Originally Posted by **Mazin** https://photos.app.goo.gl/LWzgtJsCXs4zCCbc7
Hello,
I am reading a book about Stellar Structure and Evolution and it takes an expression for the Black body function in terms of frequency and substitutes it into a 'Pressure Integral' to derive an expression for Radiation Pressure.
It just gives me the Input, and then the output and says that the Integral is easy but I can't do it.
I posted similar request for help on another forum but so far no replies so I was wondering if someone could help?
I guess it's an Integration question rather than anything else.
The problem is on the image which I linked above.
Thanks |

It's handy to know this kind of integral, so take a careful look at it.

There's a little trick to this one. First, recall that if we have a geometric series with an infinite number of terms then

$\displaystyle S = \sum_{n = 1}^{\infty} r^n = \frac{r}{1 - r}$

Keep that in your back pocket for the moment.

We need to do an integral of the form

$\displaystyle \int \frac{x^3}{e^{x} - 1} ~ dx$

$\displaystyle = \int \frac{x^3 e^{-x}}{1 - e^{-x}}~dx$

Remember the sum? Well $\displaystyle \sum_{n = 1}^{\infty} \left ( e^{-x} \right ) ^n = \frac{e^{-x}}{1 - e^{-x}}$

This means your integral becomes

$\displaystyle = \int x^3 \sum_{n = 1}^{\infty} e^{-nx} ~ dx$

To a Physicist switching sums and integrals are second nature. We leave the pesky problem of the details to the Math people. So...

$\displaystyle = \sum_{n = 1}^{\infty} \int x^3 e^{-nx} ~ dx$

That's the tricky part. The rest is integration by parts and the sum deals with the zeta function, $\displaystyle \zeta (4)$.

I leave the rest to you.

-Dan