# What is $_3F_2\left(1,\frac32,2;\ \frac43,\frac53;\ \frac4{27}\right)$ as an integral?

This was buried in a rather long question, so I'm asking it separately to give it some air. Define, $$A_3={_3F_2}\left(1,\frac{\color{blue}1}2,\frac22;\ \frac43,\frac53;\ \frac4{27}\right)$$ $$B_3={_3F_2}\left(1,\frac{\color{blue}2}2,\frac32;\ \frac43,\frac53;\ \frac4{27}\right)$$ $$C_3={_3F_2}\left(1,\frac{\color{blue}3}2,\frac42;\ \frac43,\frac53;\ \frac4{27}\right)$$

These generalized hypergeometric functions belong to infinite families that differ in the starting numerator (in blue). Given the three roots $x_n$ of $x^3-x+1=0$. Then,

$$\frac12 A_3= \int_1^\infty \frac{-3+2x}{x^3-x+1}dx =-\sum_{n=1}^3 x_n\ln(1-x_n)=0.517977\dots$$

$$\frac13 B_3 = \int_1^\infty\frac1{x(x^3-x+1)}dx=-\sum_{n=1}^3 \frac{\ln(1-x_n)}{-3+2x_n}=0.371216\dots$$

Q: But what is $\displaystyle C_3 = 3\sum_{n=1}^\infty\frac1{\binom{3n}n}$ as a similar integral?

$\color{green}{Update:}$

Courtesy of David H's answer, by generalizing we find, $$k\sum_{n=1}^\infty\frac1{\binom{kn}n}=\int_1^\infty\frac1{(x^k-x+1)^2}dx$$ hence, $$C_3 = \int_1^\infty\frac1{(x^3-x+1)^2}dx=\tfrac{12}{23}-\tfrac{6}{23}\sum_{n=1}^3 \frac{3x_n + x_n^2}{-3+2x_n}\,\ln(1-x_n)=1.242966\dots$$

and a closed-form using the roots $x_n$.

Recall that the beta function may be defined via the integral representation

$$\operatorname{B}{\left(z,w\right)}:=\int_{0}^{1}t^{z-1}\left(1-t\right)^{w-1}\,\mathrm{d}t;~~~\small{\Re{\left(z\right)}>0\land\Re{\left(w\right)}>0}.$$

Using the infinite series definition of the generalized hypergeometric function, we find

\begin{align} C_{3} &={_3F_2}{\left(1,\frac32,2;\frac43,\frac53;\frac{4}{27}\right)}\\ &=\sum_{k=0}^{\infty}\frac{\left(1\right)_{k}\,\left(\frac32\right)_{k}\,\left(2\right)_{k}\,\left(\frac{4}{27}\right)^{k}}{\left(\frac43\right)_{k}\,\left(\frac53\right)_{k}\,k!}\\ &=\sum_{k=0}^{\infty}\frac{2\,\Gamma{\left(k+2\right)}\,\Gamma{\left(2k+2\right)}}{\Gamma{\left(3k+3\right)}}\\ &=\sum_{n=1}^{\infty}\frac{2\,\Gamma{\left(n+1\right)}\,\Gamma{\left(2n\right)}}{\Gamma{\left(3n\right)}}\\ &=\sum_{n=1}^{\infty}\frac{2n\,\Gamma{\left(n\right)}\,\Gamma{\left(2n\right)}}{\Gamma{\left(3n\right)}}\\ &=2\sum_{n=1}^{\infty}n\,\operatorname{B}{\left(n,2n\right)}\\ &=2\sum_{n=1}^{\infty}n\int_{0}^{1}\mathrm{d}t\,t^{n-1}\left(1-t\right)^{2n-1}\\ &=2\int_{0}^{1}\mathrm{d}t\,\sum_{n=1}^{\infty}nt^{n-1}\left(1-t\right)^{2n-1}\\ &=2\int_{0}^{1}\frac{\mathrm{d}t}{t\left(1-t\right)}\sum_{n=1}^{\infty}n\left[t\left(1-t\right)^{2}\right]^{n}.\\ \end{align}

Then, using the infinite series,

$$\sum_{n=1}^{\infty}nz^{n}=\frac{z}{\left(1-z\right)^{2}};~~~\small{\left|z\right|<1},$$

we obtain the integral form for $C_{3}$

\begin{align} C_{3} &=2\int_{0}^{1}\frac{\mathrm{d}t}{t\left(1-t\right)}\cdot\frac{t\left(1-t\right)^{2}}{\left[1-t\left(1-t\right)^{2}\right]^{2}}\\ &=2\int_{0}^{1}\mathrm{d}t\,\frac{\left(1-t\right)}{\left[1-t\left(1-t\right)^{2}\right]^{2}}\\ &=2\int_{0}^{1}\mathrm{d}u\,\frac{u}{\left[1-\left(1-u\right)u^{2}\right]^{2}};~~~\small{\left[1-t=u\right]}\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{2u}{\left(1-u^{2}+u^{3}\right)^{2}}.\blacksquare\\ \end{align}

• Great! I'm glad I made this side-question. I knew there had to be a way to express it using the roots $x_n$ as well. – Tito Piezas III Jan 10 '17 at 12:30