# Integration by parts of $\int_0^\infty \! n^2 \ln(1-e^{-an}) \, \mathrm{d} n$

Im trying to do the following integration by hand,

$\int_0^\infty \! n^2 \ln(1-e^{-an}) \, \mathrm{d} n$

I have tried to use integration by parts and substitution, but each time it gets to complicated and messy (polylog terms etc...), is there any straight forward way to solve this?

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Try justifying the power series expansion $$\log(1-x) = -\sum_{k=1}^{\infty}\frac{x^k}{k}$$ (hint: $e^{-an}$ is small and always confined to a small positive interval when $n$ is larger than any fixed value). Now plug in the power series and you'll have a much more tractable problem. – A Blumenthal Jan 22 '13 at 22:46

I will write your integral as

$$\int_0^{\infty} dx \: x^2 \log{(1-e^{-a x})} = \frac{1}{3} \int_0^{\infty} d(x^3) \log{(1-e^{-a x})}$$

Now integrate by parts:

$$= \frac{1}{3} [x^3 \log{(1-e^{-a x})}]_0^{\infty} - \frac{1}{3} \int_0^{\infty} dx\: x^3 \frac{d}{dx} \log{(1-e^{-a x})}$$

The term in the brackets goes to zero at $\infty$ and $0$ (why?), so we get

$$= -\frac{a}{3} \int_0^{\infty} dx\: x^3 e^{-a x} (1-e^{-a x})^{-1}$$

For some values of $a$ (which ones?), we may Taylor expand the term in parentheses inside the integral and get

$$= -\frac{a}{3} \int_0^{\infty} dx\: x^3 e^{-a x} \sum_{k=0}^{\infty} e^{-k a x}$$

Because integral and sum are absolutely convergent (application of Fubini's theorem), we may interchange order of sum and integral and get:

$$-\frac{a}{3} \sum_{k=0}^{\infty} \int_0^{\infty} dx\: x^3 e^{-(k+1) a x}$$

These integrals are well-known:

$$\int_0^{\infty} dx\: x^3 e^{-(k+1) a x} = \frac{3!}{a^4 (k+1)^4}$$

so we now have

$$\int_0^{\infty} dx \: x^2 \log{(1-e^{-a x})} = -\frac{2}{a^3} \sum_{k=0}^{\infty} \frac{1}{(k+1)^4} = -\frac{\pi^4}{45 a^3}$$

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Using the power series for $\log(1-x)$: $$\int_{0}^{\infty}n^2\log(1-e^{-\alpha n})\,\mathrm{d}n=-\int_{0}^{\infty}n^2\sum_{k=1}^{\infty}\frac{e^{-k\alpha n}}{k}\,\mathrm{d}n\\=-\int_{0}^{\infty}\sum_{k=1}^{\infty}\frac{\partial^2}{\partial\alpha^2}\left(\frac{e^{-kn\alpha}}{k^3}\right)\,\mathrm{d}n\\=-\frac{\partial^2}{\partial\alpha^2}\sum_{k=1}^{\infty}\int_{0}^{\infty}\frac{e^{-kn\alpha}}{k^3}\,\mathrm{d}n\\=-\frac{\partial^2}{\partial\alpha^2}\sum_{k=1}^{\infty}\left(-\frac{e^{-kn\alpha}}{\alpha k^4}\right)_0^{\infty}\\=-\frac{\partial^2}{\partial\alpha^2}\sum_{k=1}^{\infty}\frac{1}{\alpha k^4}\\=-\sum_{k=1}^{\infty}\frac{\partial^2}{\partial\alpha^2}\left(\frac{\alpha^{-1}}{ k^4}\right)\\=-2\sum_{k=1}^{\infty}\frac{\alpha^{-3}}{k^4}=-\frac{2\zeta(4)}{\alpha^3}\\=-\frac{\pi^4}{45\alpha^3}.$$

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Integration by parts yields \begin{align} \int_0^\infty n^2 e^{-n}\,\mathrm{d}n &=\left.-n^2e^{-n}\right]_0^\infty+\int_0^\infty2n\,e^{-n}\,\mathrm{d}n\\ &=0+\left.-2n\,e^{-n}\vphantom{n^2}\right]_0^\infty+\int_0^\infty 2\,e^{-n}\,\mathrm{d}n\\ &=2\tag{1} \end{align} Apply $(1)$ and $\log(1-x)=-\sum\limits_{k=1}^\infty\dfrac{x^k}{k}$ to \begin{align} \int_0^\infty n^2\log(1-e^{-an})\,\mathrm{d}n &=-\int_0^\infty n^2\sum_{k=1}^\infty\frac1ke^{-kan}\,\mathrm{d}n\\ &=-\sum_{k=1}^\infty\frac1k\frac2{(ka)^3}\\ &=-\frac2{a^3}\zeta(4)\\ &=-\frac{\pi^4}{45a^3}\tag{2} \end{align} Using $\zeta(4)$ from this answer.

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Partial integration:

$$\int_0^{\infty}n^2\cdot \ln(1-e^{-an})dn=\underbrace{\left[\frac{x^3}{3}\ln(1-e^{-an})\right]_0^{\infty}}_{=\;0}-\frac{1}{3}\int_0^{\infty}n^3\left(\frac{a}{e^{an}-1}\right)dn$$

Let $s=an:$

$$\int_0^{\infty}n^2\cdot \ln(1-e^{-an})dn=-\frac{1}{3a^3}\int_0^{\infty}\frac{s^3}{e^{s}-1}ds$$

Using the well-known relation

$$\Gamma(z)\zeta (z)=\int_0^{\infty}\frac{s^{z-1}}{e^s-1}ds\qquad(\Re (z)>1)$$

We get,

$$\int_0^{\infty}n^2\cdot \ln(1-e^{-an})dn=-\frac{1}{3a^3}\Gamma(4)\zeta(4)=-\frac{\pi^4}{45a^3}$$

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This is a generalization. By expanding the logarithm in its Taylor series (the same method in the other answers) you'll get that $$\int_0^\infty x^{n-1}\log(1-e^{-\alpha x})dx=-\frac{1}{\alpha^n}\zeta(n+1)\Gamma(n)$$ Your case is $n=3$

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