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Deriving an expression for Radiation Pressure from Black body functionhttps://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 |

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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 |

Thank you very much for this. I'll come back to it tomorrow after sleep. I see how you re-arranged the equation by factoring out e^x and then taking it to the numerator and reversing the power. Took me a while to see why that works, but now its obvious. I will look at Geometric series piece tomorrow. It's a series of little tricks you've done I think. I'll see if I can follow. |

https://photos.app.goo.gl/3SmS28CJ49gdD5Q47 https://photos.app.goo.gl/ouPh4Rf9uyVufxqp7 Hello Dan. I found for a geometric series starting with value a, and common ratio r, the Sum is a/1-r where r < 1. I assumed that this is equivalent to r/1-r in your note, I took the Indefinite Integral but I don't know how to relate it to the Sum. I know there are formula for 1 to n, but the Sum is from 1 to Infinity. How do I get from here to the answer? I think it will relate to the Sum. Thank you for helping me. |

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Just to make sure it's been covered I needed an expression of the form $\displaystyle \frac{r}{1 - r}$, which was in the integral. So $\displaystyle \sum_{n = 0}^{\infty} r^n = 1 + \sum_{n = 1}^{\infty} r^n = \frac{1}{1 - r}$ So $\displaystyle \frac{1}{1- r} - 1 = \frac{r}{1 - r} = \sum_{n = 1}^{\infty} r^n$ I'll get back to you on the rest later today. -Dan |

Thanks. I get that aspect. r to the 0 = 1. |

Do you have any updates on this problem? Thanks |

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-Dan |

Reassuring it's on your to do list but take your time. I'd like to know how to do it, so I can do similar problems but I can wait 2-3 weeks, or however long it takes. Hope whatever's bad doesn't trouble you too much. |

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-Dan |

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