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A graduate student in physics asked me if there was a closed form for the following series, which he had found as the second virial coefficient for a gas in an exponential potential: $$\sum_{k=0}^\infty \frac{x^k}{(k+1)^4 k!}$$ where $x < 0$. This is the generalized hypergeometric function $_4 F_4(1, 1, 1, 1; 2, 2, 2, 2; x)$ by definition. We were skeptical that it could be simplified, as neither the series nor any of its first few derivatives and antiderivatives is the Taylor series of a function that we could recognize. Even if there's not a truly closed form, we're still curious whether it has a formula with special functions other than the generalized hypergeometric.

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    $\begingroup$ the site suggested an analogous math.stackexchange.com/questions/2005837/… im not too optimistic of a more closed form then $\endgroup$ – user499752 Nov 8 '17 at 15:24
  • $\begingroup$ Even better, look for the commentary on the first answer math.stackexchange.com/questions/2006799/… $\endgroup$ – user499752 Nov 8 '17 at 15:27
  • $\begingroup$ The given series equals $$-\frac{1}{6}\int_{0}^{1}e^{xy}\log^3(y)\,dy$$ so its asymptotic behaviour is pretty simple to estimate, but a nice closed form... I won't bet on that. $\endgroup$ – Jack D'Aurizio Nov 8 '17 at 18:35
  • $\begingroup$ @JackD'Aurizio I don't immediately see where that integral comes from; can you give a justification? $\endgroup$ – Connor Harris Nov 8 '17 at 19:22
  • $\begingroup$ @ConnorHarris: $$\frac{1}{(k+1)^4}=-\frac{1}{6}\int_{0}^{1}y^k\log^3(y)\,dy,$$ hence by multiplying both sides by $\frac{x^k}{k!}$ and summing over $k\geq 0$ we get the previous integral representation. $\endgroup$ – Jack D'Aurizio Nov 8 '17 at 19:29
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Let us denote \begin{equation} S_q(x) := \sum\limits_{k=0}^\infty \frac{x^k}{(k+1)^q k!} \end{equation} and $y=-x$.

Using JackD'Azurio's integral representation we get: \begin{eqnarray} S_4(x) &=& -\frac{1}{3!}\left. \frac{d^3}{d \theta^3} \int\limits_0^1 e^{x y} y^\theta dy \right|_{\theta=0} \\ &=&-\frac{1}{3!} \left. \frac{d^3}{d \theta^3} \frac{\Gamma(1+\theta) - \Gamma(1+\theta,-x)}{(-x)^{1+\theta}} \right|_{\theta=0} \\ &=& \frac{4 \zeta(3)+2 \gamma^3+ \gamma \pi^2+(6 \gamma^2+\pi^2)\log(y)+6 \gamma \log(y)^2+2 \log(y)^3+12 {\mathfrak f}(y)}{12 y} \end{eqnarray} where \begin{equation} {\mathfrak f}(y) := \int\limits_1^\infty \frac{e^{-y t}}{t} \frac{\log(t)^2}{2!} dt = G_{3,4}^{4,0}\left(y\left| \begin{array}{c} 1,1,1 \\ 0,0,0,0 \\ \end{array} \right.\right) \end{equation} where the function above is expressed through the Meijer-G function and $\gamma$ is the Euler's gamma constant.

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