@ So it’s called because it’s associated with The “Planck spectrum” which can be used to derive both the Stefan-Boltzmann law and Wien’s law, and these results can be used, in turn, to calculate the value of h(Planck’s constant). If one needs to calculate total energy density of blackbody radiation then this is simply the integral of the Planck spectrum over all frequencies: \int_0 ^ \infty {8πh \over c^3} { u^3 \over e^{h u \over kT} -1 } d u We make a change of variables: x = {h u \over kT} then we get something like integral (0, infinity) x^ 3 over e^x -1 dx which gives π ^4 / 15 as the answer. I’ve done this for the class a while ago because undergrad texts don’t actually show the workout with gamma and zeta functions , they give the answer straight out. So hence a workout was needed: www.nebo.app/page/8d6abb24-49c3-48b9-a447-679dfaf9156e
@@owl3math “Planck Integral”, so it’s called and it’s associated with the Planck Spectrum of radiation from which Wien’s Law and Stefan-Boltzmann’s law can be derived. Here we make a variable change: x={h u \over kT} and the integral which renders π^4 over 15 is \int_0 ^ \infty {8πh \over c^3}{x^3 \over e^x -1}dx
@@owl3mathbut don't ypubagree it's contrived.im. kt sure why anyone wluld.think to multiply and divide by e^x like you did? Wouldn't ypu agree? I think you can solve without it can't you? Like rewriting e in terms of it's power series x^2/2! Etc. Or by integration by parts, right?
Nice solution. However, I would have skipped the Laplace transform and seen that the integral was very similar to the definition of the Gamma function, just a one change of variable u=nx, dx=du/n
I'm not sure. It seems like it would need to be a different method because in this one I think it relies on the idea that the power series works well with the x^(s-1)
For those of you wondering, that’s where the *famous “Planck Integral” pops up! And many texts leave it to us to somehow wrap it up!
Hi. What's the Planck Integral? Is it the same as this integral?
@ So it’s called because it’s associated with The “Planck spectrum” which can be used to derive both the Stefan-Boltzmann law and Wien’s law, and these results can be used, in turn, to calculate the value of h(Planck’s constant). If one needs to calculate total energy density of blackbody radiation then this is simply the integral of the Planck spectrum over all frequencies:
\int_0 ^ \infty {8πh \over c^3} {
u^3 \over e^{h
u \over kT} -1 } d
u
We make a change of variables: x = {h
u \over kT} then we get something like integral (0, infinity) x^ 3 over e^x -1 dx which gives π ^4 / 15 as the answer. I’ve done this for the class a while ago because undergrad texts don’t actually show the workout with gamma and zeta functions , they give the answer straight out. So hence a workout was needed: www.nebo.app/page/8d6abb24-49c3-48b9-a447-679dfaf9156e
@@owl3math “Planck Integral”, so it’s called and it’s associated with the Planck Spectrum of radiation from which Wien’s Law and Stefan-Boltzmann’s law can be derived. Here we make a variable change: x={h
u \over kT} and the integral which renders π^4 over 15 is \int_0 ^ \infty {8πh \over c^3}{x^3 \over e^x -1}dx
@@dean532 oh i see! I'm not familiar with that but thanks for clarifying.
@@owl3math Yup 👍
Perfecto! Love the geometric sum and Laplace🍻
thanks! This was a fun one :) 🍻
@@owl3mathbut don't ypubagree it's contrived.im. kt sure why anyone wluld.think to multiply and divide by e^x like you did? Wouldn't ypu agree? I think you can solve without it can't you? Like rewriting e in terms of it's power series x^2/2! Etc. Or by integration by parts, right?
Fantastic
Thanks!
Awesome video!
Thanks!!! 🙏
infinity
Nice solution. However, I would have skipped the Laplace transform and seen that the integral was very similar to the definition of the Gamma function, just a one change of variable u=nx, dx=du/n
Makes sense. It’s a good way to go.
Can you do it with sin(x) or ln(1-x) in the numerator?
I=int[0,♾️](sin(x)/(e^x+1))dx
I=int[0,♾️](sin(x)e^-x/(1+e^-x))dx
I=int[0,♾️](sin(x)sum[k=1,♾️]((-1)^(k-1)•e^-kx))dx
I=sum[k=1,♾️]((-1)^(k-1)•L{sin(x)}|k)
L{sin(x)}=1/(k^2+1)
I=sum[k=1,♾️]((-1)^(k-1)/(1+k^2))
1/(1+k^2)=i/2(k+i)-i/2(k-i)
I=-i/2•sum[k=1,♾️]((-1)^k•(1/(k+i)-1/(k-i)))
I=i/2•sum[k=1,♾️](1/(2k-1+i)-1/(2k-1-i)-1/(2k+i)+1/(2k-i))
I=i/4•sum[k=1,♾️](1/(k-(1-i)/2)-1/(k-(1+i)/2)-1/(k+i/2)+1/(k-i/2))
¥(a+1)-¥(b+1)=sum[k=1,♾️](1/(k+b)-1/(k+a)
I=i/4•(¥((-3-i)/2)-¥((-3+i)/2)+¥((-1+i)/2)-¥((-1-i)/2))
¥(x+1)-¥(x)=1/x
I=i/4•(2/(3+i)-2/(3-i))
I=i/2•(-2i/(9+1))
I=1/10
I'm not sure. It seems like it would need to be a different method because in this one I think it relies on the idea that the power series works well with the x^(s-1)
@@owl3math I was thinking if you expand them as a power series so you get a double sum? Although you get issues with convergence for ln(1-x).