# Asymptotic expansion for $\frac{1}{2\zeta(3)}\int_x^\infty \frac{u^2}{e^u - 1} du$?

Is there an asymptotic expansion for the function: $$g(x)=\frac{1}{2\zeta(3)}\int_x^\infty \frac{u^2}{e^u - 1} du,$$ over the domain $x\in [0,\infty)$ in terms of elementary functions? Here $\zeta$ is the Riemann zeta function and $1/2\zeta(3)$ is a normalization factor included to ensure $g(0)=1$.

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This is sometimes referred to as the incomplete Bose-Einstein integral. –  Ｊ. Ｍ. Apr 4 '13 at 15:13

We have $$g(x) = \dfrac1{2 \zeta(3)}\underbrace{\int_x^{\infty} \dfrac{u^2}{e^u-1} du}_{I(x)}$$ We will now obtain a series expansion for $I(x)$. We have $$I(x) = \int_x^{\infty} \dfrac{u^2}{e^u-1} du = \int_x^{\infty} \dfrac{u^2 e^{-u}}{1-e^{-u}} du = \int_x^{\infty} \sum_{k=0}^{\infty} u^2 e^{-(k+1)u} du = \sum_{k=1}^{\infty} \int_x^{\infty} u^2 e^{-ku} du$$ We now have that $$\int_x^{\infty} u^2 e^{-ku} du = \dfrac{e^{-kx} \left(k^2 x^2 + 2kx + 2\right)}{k^3}$$ We have have $$I(x) = \sum_{k=1}^{\infty} \dfrac{e^{-kx} \left(k^2 x^2 + 2kx + 2\right)}{k^3}\tag{\star}$$ For a given $x$, truncating $(\star)$ will give you an exponentially converging approximation. Note that $$I(0) = \displaystyle \sum_{k=1}^{\infty} \dfrac2{k^3} = 2 \zeta(3)$$

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Great! But surely you should substitute $k+1\rightarrow k$ to simplify $(\star)$? Also, the sum is missing in $l(0)$. –  Douglas B. Staple Apr 4 '13 at 15:58
@DouglasB.Staple Oh yes:)!. Thanks. Done. –  user17762 Apr 4 '13 at 16:00
Beautiful. Also, $(\star)$ gives $l(x)=2\operatorname{polylog}(3,e^{-x}) + 2x\operatorname{polylog}(2,e^{-x}) - x^2\ln(1-e^{-x})$ using the definition of the $\operatorname{polylog}$ function. –  Douglas B. Staple Apr 4 '13 at 16:15
@DouglasB.Staple Yes. I always forget the polyLog expansions, though I am encountering it frequently in recent times. –  user17762 Apr 4 '13 at 16:18

yes is equal to the incomplete gamma function http://mathworld.wolfram.com/IncompleteGammaFunction.html

$$\Gamma (3,x)$$

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No, it's not.$\phantom{}$ –  Ｊ. Ｍ. Apr 4 '13 at 15:39