# STEP III 1998 Q7

I am stuck in the last part. I might be missing something, as I cannot see how to get $B=KT^n$ .

The integral of b(x) from 0 to infinity seems hard to evaluate: I have tried by parts so far. I do not know if I have to actually integrate it or if there is another way to reach the given result for B??

### You can rewrite it like so $$You can rewrite it like so$$\int_{0}^{\infty} \frac{x^3}{e^{x/T} - 1} \, \mathrm{d}x = \int_0^{\infty} \frac{x^3e^{-x/T}}{1 - e^{-x/T}} \, \mathrm{d}x = \int_0^{\infty} x^3 e^{-x/T} \sum_{k\geq 0} e^{-kx/T} \, \mathrm{d}x then wave your hands and scream dominated convergence theorem and interchange the integral and summation.

NB: The third equality arises from recognising that $\frac{1}{1-a}$ is a geometric series. Here we have $a = e^{-x/T}$ and it converges since $e^{-x} < 1$ for $x > 0$. This is a very standard technique when dealing with integrals that isn't quite taught at A-Level.

### Nice

i would never have tried that, though. Thanks!

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