pi, but there wasn't any discussion about where the pi came from, but the first few steps you introduced are brilliant!
Thanks so much! That sounds great, I'll try and find that video and check it out.
Amazing video,pls make more videos like this and also please do some step integrals please
Okay! All of these are great for STEP preparation but I'd recommend you take a look at my most recent video because recursive integrals of that sort come up all the time.
Very nice trick with that geometric series there. Funnily enough I actually have a video which answers the challenge called "Bose Integral" back when I had my small whiteboard and my parents were still filming my videos lmao.
Also 1k subs incoming soon?!
Nice!! Bose integral is so good. Hopefully got the 1k coming soon 🤞🤞🤞.
really enjoyable and interesting 🥰🥰keep going 😘
Thanks so much!! Really appreciate the comment.
Challenge problem :
let's take e^x common and try to use the same gp idea
integral(x^5 e^-x (1 + e^-x + e^-2x + e^-3x + ...) dx) from 0 to infinity
= integral(x^5 sum(e^-nx) dx) n from 1 to infinity, integral from 0 to infinity
= sum (integral(x^5 e^-nx) dx) from 0 to infinity) n from 1 to infinity
When we integrate this by parts, we'll get both power of x varying and power so n showing up in the denominator, so to avoid that creating t^5 e^-t dt will be a better bet
nx = t
dx = dt/n
x^5 = t^5/n^5
sum(integral(t^5 e^-t /n^6 dt) = sum(1/n^6 integral(t^5 e^-t dt))
integral(t^5 e^-t dt) :
D I
+ t^5 e^-t
- 5t^4 -e^-t
+ 20t^3 e^-t
- 60t^2 -e^-t
+ 120t e^-t
- 120 -e^-t
+ 0 +e^-t
Because the limits are 0 and infinity, any term having both t and e^-t will invariably give 0. So the integral will simply collapse down to -120 e^-t (limits from 0 to infinity). Plugging in infinity this will give 0, so the answer is 120 (from plugging in 0).
back to the summation :
sum(120/n^6) [from n = 1 to infinity] = 120 zeta(6)
btw I absolutely loved the video, the summation step is pure genius!
My proof for the challenge problem will use a bit more rigor than some may be used to, namely a bit of introductory special function theory and some measure theory (I'm sure that there are other methods, this is just the first that came to my mind). This is mainly for didactic reasons, but I also wish to lay a firm foundation for the bound on the main result. Another commenter multiplied both the numerator and denominator by e⁻ˣ and used the geometric series to turn the target integral into an integral of sums; however, they stopped short of justifying the interchange of summation and integration and did not give the answer. I will take a similar approach, using a substitution to make the crux of my rigorous argument for interchange more readily apparent. Our goal is to evaluate
I(s)=∫₀ ᪲xˢ/(eˣ-1)dx. (1)
We start by making the substitution x=ln(1/u), so we have dx=-1/u du, e⁻ˣ=u, and 1/(eˣ-1)dx=-1/(1-u)du. For the bounds of integration, we obtain u=0 as x approaches infinity and u=1 as x approaches 0. Therefore, we have
∫₀ ᪲xˢ/(eˣ-1)dx=∫₀¹lnˢ(1/u)/(1-u)du. (2)
Since |u|0 (k=0 coincides with an alternate definition for the gamma function, Γ(s)=∫₀¹lnˢ(1/t)dt for Re(s)>-1, and thus still causes no issue); however, it is trivial to show that lim_(u→1⁻) (uᵏlnˢ(1/u))=0 for all k∈ℂ and Re(s)>0. For s along the imaginary axis, we have
∫₀¹Σ_(k≥0) |uᵏ(ln(1/u))ᵇⁱ|du=∫₀¹Σ_(k≥0) uᵏ|(ln(1/u))ⁱ|ᵇdu
=∫₀¹Σ_(k≥0) uᵏ·1ᵇdu=∫₀¹1/(1-u)du. (3.1)
This diverges, so (3) is not Lebesgue integrable for s with Re(s)=0 (furthermore, (2) diverges). Next, for Re(s)cu for all u∈(0,1). With this, we can write
∫₀¹Σ_(k≥0) |uᵏlnˢ(1/u)|du=∫₀¹Σ_(k≥0) uᵏ(ln(1/u))ᴿᵉ⁽ˢ⁾du
>c∫₀¹Σ_(k≥0) uᵏ⁺¹du=c∫₀¹u/(1-u)du. (3.2)
This also diverges; combining this with our results from (3.1), (3) is not Lebesgue integrable for any s with Re(s)≤0, and (2) diverges for the same s. Since our integral is well-behaved for all other s and has no other singularities within our domain of (0,1), it follows that it is bounded for any s such that Re(s)>0. Therefore, I(s) is Lebesgue integrable on the product measure of the counting measure on ℕ₀ and the Lebesgue measure on (0,1) if and only if Re(s)>0. By Fubini's theorem, we can freely swap integration and summation, and we have the following:
I(s)=Σ_(k≥0) ∫₀¹uᵏ(ln(1/u))ˢdu. (4)
To proceed, we can introduce the upper incomplete gamma function (a generalization of the usual gamma function Γ(s):=∫₀ ᪲tˢ⁻¹e⁻ᵗdt for Re(s)>0), defined as Γ(s,x):=∫ₓ ᪲tˢ⁻¹e⁻ᵗdt for complex s and x with positive real parts. By the fundamental theorem of calculus, ∂/∂x Γ(s,x)=-xˢ⁻¹e⁻ˣ. Examining this derivative and the new form of I(s) in (4), we can deduce
I(s)=Σ_(k≥0) (k+1)⁻ˢ⁻¹[Γ(s+1,(k+1)ln(1/u))]₀¹
=Σ_(k≥1) k⁻ˢ⁻¹[Γ(s+1,0)-lim_(v→∞) Γ(s+1,v)], (5)
where v=k ln(1/u). By the definitions given above, we have Γ(s+1,0)=Γ(s+1) and lim_(v→∞) Γ(s+1,v)=0. Therefore, we can conclude
∫₀ ᪲xˢ/(eˣ-1)dx=Γ(s+1)Σ_(k≥1) k⁻ˢ⁻¹=Γ(s+1)ζ(s+1). □
This result holds for any complex s satisfying Re(s)>0, as shown above.
Wow!!! That's amazing. Fantastic proof, love the rigour. Thanks so much for sharing this, it's brilliant.
@@OscgrMaths Thank you! Loved working on this one. We take a lot of mathematical results for granted, but it's super instructive to look at them and figure out exactly how and why they work.
Cool video :)
Thanks!
So it wasn't Basil Faulty the? dohhh! :-)
Bose-Einstein!
8*pi^6/63 🫡
Yes! For the challenge you might recognise from the bose einstein equation that it's equal to zeta (s+1) gamma(s+1). Which one did you get an answer of 8pi^6/63 for?
@@OscgrMaths Yes! For the challenge, I used the specific case of the Bose-Einstein integral where s = 5, the integral becomes gamma(6)zeta(6) and I estimated zeta(6) by applying the formula involving Bernoulli numbers, which combined with gamma(6) = 5! = 120, leads to the result of 8pi^6/63!
Also, thanks a lot for your high-quality videos! Keep up the good work:)
@@clementp7648 Thats great!! I'm really glad you're enjoying.
❤
Wow genial
Thank you!
What does a man do when he feels like all the math he learns is for nothing of a use in his life?
Don't forget... the great thing about maths is that if you can prove it then you've proven something that is always true!! Unlike art or science which are changing and developing, maths may become more and more complex - but something true won't ever be untrue. So I guess you're discovering universal truths... But if that's not enough it's also extremely useful in everything (physics, engineering, economics etc...)
So it's all about the sensation of happines this gives when it's deeply connected to something usefull
“The scientist does not study nature because it is useful to do so. He studies it because he takes pleasure in it, and he takes pleasure in it because it is beautiful."
― Henri Poincaré, Science and Method