Simple integral

1. Sep 14, 2004

P3X-018

"Simple" integral

Hey. I'm having problems with integrating the following function:

I used partial integration (if it's called that in english?), and end up with the same term. In equation (1.2) I just tried to integrate the middle term in (1.1), and as you can see it didn't work with parial integration. How do I then solve it?
By the way, I'm not on higher education, only in "senior high school". So it would best if you could show how you get to the solution.

Last edited by a moderator: May 1, 2017
2. Sep 14, 2004

mathwonk

well you alreaedy beat me by integrating 1/(e^x-1) but axxording to my basic philosophy of "make the rpoblem easier" you might try just doing x/[e^x-1].

or use infinite series as a last resort.

3. Sep 14, 2004

P3X-018

No, I can't just do x/((e^x) -1)). I need the first value, in calculating stefan-boltzmanns law. But I don't know anything about integrating by use of the infinite series.

4. Sep 14, 2004

Dr Transport

The integral is called a Fermi integral. Look up the general form, the solution is a product of the Riemann zeta function and the gamma function.

$$\int^{\inf} _{0} \frac{x^{n-1}}{e^{x} - 1} dx = \Gamma (n) \zeta (n)$$

5. Sep 15, 2004

P3X-018

Why am I unable to see the latex graphic? It keeps saying: "LaTeX graphic is being generated. Reload this page in a moment."
And it's not helping by refreshing the page.

6. Sep 15, 2004

da_willem

My phyiscs professor in my first year at the university derived the Stefan-Boltzman law the same way you are trying to do from the Planck's BB radiation formula. He made the same substitution but on encountering the integral you mention he said "this is just the area undre a certain curve wich can be shown to be $\pi^4 / 15$". I guess it's quite a tricky integral...

7. Sep 15, 2004

P3X-018

Yea, I know it's 6,49 (pi^4/15). You can calculate the area, on the calculator, but I just wanted to know how you get to the solution. Well Dr Transport tells also how to solve it. But I can't see the latex graphic. Why? Is it because you've writin the command wrong?? Would you pls rewrite it again (not the last one, which is pi^4/15). I would appreciate that!

8. Sep 15, 2004

P3X-018

Now we are talking about the substitution, I got the value -2pi(kT)^4. but shouldn't it be with af positive sign??

9. Sep 15, 2004

da_willem

I'm not sure what you mean, what value is that you got there?

The final answer is the intensity of the radiation is: $I=\sigma T^4$ (with $\sigma$ the stefan Boltzman constant. This constant turns out to be not fundamental if you work everything out:

$$\sigma = \frac{2 k^4 \pi^5}{15 h^3 c^2}$$

The easiest way is to substitute $x= hc/ \lambda kT$ when integrating Plancks law over all wavelengths (now x) to get the total intensity (Stefan-Boltzmann law).

Note: You can see the latex code if you click on the (nongenerated) latex image

10. Sep 15, 2004

P3X-018

Yes I know that da willem (check the link at #1). But I mean that after the substitution i get a negative value. (See the attached document). Can you tell why I get a negativ value (-kT)?
Or is it supposed to be negativ?

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11. Sep 15, 2004

da_willem

You get the minus sign from the differential d(lambda)=-(hc/kT)x^-2 dx. But you forgot to cancel it with the minus introduced by changing your integration limits.

12. Sep 15, 2004

Dr Transport

The LateX graphics have had some trouble since the server switch the other day. If you look at what was written, the general formula has been given. The problem is being fixed.

dt

13. Sep 15, 2004

Theelectricchild

Phew I was wondering about what was going wrong with the TeX--- I thought it was just something I was experiencing, nice to know that that is indeed not the case.

14. Sep 16, 2004

P3X-018

Ok. I thougt the same as Theelectricchild, that it was only me who was experiencing that!

By the way "da willem" what do you mean by "changing the integration limits"? Where do you change it? I mean you can't just change them.

15. Sep 16, 2004

da_willem

You have to integrate the wavelength from 0 to infinity. You made the substitution x= hc/ (wavelength)kT. So when the wavelength goes from 0 to infinity, x goes from infinity to 0 (because of the inverse relation between x and the wavelength). So you have to change your integration limits from (from 0 to infinity) to (from infinity to zero). You evaluated the integral from 0 to infinity where you should have evaluated from infinity to zero...

Thisis allowed; but you have to take into account another minus sign wich you forgot. Because:

$$\int _a^b f(t) dt = - \int _b^a f(t) dt$$

Got it?

16. Sep 16, 2004

P3X-018

Yea I got ya. Well, thanks for help :)

17. Sep 16, 2004

Dr Transport

let's try this again

$$\int^{\infty} _{0} \frac{x^{n-1}}{e^{x} - 1} dx = \Gamma (n) \zeta (n)$$