We start this section with noticing that for \(r = 2\), the value of \(W_2(k;s)\) coincides with \(Z(k + x + x^{-1} + y + y^{-1}; s)\) (see (1)). Indeed, the substitution \(x = x_1x_2\) and \(y = x_1x_2^{-1}\) in the latter leads to
$$\begin{aligned} k + x_1x_2 + (x_1x_2)^{-1} + x_1x_2^{-1} + \left( x_1x_2^{-1}\right) ^{-1} = k + \left( x_1 + x_1^{-1}\right) \left( x_2 + x_2^{-1}\right) . \end{aligned}$$
For \(r \ge 3\), the value of (2) is different from \(Z(k + x_1 + x_1^{-1} + \dots + x_r + x_r^{-1};s)\).
In this section we discuss the densities \(p_r(k;-)\).
The probability densities \(p_r(k;-)\)
In this section we explicitly compute the probability distributions \(p_r(k;-)\) for \(r = 1,2,3\) and discuss how to obtain them for any r.
Define \(\hat{p}_r\) to be the density of the random variable \((X_1 + X_1^{-1}) \cdots (X_r + X_r^{-1})\) on \(\mathbb {C}^r\); note that the quantity assumes real values only. We can relate the density of \(\hat{p}_r\) to \(p_r(k;-)\) in the following way (compare with Sect. 2.2).
Lemma 4.1
For \(|k| < 2^r\), we have
$$\begin{aligned} p_r(k;x) = {\left\{ \begin{array}{ll} \hat{p}_r(x - |k|)&{}\text {for } 2^r - |k| \le x< 2^r+|k|,\\ \hat{p}_r(x - |k|) + \hat{p}_r(x + |k|)&{}\text {for } 0 \le x<2^r-|k|,\\ 0 &{}\text {for }x<0 \text { or } x \ge 2^r +|k|. \end{array}\right. } \end{aligned}$$
For \(|k| \ge 2^r\), we have \(p_r(k;x) = \hat{p}_r(x - |k|)\).
Proof
We have
$$\begin{aligned} W_r(k;s)&= \int \limits _{\mathbb {R}} |z|^s \hat{p}_r(z-|k|) \, \mathrm {d}z\\&= \int \limits _{2^r - |k|}^{2^r + |k|} |z|^s \hat{p}_r(z-|k|) \, \mathrm {d}z. \end{aligned}$$
If \(|k| \ge 2^r\), then
$$\begin{aligned} W_r(k;s) = \int \limits _{2^r - |k|}^{2^r + |k|} z^s \hat{p}_r(z-|k|) \, \mathrm {d}z, \end{aligned}$$
so that \(p_r(k;x) = \hat{p}_r(x - |k|)\).
If \(|k| < 2^r\), then
$$\begin{aligned} W_r(k;s)&= \int \limits _{2^r - |k|}^{2^r + |k|} |z|^s\hat{p}_r(z - |k|) \, \mathrm {d}z\\&= \int \limits _0^{2^r + |k|} z^s\hat{p}_r(z - |k|) \, \mathrm {d}z + \int \limits _0^{2^r - |k|} z^s\hat{p}_r(- z - |k|) \, \mathrm {d}z \\&= \int \limits _0^{2^r + |k|} z^s\hat{p}_r(z - |k|) \, \mathrm {d}z + \int \limits _0^{2^r - |k|} z^s\hat{p}_r(z + |k|) \, \mathrm {d}z\\&= \int \limits _0^{2^r - |k|} z^s \left( \hat{p}_r(z - |k|) \, \mathrm {d}z + \hat{p}_r(z + |k|) \right) \, \mathrm {d}z + \int \limits _{2^r - |k|}^{2^r + |k|} z^s\hat{p}_r(z - |k|) \, \mathrm {d}z, \end{aligned}$$
where the symmetry \(\hat{p}_r(-z) = \hat{p}_r(z)\) was employed. \(\square \)
Using Lemma 4.1, it is clear that it suffices to compute the distributions \(\hat{p}_r\).
Computation of \(\hat{p}_r\)
We will now compute \(\hat{p}_r\) explicitly for \(r = 1,2\) and 3.
It follows from (9) that
$$\begin{aligned} \hat{p}_1(x) = \frac{1}{2\pi } \frac{1}{\sqrt{1 - \frac{x^2}{4}}}, \end{aligned}$$
for \(|x| < 2\) and \(\hat{p}_1(x) = 0\) otherwise.
For \(r \ge 1\), define \(G_r(y) = \hat{p}_r(2^r \sqrt{1-y})\) for \(0 < y \le 1\) and \(G_r(y) = 0\) otherwise, so that \(\hat{p}_r(y) = G_r(1-x^2/4^r)\). By the above,
$$\begin{aligned} G_1(y) = \frac{1}{2 \pi } \frac{1}{\sqrt{y}} \end{aligned}$$
for \(0 < y \le 1\) and \(G_1(y) = 0\) otherwise.
Further, using basic properties of the Mellin transform, it follows that the \(G_r\) satisfy a recurrence for \(r \ge 2\):
$$\begin{aligned} G_r \left( 1 - \frac{x^2}{4^r} \right)&= \int \limits _{\mathbb {R}} G_{r-1} \left( 1-\frac{t^2}{4^{r-1}} \right) G_{1}\left( 1 - \frac{x^2}{4t^2} \right) \frac{\mathrm {d}t}{|t|} \nonumber \\&= 2 \int \limits _{x/2}^{2^{r-1}} G_{r-1} \left( 1-\frac{t^2}{4^{r-1}} \right) G_{1}\left( 1 - \frac{x^2}{4t^2} \right) \frac{\mathrm {d}t}{t} \nonumber \\&= \frac{1}{\pi } \int \limits _{x/2}^{2^{r-1}} G_{r-1} \left( 1 - \frac{t^2}{4^{r-1}} \right) \frac{\mathrm {d}t}{\sqrt{t^2 - \frac{x^2}{4}}} . \end{aligned}$$
(16)
Then using the substitution \(u = t^2/4^{r-1}\) in (16), we find
$$\begin{aligned} G_r \left( 1 - \frac{x^2}{4^r} \right)&= \frac{1}{2 \pi } \int \limits _{x^2/4^r}^{1} G_{r-1}(1 - u) \frac{\mathrm {d}u}{\sqrt{u(u - \frac{x^2}{4^r})}},\\&= \frac{1}{2 \pi } \int \limits _{0}^{1 - x^2/4^r} G_{r-1}(u) \frac{\mathrm {d}u}{\sqrt{(1-u)(1 - \frac{x^2}{4^r} - u)}}. \end{aligned}$$
Finally, let \(y = 1 - x^2/4^r\) and \(v = u/y\) to arrive at the following result.
Theorem 4.2
(Recursive formula for the \(G_r\)) For \(G_r\) with \(r \ge 2\) we have the following recursion:
$$\begin{aligned} G_r(y)&= \frac{\sqrt{y}}{2\pi } \int _{0}^1 G_{r-1}(yv) \frac{\mathrm {d}v}{\sqrt{(1-v)(1-yv)}}. \end{aligned}$$
Applying this recursion with \(r = 2\) we obtain for \(0< y < 1\),
$$\begin{aligned} G_2(y)&= \frac{\sqrt{y}}{2\pi } \int \limits _{0}^1 G_1 (yv) \frac{\mathrm {d}v}{\sqrt{(1-v)(1-yv)}}\\&= \frac{1}{4 \pi ^2} \int \limits _{0}^1 \frac{\mathrm {d}v}{\sqrt{v(1-v)(1-yv)}} \\&= \frac{1}{4 \pi } \cdot {}_2 F_{1} \left( \frac{1}{2}, \frac{1}{2};1;y \right) . \end{aligned}$$
In the last step we use the integral representation (10). This shows that
$$\begin{aligned} \hat{p}_2(x) = \frac{1}{4 \pi } \cdot {}_2 F_{1} \left( \frac{1}{2}, \frac{1}{2};1;1 - \frac{x^2}{16} \right) , \end{aligned}$$
for \(0 < |x| \le 4\) and \(\hat{p}_2(x) = 0\) otherwise.
In the case \(r = 3\) we proceed similarly. For \(0< y < 1\),
$$\begin{aligned} G_3(y)&= \frac{\sqrt{y}}{2 \pi } \int \limits _0^1 G_2(yv) \frac{\mathrm {d}v}{\sqrt{(1-v)(1-yv)}} \\&= \frac{\sqrt{y}}{8 \pi ^2} \int \limits _{0}^1 {}_2 F_{1} \left( \frac{1}{2}, \frac{1}{2};1; yv \right) \frac{\mathrm {d}v}{\sqrt{(1 - v)(1 - yv)}}. \end{aligned}$$
The expression
$$\begin{aligned} Y(a)&:= \int _{0}^1 {}_2 F_{1} \left( \frac{1}{2}, \frac{1}{2};1; av \right) \frac{\mathrm {d}v}{\sqrt{(1 - v)(1 - av)}}\\&= \sum _{r \ge 0} \frac{r! \Gamma (1/2)}{\Gamma (r+3/2)} \left( \sum _{n + m = r} \frac{(\frac{1}{2})_n^2 (\frac{1}{2})_m}{(n!)^2 m!}\right) a^r \end{aligned}$$
satisfies a third-order linear differential equation
$$\begin{aligned} 8a^2(a-1)^2\frac{d^3Y}{da^3} + 24a(a-1)(2a-1)\frac{d^2Y}{da^2} + (56a^2 - 54a + 6) \frac{dY}{da} + (8a - 3)Y = 0. \end{aligned}$$
After solving this differential equation (for example, with Mathematica [12]) and checking the initial conditions we find out that
$$\begin{aligned} G_3(y) = \frac{\sqrt{y}}{4 \pi ^2} {}_2 F_{1} \left( \frac{1}{4}, \frac{1}{4} ;\frac{1}{2};y\right) {}_2 F_{1} \left( \frac{3}{4}, \frac{3}{4} ;\frac{3}{2};y\right) , \end{aligned}$$
so that
$$\begin{aligned} \hat{p}_3(x) = \frac{\sqrt{1 - \frac{x^2}{64}}}{4 \pi ^2} {}_2 F_{1} \left( \frac{1}{4}, \frac{1}{4} ;\frac{1}{2};1 - \frac{x^2}{64}\right) {}_2 F_{1} \left( \frac{3}{4}, \frac{3}{4} ;\frac{3}{2};1 - \frac{x^2}{64}\right) \end{aligned}$$
for \(0< |x| \le 8\) and \(\hat{p}_3(x) = 0\) for \(|x| > 8\).
Already at this stage it is pretty suggestive that \(G_r(1-y)\) satisfies the same differential equation as
$$\begin{aligned} {}_r F_{r-1} \left( \frac{1}{2}, \dots , \frac{1}{2};1, \dots , 1; y \right) . \end{aligned}$$
We discuss this in the next subsection.
Mellin Transform of \(\hat{p}_r\)
In this section we find an expression for
$$\begin{aligned} \int _{-2^r}^{2^r} x^{v} \hat{p}_r(x) \, \mathrm {d}x \end{aligned}$$
when \(v \in \mathbb {C}\). This is the v-th moment of the random variable \((X_1 + X_1^{-1}) \dots (X_r + X_r^{-1})\).
Lemma 4.3
For \(\text {Re}(v) > -1\), we have
$$\begin{aligned} \int \limits _0^{2^r} x^v \hat{p}_r(x) \, \mathrm {d}x =\frac{1}{2}\left( \frac{2^v}{\pi } \frac{\Gamma \left( \frac{1}{2} \right) \Gamma (\frac{v+1}{2})}{\Gamma \left( 1 + \frac{v}{2} \right) } \right) ^r \end{aligned}$$
and
$$\begin{aligned} \int _{-2^r}^{2^r} x^v \hat{p}_r(x) \, \mathrm {d}x = \frac{1+e^{\pi i v}}{2}\left( \frac{2^v}{\pi } \frac{\Gamma \left( \frac{1}{2} \right) \Gamma (\frac{v+1}{2})}{\Gamma \left( 1 + \frac{v}{2} \right) } \right) ^r. \end{aligned}$$
Proof
Since
$$\begin{aligned} |(X_1 + X_1^{-1}) \cdots (X_r + X_r^{-1})|^v =|X_1 + X_1^{-1}|^v \cdots |X_r + X_r^{-1}|^v \end{aligned}$$
and for the expected value
$$\begin{aligned} \mathsf {E}[|(X_1 + X_1^{-1}) \cdots (X_r + X_r^{-1})|^v] = 2 \int _0^{2^r} x^v \hat{p}_r(x) \, \mathrm {d}x, \end{aligned}$$
we have
$$\begin{aligned} 2 \int \limits _{0}^{2^r} x^v \hat{p}_r(x) \, \mathrm {d}x = \left( 2 \int _{0}^{2} x^v \hat{p}_1(x) \, \mathrm {d}x \right) ^r. \end{aligned}$$
For \(r = 1\), we obtain
$$\begin{aligned} \int \limits _{0}^2 x^v \hat{p}_1(x) \, \mathrm {d}x&= \frac{1}{2 \pi } \int \limits _{0}^2 \frac{x^v}{\sqrt{1 - \frac{x^2}{4}}} \, \mathrm {d}x \\&=\frac{2^{v}}{\pi } \int \limits _{0}^1 \frac{x^v}{\sqrt{1 - x^2}} \, \mathrm {d}x\\&=\frac{2^{v}}{\pi } \frac{\Gamma (v+1)\Gamma (\frac{1}{2})}{\Gamma (\frac{3}{2}+v)} \cdot {}_2 F_{1} \left( \frac{1}{2} ,v+1;\frac{3}{2} + v; -1 \right) \\&= \frac{ 2^{v}}{\pi } \frac{\Gamma (\frac{1}{2}) \Gamma (1 + \frac{v+1}{2})}{(v+1) \Gamma (1 + \frac{v}{2})}. \end{aligned}$$
Therefore,
$$\begin{aligned} \int \limits _0^{2^r} x^v \hat{p}_r(x) \, \mathrm {d}x = \frac{1}{2} \left( \frac{2^{v+1}}{\pi } \frac{\Gamma (\frac{1}{2}) \Gamma (1 + \frac{v+1}{2})}{(v+1) \Gamma (1 + \frac{v}{2})} \right) ^r \end{aligned}$$
and, clearly,
$$\begin{aligned} \int _{-2^r}^{2^r} x^v \hat{p}_r(x) \, \mathrm {d}x = \left( 1 + e^{\pi i v} \right) \int _{0}^{2^r} x^v \hat{p}_r(x) \, \mathrm {d}x. \end{aligned}$$
\(\square \)
Let \(f :(0, \infty ) \rightarrow \mathbb {R}\) be continuously differentiable and \(s \in \mathbb {C}\). Denote by M(f; s) its Mellin transform
$$\begin{aligned} M(f;s) := \int \limits _0^\infty x^s f(x) \, \mathrm {d}x. \end{aligned}$$
Note that our definition of the Mellin transform differs slightly from the standard one, this does not affect the properties discussed below. Further, let \(\theta \) be the differential operator \(x \frac{\mathrm {d}}{\mathrm {d}x}\).
Proposition 4.4
(General properties of Mellin transforms; see [13, Sect. 3.1.2])
-
(i)
\(M(\theta f(x);s) = -(s+1) M(f(x);s)\);
-
(ii)
\(M(xf(x);s) = M(f(x);s+1) \).
Let \(H_r(y) := G_r(1-y)\) and consider the Mellin transform
$$\begin{aligned} M(H_r;s) = \int \limits _0^\infty y^s H_r(y) \, \mathrm {d}y. \end{aligned}$$
Substituting \(y = x^2/4^r\) and using Lemma 4.3, we obtain
$$\begin{aligned} M(H_r;s)&=\int \limits _0^1 y^s H_r(y) \, \mathrm {d}y = \frac{2}{4^{r(s+1)}} \int _0^{2^r} x^{2s+1} \hat{p}_r(x) \, \mathrm {d}x \\&=\frac{1}{4^{r(s+1)}} \left( \frac{1}{\pi } 2^{2s+1} \frac{\Gamma \left( \frac{1}{2} \right) \Gamma \left( \frac{(2s+1)+1}{2}\right) }{\Gamma \left( 1 + \frac{2s+1}{2} \right) } \right) ^r\\&= \left( \frac{\Gamma \left( \frac{1}{2} \right) \Gamma (s+1)}{2\pi \Gamma \left( s+\frac{3}{2}\right) } \right) ^r . \end{aligned}$$
It is clear that \(M(H_r;s)\) satisfies a recursion:
$$\begin{aligned} \left( s+\frac{1}{2}\right) ^r M(H_r;s) = s^r M(H_r;s-1). \end{aligned}$$
If \(H_r\) were continuously differentiable, it would follow that
$$\begin{aligned} s^r M(H_r;s-1) = (-1)^r M(\theta ^r H_r;s-1) \end{aligned}$$
and
$$\begin{aligned} \left( s+\frac{1}{2} \right) ^r M(H_r;s) = (-1)^r M \left( \left( \theta + \frac{1}{2} \right) ^r H_r;s \right) . \end{aligned}$$
In other words,
$$M(\theta ^r H_r;s-1) = M \left( \left( \theta + \frac{1}{2} \right) ^r H_r;s \right) ,$$
so that \(H_r(y)\) is annihilated by the differential operator \(\theta ^{r} - y(\theta + 1/2)^r\). This means that \(H_r\) satisfies the same differential equation as \({}_r F_{r-1} \left( \frac{1}{2}, \dots , \frac{1}{2};1, \dots , 1; y \right) \). Though \(H_r\) is not continuously differentiable, giving a similar argument as in the proof of [14, Theorem 2.4] it follows rigorously that \(H_r\) satisfies the same differential equation in a distributional sense.
\(W_r\) for \(|k| \ge 2^r\)
In this part, we compute the value of \(W_r(k;s)\) for \(|k| \ge 2^r\).
Define for \(s \in \mathbb {C}\) the following function of \(z \in \mathbb {C}\):
$$\begin{aligned} F_{r,s}(z) := \int _{z - 2^r}^{z + 2^r} x^s \hat{p}_r(x- z) \, \mathrm {d}x. \end{aligned}$$
Proposition 4.5
(Properties of \(F_{r,s}\))
-
(i)
\(F_{r,s}\) is analytic on \(\mathbb {C} \setminus (-\infty , 2^r]\).
-
(ii)
\(F_{r,s}\) is continuous on \(\{ z \in \mathbb {C} \ :{\text {Im}}(z) \ge 0\}\); in other words for real \(z \ge 0\).
Proof
This follows immediately as we can write
$$\begin{aligned} F_{r,s}(z) = \int \limits _{- 2^r}^{2^r} (x + z)^s \hat{p}_r(x) \, \mathrm {d}x \end{aligned}$$
and notice that \(z \mapsto (x + z)^s\) is holomorphic for \(z \not \in (-\infty , 2^r]\). \(\square \)
First of all, we can find an explicit expression for \(F_{r,s}(z)\) if \(|z|>2^r\). The value of \(F_{r,s}(z)\) when \(|z| \le 2^r\) will follow from the analytic continuation of \(F_{r,s}\), with the help of Proposition 4.5.
Lemma 4.6
For \(|z| > 2^r\), we have
$$F_{r,s}(z) = z^s \cdot {}_{r+1} F_{r} \left( \frac{-s}{2}, \frac{1-s}{2}, \frac{1}{2}, \dots , \frac{1}{2}; 1, \dots , 1; \frac{4^r}{z^2} \right) . $$
Proof
Note that
$$\begin{aligned} F_{r,s}(z)&= \int \limits _{-2^r}^{2^r} (x+z)^s \hat{p}_r(x) \, \mathrm {d}x, \\&= z^s\int \limits _{-2^r}^{2^r} \left( 1+ \frac{x}{z} \right) ^s \hat{p}_r(x) \, \mathrm {d}x, \end{aligned}$$
as \({\text {Re}}\left( 1+ \frac{x}{z} \right) > 0\). Since \(|x/z|<1\), we can write
$$\begin{aligned} \left( 1 + \frac{x}{z} \right) ^s = \sum _{n \ge 0} \left( {\begin{array}{c}s\\ n\end{array}}\right) \frac{x^n}{z^n}, \end{aligned}$$
so that
$$\begin{aligned} F_{r,s}(z) = z^s \sum _{n \ge 0} \left( {\begin{array}{c}s\\ n\end{array}}\right) \frac{1}{z^n} \int \limits _{-2^r}^{2^r} x^n \hat{p}_r(x) \, \mathrm {d}x. \end{aligned}$$
Now using Lemma 4.3, we find
$$\begin{aligned} \int \limits _{-2^r}^{2^r} x^n \hat{p}_r(x) \, \mathrm {d}x&= \frac{1+(-1)^n}{2}\left( \frac{2^n}{\pi } \frac{\Gamma (\frac{1}{2}) \Gamma (\frac{n+1}{2})}{\Gamma (1 + \frac{n}{2})} \right) ^r \\&= {\left\{ \begin{array}{ll} \left( {\begin{array}{c}n\\ n/2\end{array}}\right) ^r \qquad &{}\hbox {if } n\hbox { is even,} \\ 0 \qquad &{}\text {otherwise.} \end{array}\right. } \end{aligned}$$
Thus,
$$\begin{aligned} F_{r,s}(z)&= z^s \sum _{n \ge 0} \left( {\begin{array}{c}s\\ 2n\end{array}}\right) \left( {\begin{array}{c}2n\\ n\end{array}}\right) ^r \frac{1}{z^{2n}}\\&= z^s \sum _{n \ge 0} \left( {\begin{array}{c}s\\ 2n\end{array}}\right) \left( \frac{(\frac{1}{2})_n \cdot 4^n}{n!} \right) ^r \frac{1}{z^{2n}}\\&= z^s \sum _{n \ge 0} \frac{(\frac{1-s}{2})_n (\frac{-s}{2})_n}{n!^2} \left( \frac{(\frac{1}{2})_n}{n!} \right) ^{r-1} \left( \frac{4^r}{z^{2}}\right) ^n\\&= z^s \cdot {}_{r+1} F_{r} \left( \frac{-s}{2}, \frac{1-s}{2}, \frac{1}{2}, \dots , \frac{1}{2}; 1, \dots , 1; \frac{4^r}{z^2} \right) . \end{aligned}$$
\(\square \)
Now for \(|k| > 2^r,\) \(k \in \mathbb {R}\) we can deduce an explicit formula for \(W_r\).
Proof of Theorem 1.4.(i)
Clearly,
$$\begin{aligned} W_r(k;s) = F_{r,s}(|k|), \end{aligned}$$
so that
$$\begin{aligned} W_r(k;s) = |k|^s \cdot {}_{r+1} F_{r} \left( \frac{-s}{2}, \frac{1-s}{2}, \frac{1}{2}, \dots , \frac{1}{2}; 1, \dots , 1; \frac{4^r}{k^2} \right) , \end{aligned}$$
where Lemma 4.6 was used. \(\square \)
We see that, for a fixed \(|k| > 2^r\), the mapping \(s \mapsto W_r(k;s)\) defines an entire function.
Remark 1
If s is a non-negative integer, the hypergeometric sum is terminating. Hence, \(W_r(k;s)\) becomes a polynomial in |k|.
The value of \(W_r(k;s)\) for \(k = \pm 2^r\) when \(r>1\) can be found by taking the corresponding limit.
Proof of Theorem 1.4.(ii)
This follows from the absolute convergence of
$$\begin{aligned} {}_{r+1} F_r \left( a_1, \dots , a_{r+1};b_1, \dots , b_r;z \right) \end{aligned}$$
on the domain \(|z| \le 1\) if
$$\begin{aligned} {\text {Re}}\left( \sum _{i =1}^{r} b_i - \sum _{j=1}^{r+1} a_j \right) >0 \end{aligned}$$
(see [15, p. 156]).
Example 4.7
For \(r = 1\) and \(|k|>2\), Eq. (6) gives
$$\begin{aligned} W_1(k;s) = |k|^s \cdot {}_{2} F_{1} \left( \frac{-s}{2}, \frac{1-s}{2} ; 1; \frac{4}{k^2} \right) \end{aligned}$$
and for \(|k| = 2\), Eq. (7) gives
$$\begin{aligned} W_1(k;s) = \frac{2^s \Gamma (\frac{1}{2} + s)}{\Gamma (1+\frac{s}{2}) \Gamma (\frac{1+s}{2})}, \end{aligned}$$
for \({\text {Re}}(s) > -\frac{1}{2}\). This leads to a second proof of Theorem 1.1.
\(W_r\) for \(|k| < 2^r\)
For \(|k|<2^r\), we write
$$\begin{aligned} W_r(k;s) = \int \limits _{0}^{|k|+2^r} x^s \hat{p}_r(x-|k|) \, \mathrm {d}x + \int _{0}^{-|k| + 2^r} x^s \hat{p}_r(x+|k|) \, \mathrm {d}x. \end{aligned}$$
Note that
$$\begin{aligned} \int _{|k|- 2^r}^{|k|+2^r} x^s \hat{p}_r(x-|k|) \, \mathrm {d}x = \int _{0}^{|k|+2^r} x^s \hat{p}_r(x-|k|) \, \mathrm {d}x + e^{\pi i s} \int _{0}^{-|k|+2^r} x^s \hat{p}_r(x+|k|) \, \mathrm {d}x, \end{aligned}$$
hence
$$\begin{aligned} W_r(k;s)&= \frac{1}{1+e^{\pi i s}} \left( \int \limits _{|k| - 2^r}^{|k|+2^r} x^s \hat{p}_r(x-|k|) \, \mathrm {d}x + \int \limits _{-|k| - 2^r}^{-|k|+2^r} x^s \hat{p}_r(x+|k|) \, \mathrm {d}x \right) \\&= \frac{1}{1+e^{\pi i s}} \left( F_{r,s}(|k|) + F_{r,s}(-|k|) \right) . \end{aligned}$$
We now define the following analytic continuations \(F_{r,s}^{+}\) and \(F_{r,s}^{-}\). Let \(F_{r,s}^{+}(z)\) be the analytic continuation of the function
$$\begin{aligned} z^s \cdot {}_{r+1} F_{r} \left( \frac{-s}{2}, \frac{1-s}{2}, \frac{1}{2}, \dots , \frac{1}{2}; 1, \dots , 1; \frac{4^r}{z^2} \right) \end{aligned}$$
on the complement of the upper half-disk \(D(0,2^r)^\mathsf {c} \cap \mathbb {H}^{+}\) to \(\mathbb {C} \setminus ((-\infty ,0] \cup [2^r, \infty ))\), while \(F_{r,s}^{-}(z)\) is the analytic continuation of the function
$$(-z)^s \cdot {}_{r+1} F_{r} \left( \frac{-s}{2}, \frac{1-s}{2}, \frac{1}{2}, \dots , \frac{1}{2}; 1, \dots , 1; \frac{4^r}{z^2} \right) $$
on the complement of the lower half-disk \(D(0,2^r)^\mathsf {c} \cap \mathbb {H}^{-}\) to \(\mathbb {C} \setminus ((-\infty ,0] \cup [2^r, \infty ))\). Here \(\mathbb {H}^+\) and \(\mathbb {H}^{-}\) denote the (strict) upper- and lower-half planes, respectively.
Using Proposition 4.5, it is clear that for \(|k| < 2^r\) we have
$$\begin{aligned} F_{r,s}(|k|) = F_{r,s}^{+}(|k|) \quad \text { and } \quad F_{r,s}(-|k|) = F_{r,s}^{-}(|k|). \end{aligned}$$
Define
$$\begin{aligned} H_{r,s}(z) := \frac{1}{1+e^{\pi i s}} \left( F_{r,s}^{+}(z) + F_{r,s}^{-}(z) \right) , \end{aligned}$$
which is analytic on \(\mathbb {C} \setminus ((-\infty ,0] \cup [2^r, \infty ))\), with the following motivation.
Theorem 4.8
For \(|k| < 2^r\),
$$\begin{aligned} W_r(k;s) = H_{r,s}(|k|). \end{aligned}$$
In fact, in the case of real s, we have an additional structure.
Lemma 4.9
For real positive s, we have
$$\begin{aligned} \overline{F_{r,s}(-|k|)} = e^{-\pi i s}F_{r,s}(|k|). \end{aligned}$$
Proof
For notational convenience assume \(0 \le k < 2^r\). Then
$$\begin{aligned} F_{r,s}(-k) = \int \limits _{k}^{2^r} (x-k)^s \hat{p}_r(x) \,\mathrm {d}x + e^{\pi i s} \int _{-k}^{2^r} (x+k)^s \,\mathrm {d}x \end{aligned}$$
and taking the complex conjugate, we deduce that
$$\begin{aligned} \overline{F_{r,s}(-k)}&= e^{-\pi i s}\int \limits _{-2^r}^{-k} (x+k)^s \hat{p}_r(x) \,\mathrm {d}x + e^{-\pi i s} \int \limits _{-k}^{2^r} (x+k)^s \,\mathrm {d}x\\&= e^{-\pi i s}F_{r,s}(k), \end{aligned}$$
which is the desired claim. \(\square \)
Now we establish Theorem 1.5.
Proof of Theorem 1.5
Observe that for real s Theorem 1.4 coincides with the statement of Theorem 1.5, as
$$\begin{aligned} {}_{r+1} F_{r} \left( \frac{-s}{2}, \frac{1-s}{2}, \frac{1}{2}, \dots , \frac{1}{2}; 1, \dots , 1; \frac{4^r}{k^2} \right) \end{aligned}$$
is real-valued for \(|k| \ge 2^r\). Therefore, we can assume \(|k| < 2^r\). Then by Lemma 4.9 and Theorem 4.8 we obtain
$$\begin{aligned} W_r(k;s)&= \frac{1}{1+e^{\pi i s}} (F_{r,s}(|k|) + e^{\pi i s}\overline{F_{r,s}(|k|)}) \\&= {\text {Re}} F_{r,s}(|k|) + \tan \left( \frac{\pi s}{2} \right) {\text {Im}} F_{r,s}(|k|). \end{aligned}$$
\(\square \)
Remark 2
Note that \(F_{r,s}^{+}(z), F_{r,s}^{-}(z)\) and hence also \(H_{r,s}(z)\) satisfy the same differential equation as
$$\begin{aligned} {}_{r+1} F_{r} \left( \frac{-s}{2}, \frac{1-s}{2}, \frac{1}{2}, \dots , \frac{1}{2}; 1, \dots , 1; \frac{4^r}{z^2} \right) . \end{aligned}$$
It is clear from the definition that \(F_{r,s}^{+}(z) = F_{r,s}^{-}(-z)\) for \(z \in D(0,2^r)^\mathsf {c} \cap \mathbb {H}^{+}\); therefore, from the theory of analytic continuation we conclude with the following.
Proposition 4.10
For all \(z \in \mathbb {H}^{+},\)
$$\begin{aligned} F_{r,s}^{+}(z) = F_{r,s}^{-}(-z). \end{aligned}$$
Proposition 4.11
(Real differentiability at \(k = 2^r\)) Suppose \({\text {Re}}(s) > n - \frac{r}{2}\). Then \(k \mapsto W_r(k;s)\) is n times (real) differentiable at \(k = 2^r\).
Proof
We will prove that \(k \mapsto W_r(k;s)\) is differentiable at \(k = 2^r \) if \({\text {Re}}(s) > \frac{1}{2}\). The general case follows by induction. For \(k > 2^r\), we have \(W_r(k;s) = F_{r,s}(k)\), so that the right derivative at \(k = 2^r\) is given by
$$\begin{aligned} \int _{-2^r}^{2^r} s(x + 2^r)^{s-1} \hat{p}_r(x) \, \mathrm {d}x = s F_{r,s-1}(2^r) \end{aligned}$$
when \({\text {Re}}(s) > 1 - \frac{r}{2}\). For \(|k|< 2^r\), we have
$$\begin{aligned} W_r(k;s) = \frac{1}{1 + e^{\pi i s}} \left( F_{r,s}(k) + F_{r,s}(-k) \right) . \end{aligned}$$
Therefore, the left derivative at \(k = 2^r\) is given by
$$\begin{aligned} \frac{1}{1+e^{\pi i s}} \left( s F_{r,s-1}(2^r) + e^{\pi i s} sF_{r,s-1}(2^r) \right) = s F_{r,s-1}(2^r). \end{aligned}$$
This means that at \(k = 2^r\) the left and right derivatives coincide. \(\square \)
Proposition 4.12
(Real derivatives at \(k = 0\)) The right real derivatives \((\mathrm {d}^j/\mathrm {d}k^j)^{+}\) for \(0 \le j \le \lfloor {{\text {Re}}(s)}\rfloor \) of \(k \mapsto W_r(k;s)\) at \(k=0\) are given by
$$\begin{aligned} \frac{\mathrm {d}^jW_r(k;s)}{\mathrm {d}k^j}^{+}\Biggr |_{k = 0} = {\left\{ \begin{array}{ll} 0 \qquad &{}\hbox {if }j\hbox { is odd,}\\ \frac{\Gamma (s+1)^{r+1} }{\Gamma (s-j+1) \Gamma (1+\frac{s}{2})^{2r}} \qquad &{}\hbox {if}\, j\hbox { is even.} \end{array}\right. } \end{aligned}$$
Proof
For \(k \ge 0\), we have
$$\begin{aligned} \frac{\mathrm {d}^jW_r(k;s)}{\mathrm {d}k^j}^{+} = s(s-1)\frac{\mathrm {d}^{j-2} W_r(k;s)}{\mathrm {d}k^{j-2}}^{+}. \end{aligned}$$
By induction,
$$\begin{aligned} \frac{\mathrm {d}^jW_r(k;s)}{\mathrm {d}k^j}^{+}&= s(s-1) \cdots (s-j+1) W_r(k;s-j)\\&= \frac{\Gamma (s+1)}{\Gamma (s-j+1)}W_r(k;s-j) \end{aligned}$$
for even j. Finally, notice that
$$\begin{aligned} \frac{\mathrm {d}W_r(k;s)}{\mathrm {d}k}^{+}\Bigr |_{k = 0} = 0. \end{aligned}$$
\(\square \)
We now return to Theorem 1.1.(iii) and give another proof of it.
Proof of Theorem 1.1.(iii)
For \(|z|>2\), we have that
$$\begin{aligned} F_{1,s}(z) = z^s \cdot {}_{2} F_{1} \left( \frac{-s}{2}, \frac{1-s}{2} ; 1; \frac{4}{z^2} \right) \end{aligned}$$
implying that \(F_{1,s}\) is a solution to the differential equation
$$\begin{aligned} \left( -\frac{1}{4}z^2+1 \right) \frac{dY^2}{dz^2}+\left( \frac{1}{2}sz - \frac{1}{4}z\right) \frac{dY}{dz} -\frac{1}{4}s^2Y = 0. \end{aligned}$$
(17)
Equation (17) has a basis of solutions
$$\begin{aligned} Y_0(z)&= {}_{2} F_{1} \left( \frac{-s}{2}, \frac{-s}{2} ; \frac{1}{2}; \frac{z^2}{4} \right) ,\\ Y_1(z)&= z \cdot {}_{2} F_{1} \left( \frac{1-s}{2}, \frac{1-s}{2} ; \frac{3}{2}; \frac{z^2}{4} \right) , \end{aligned}$$
so that \(F_{1,s}^{+}(z) = C_0 Y_0(z) + C_1 Y_1(z)\), \(F_{1,s}^{-}(z) = \tilde{C_0} Y_0(z) + \tilde{C_1} Y_1(z)\) and \(H_{1,s}(z) = D_0 Y_0(z) + D_1 Y_1(z)\) for constants \(C_0, C_1, \tilde{C_0}, \tilde{C_1}, D_0\) and \(D_1\) depending only on s. Using Proposition 4.10, it follows that \(C_1 = -\tilde{C_1}\), hence \(D_1 = 0\).
As \(\lim _{z \rightarrow 0} H_{1,s}(z) = W_1(0;s)\) and we have
$$W_1(0;s) = \frac{2^{s} \Gamma (\frac{1}{2}) \Gamma (\frac{s+1}{2})}{\pi \Gamma (1+\frac{s}{2})},$$
it follows that
$$\begin{aligned} H_{1,s}(z) = \frac{2^{s} \Gamma \left( \frac{1}{2}\right) \Gamma \left( \frac{s+1}{2}\right) }{\pi \Gamma \left( 1+\frac{s}{2}\right) } \cdot {}_{2} F_{1} \left( \frac{-s}{2}, \frac{-s}{2} ; \frac{1}{2}; \frac{z^2}{4} \right) . \end{aligned}$$
Thus,
$$\begin{aligned} W_{1}(k;s) = \frac{4^s \Gamma (\frac{1 + s}{2})^2}{\pi \Gamma (1 + s)} \cdot {}_{2} F_{1} \left( \frac{-s}{2}, \frac{-s}{2} ; \frac{1}{2}; \frac{k^2}{4} \right) \end{aligned}$$
for \(|k|<2^r\).
Remark 3
In the proof above we see that \(H_{1,s}(z)\) extends to an analytic function in a neighborhood of \(z = 0\). We expect that this will only happen in the case \(r = 1\).
Remark 4
We use the symmetry of \(H_{1,s}\) to show that \(D_1 = 0\). Clearly, using
$$\begin{aligned} \lim _{z \rightarrow 2} H_{1,s}(z) = W_1(2;s) = 2^s \cdot {}_{2} F_{1} \left( \frac{-s}{2}, \frac{1-s}{2}; 1 ;1 \right) \end{aligned}$$
would lead to the same conclusion.
For \(r=2\) we follow the above strategy of the case \(r = 1\).
Proof of Theorem 1.6
We have, for \(|z|>4\),
$$\begin{aligned} F_{2,s}(z) = z^s \cdot {}_{3} F_{2} \left( \frac{-s}{2}, \frac{1-s}{2}, \frac{1}{2} ; 1, 1; \frac{16}{z^2} \right) , \end{aligned}$$
hence \(F_{2,s}\) is a solution to the differential equation
$$\begin{aligned}&\left( -\frac{1}{8}z^3+2z\right) \frac{\mathrm {d}^3Y}{\mathrm {d}z^3} + \left( \frac{3}{8}sz^2 - \frac{3}{8}z^2 - 2s + 2 \right) \frac{\mathrm {d}^2Y}{\mathrm {d}z^2}\nonumber \\&\quad +\left( -\frac{3}{8}s^2z + \frac{3}{8}sz - \frac{1}{8}z \right) \frac{\mathrm {d}Y}{\mathrm {d}z} + \frac{1}{8}s^3Y =0. \end{aligned}$$
(18)
A basis of solutions for (18) is given by
$$\begin{aligned} Y_0(z)&= {}_{3} F_{2} \left( \frac{-s}{2}, \frac{-s}{2}, \frac{-s}{2} ; \frac{1-s}{2}, \frac{1}{2}; \frac{z^2}{16} \right) ,\\ Y_1(z)&= z \cdot {}_{3} F_{2} \left( \frac{1-s}{2}, \frac{1-s}{2}, \frac{1-s}{2} ; 1-\frac{s}{2}, \frac{3}{2}; \frac{z^2}{16} \right) , \\ Y_2(z)&= z^{1+s} \cdot {}_{3} F_{2} \left( \frac{1}{2}, \frac{1}{2}, \frac{1}{2} ; 1+\frac{s}{2}, \frac{3}{2} + \frac{s}{2}; \frac{z^2}{16} \right) . \end{aligned}$$
It follows that \(H_{2,s}(z) = D_0 Y_0(z) + D_1 Y_1(z) + D_2 Y_2(z)\) for some constants \(D_0,D_1\) and \(D_2\) depending only on s. Using the same argument as in the second proof of Theorem 1.5, it follows that \(D_1 = 0\). Since
$$\begin{aligned} \lim _{z \rightarrow 0}H_{2,s}(z) = W_{2}(0;s) = \frac{\Gamma (s+1)^2}{\Gamma (\frac{s}{2}+1)^4} \end{aligned}$$
and
$$\begin{aligned} \lim _{z \rightarrow 4}H_{2,s}(z) = W_{2}(4;s) = 4^s \cdot {}_{3} F_{2} \left( \frac{-s}{2}, \frac{1-s}{2}, \frac{1}{2} ; 1, 1; 1 \right) , \end{aligned}$$
we find out that
$$\begin{aligned} {\begin{matrix} H_{2,s}(z) =&{}\frac{\Gamma (s+1)^2}{\Gamma (\frac{s}{2}+1)^4} \cdot Y_0(z) \\ {} &{}-\frac{4^s {}_3 F_{2} \left( \frac{1}{2},\frac{-s}{2},\frac{1-s}{2};1,1;1 \right) - \frac{\Gamma (s+1)^2}{\Gamma (\frac{s}{2}+1)^4} {}_3 F_{2} \left( \frac{-s}{2} ,\frac{-s}{2},\frac{-s}{2}; \frac{1-s}{2},\frac{1}{2}; 1 \right) }{4^{1+s} \cdot {}_3 F_{2} \left( \frac{1}{2},\frac{1}{2} ,\frac{1}{2};1 +\frac{s}{2}, \frac{3}{2} + \frac{s}{2} ; 1 \right) } \cdot Y_2(z).\\ \end{matrix}} \end{aligned}$$
We can further simplify the coefficient in front of \(Y_2\). We have the following hypergeometric identity for the special value at \(z=1\) (see [8]):
$$\begin{aligned} {}_3 F_{2} \left( a_1,a_2,a_3;b_1,b_2;1 \right) =&\frac{\Gamma (1-a_2)\Gamma (a_3-a_1)\Gamma (b_1)\Gamma (b_2)}{\Gamma (a_1-a_2+1)\Gamma (a_3)\Gamma (b_1-a_1)\Gamma (b_2-a_1)} \\ {}&\quad \times {}_3 F_{2} \left( a_1,a_1-b_1+1,a_1-b_2+1;a_1-a_2+1,a_1-a_3+1;1 \right) \\&+\frac{\Gamma (1-a_2)\Gamma (a_1-a_3)\Gamma (b_1)\Gamma (b_2)}{\Gamma (a_1)\Gamma (a_3-a_2+1)\Gamma (b_1-a_3)\Gamma (b_2-a_3)} \\ {}&\quad \times {}_3 F_{2} \left( a_3,a_3-b_1+1,a_3-b_2+1;a_3-a_1+1,a_3-a_2+1;1 \right) . \end{aligned}$$
Applying it with \(a_1 = 1/2, a_2 = 1/2-s/2, a_3 = -s/2, b_1 = 1\) and \( b_2 = 1\) gives
$$\begin{aligned} {\begin{matrix} {}_3 F_{2} \left( \frac{1}{2},\frac{1-s}{2},\frac{-s}{2};1,1;1 \right) =\frac{\Gamma (\frac{1}{2}+\frac{s}{2})\Gamma (\frac{-s}{2}-\frac{1}{2})}{\Gamma (\frac{s}{2}+1)\Gamma (\frac{-s}{2}) \pi } &{} {}_3 F_{2} \left( \frac{1}{2},\frac{1}{2},\frac{1}{2};1+\frac{s}{2},\frac{3}{2}+\frac{s}{2};1 \right) \\ +\frac{\Gamma (\frac{1}{2}+\frac{s}{2})^2}{\Gamma (1+\frac{s}{2})\pi } &{} {}_3 F_{2} \left( \frac{-s}{2},\frac{-s}{2},\frac{-s}{2};\frac{1-s}{2},\frac{1}{2};1\right) . \end{matrix}} \end{aligned}$$
Hence \(H_{2,s}\) can be written as
$$\begin{aligned} H_{2,s}(z)&= \frac{1}{2 \pi } \frac{\tan (\frac{\pi s}{2})}{s+1} z^{1+s} \cdot {}_3 F_{2} \left( \frac{1}{2},\frac{1}{2} ,\frac{1}{2};1 +\frac{s}{2}, \frac{3}{2} + \frac{s}{2} ; \frac{z^2}{16} \right) \\&\quad + \frac{\Gamma (s+1)^2}{\Gamma (\frac{s}{2}+1)^4} \cdot {}_3 F_{2} \left( -\frac{s}{2} ,-\frac{s}{2},-\frac{s}{2}; \frac{1-s}{2},\frac{1}{2}; \frac{z^2}{16} \right) . \end{aligned}$$
It remains to apply Theorem 4.8 to arrive at the formula for \(W_2(k;s)\). \(\square \)
Remark 5
Notice that \(H_{2,s}(z)\) is not anymore analytic at \(z = 0\).
For odd positive values of s we need to compute the corresponding limits. We present an explicit formula, which can no longer be written in terms of hypergeometric functions.
Theorem 4.13
For odd positive integers n and \(|k|<4\),
$$\begin{aligned} W_2(k;n) = (-1)^{\frac{n+1}{2}} \frac{2^n n!}{\pi ^3} \cdot G_{3,3}^{2,3} \left( 1 + \frac{n}{2}, 1 + \frac{n}{2},1 + \frac{n}{2};0,\frac{n+1}{2},\frac{1}{2};\frac{k^2}{16}\right) . \end{aligned}$$
(19)
Proof
For \({\text {Re}}(s)>-1\), not an odd integer, write
$$\begin{aligned} \cos \left( \frac{\pi s}{2} \right) W_2(k;s) = \frac{1}{2 \pi } \frac{\sin \left( \frac{\pi s}{2} \right) }{s+1} F_1 + \frac{\Gamma (s+1)^2 \pi }{\Gamma (\frac{s}{2} + 1)^4 \Gamma (\frac{1+s}{2})} \frac{F_2}{\Gamma (\frac{1-s}{2})}, \end{aligned}$$
where
$$\begin{aligned} F_1 = |k|^{1+s} \cdot {}_3 F_{2} \left( \frac{1}{2},\frac{1}{2} ,\frac{1}{2};1 +\frac{s}{2}, \frac{3}{2} + \frac{s}{2} ; \frac{k^2}{16} \right) \end{aligned}$$
and
$$\begin{aligned} F_2 = {}_3 F_{2} \left( -\frac{s}{2} ,-\frac{s}{2},-\frac{s}{2}; \frac{1-s}{2},\frac{1}{2}; \frac{k^2}{16} \right) . \end{aligned}$$
Using Mathematica, we have
$$\begin{aligned} \cos \left( \frac{\pi s}{2} \right)&G = -\frac{2^{-1-s} \pi ^2}{\Gamma (2+s)} F_1 + \sqrt{\pi } \Gamma \left( -\frac{s}{2} \right) ^3 \frac{F_2}{\Gamma (\frac{1-s}{2})}, \end{aligned}$$
where
$$\begin{aligned} G = G_{3,3}^{2,3} \left( 1 + \frac{s}{2}, 1 + \frac{s}{2},1 + \frac{s}{2};0,\frac{s+1}{2},\frac{1}{2};\frac{k^2}{16}\right) . \end{aligned}$$
Thus,
$$\begin{aligned} W_2(k;s)&= \left( \frac{\tan \left( \frac{\pi s}{2} \right) }{2 \pi (s+1)} + \frac{2^{-1-s} \pi ^{5/2} \Gamma (s+1)^2}{\cos \left( \frac{\pi s}{2} \right) \Gamma (2+s)\Gamma (\frac{s}{2}+1)^4 \Gamma (\frac{1+s}{2}) \Gamma (-\frac{s}{2})^3}\right) F_1\\&\quad + \left( \frac{\Gamma (s+1)^2 \sqrt{\pi }}{\Gamma \left( \frac{s}{2} + 1\right) ^4 \Gamma \left( \frac{1+s}{2} \right) \Gamma \left( -\frac{s}{2}\right) ^3}\right) G. \end{aligned}$$
Taking the limit \(s \rightarrow n\), for n odd and positive, gives the result. \(\square \)
Using the explicit expression for \(W_2(k;s)\) in Theorem 1.6, the Mahler measure of the Laurent polynomial \(k + (x+x^{-1})(y+y^{-1})\) can be computed for \(|k|<4\).
Corollary 4.14
[[16, Theorem 3.1]] For \(|k| < 4\),
$$\begin{aligned} {\text {m}} (k + (x+x^{-1})(y+y^{-1})) = \frac{|k|}{4} {}_3 F_{2} \left( \frac{1}{2},\frac{1}{2} ,\frac{1}{2};1, \frac{3}{2} ; \frac{k^2}{16} \right) . \end{aligned}$$
(20)
Proof
The Mahler measure of \(k + (x+x^{-1})(y+y^{-1})\) can be recovered as
$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}s} W_2(k;s) |_{s = 0}. \end{aligned}$$
Note that in the neighborhood of \(s=0\),
$$\begin{aligned}&\frac{\Gamma (s+1)^2}{\Gamma (\frac{s}{2}+1)^4} = 1 + \mathcal {O}(s^2),\\&{}_3 F_{2} \left( \frac{-s}{2} ,\frac{-s}{2},\frac{-s}{2}; \frac{1-s}{2},\frac{1}{2}; \frac{k^2}{16} \right) = 1 + \mathcal {O}(s^2) \end{aligned}$$
and
$$\begin{aligned} \frac{\tan (\frac{\pi s}{2})}{s+1} = \frac{\pi s}{2}+\mathcal {O}(s^2). \end{aligned}$$
Thus,
$$\begin{aligned}\frac{\mathrm {d}W_{2}(k;s)}{\mathrm {d}s}\Bigr |_{s = 0}&= \frac{|k|}{4} {}_3 F_{2} \left( \frac{1}{2},\frac{1}{2} ,\frac{1}{2};1, \frac{3}{2} ; \frac{k^2}{16} \right) . \end{aligned}$$
\(\square \)
Remark 6
The Mahler measure \({\text {m}} (k + (X+X^{-1})(Y+Y^{-1}))\) can also be written as the double integral
$$\begin{aligned} {\text {m}} (k + (x+x^{-1})(y+y^{-1})) = \frac{|k|}{8 \pi }\int _{[0,1]^2} \frac{\mathrm {d}x_1 \, \mathrm {d}x_2}{\sqrt{x_1 x_2(1-x_2)(1-x_1x_2\frac{k^2}{16})}} \end{aligned}$$
(21)
using Corollary 4.14 and [7, Eq. (16.5.2)].
For the case \(r \ge 3\) and \(|k|<2^r\) we expect that \(W_r(k;s)\) cannot be written anymore as a linear combination of hypergeometric functions (not their real and imaginary parts as in Theorem 1.5). We now give the proof of Theorem 1.7.
Proof of Theorem 1.7
For \(|k| < 8\) and real \(s>1\), we have
$$\begin{aligned} W_3(k;s)&= |k|^s{\text {Re}} \left( {}_{4} F_{3} \left( \frac{-s}{2}, \frac{1-s}{2}, \frac{1}{2}, \dots , \frac{1}{2}; 1, \dots , 1; \frac{64}{k^2} \right) \right) \nonumber \\&\quad + \tan \left( \frac{ \pi s}{2} \right) |k|^s {\text {Im}} \left( {}_{4} F_{3} \left( \frac{-s}{2}, \frac{1-s}{2}, \frac{1}{2}, \dots , \frac{1}{2}; 1, \dots , 1; \frac{64}{k^2} \right) \right) . \end{aligned}$$
(22)
Let \(\epsilon > 0\). Then for \(|k|<8\) we have
$$\begin{aligned} |k|^s {}_{4} F_{3} \left( \frac{-s}{2}, \frac{1-s}{2}, \frac{1}{2},\frac{1}{2} + \epsilon ; 1, 1 , 1; \frac{4^r}{k^2} \right) = \alpha _1(s)Y_1(k) + \dots + \alpha _4(s)Y_4(k), \end{aligned}$$
where
$$\begin{aligned} Y_1(k)&= {}_4 F_{3} \left( \frac{-s}{2},\frac{-s}{2},\frac{-s}{2},\frac{-s}{2};\frac{1-s}{2}-\epsilon ,\frac{1-s}{2},\frac{1}{2};\frac{k^2}{64}\right) ,\\ Y_2(k)&= |k| \cdot {}_4 F_{3} \left( \frac{1-s}{2},\frac{1-s}{2},\frac{1-s}{2},\frac{1-s}{2};\frac{3}{2},1 - \frac{s}{2},1 - \epsilon - \frac{s}{2};\frac{k^2}{64}\right) ,\\ Y_3(k)&= |k|^{1+s} {}_4 F_{3} \left( \frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1-\epsilon ,1+\frac{s}{2},\frac{3+s}{2};\frac{k^2}{64}\right) ,\\ Y_4(k)&= |k|^{1+s+2 \epsilon } {}_4 F_{3} \left( \frac{1}{2} + \epsilon ,\frac{1}{2} + \epsilon ,\frac{1}{2} + \epsilon ,\frac{1}{2} + \epsilon ;1+\epsilon ,1+\frac{s}{2} + \epsilon ,\frac{3+s}{2}+\epsilon ;\frac{k^2}{64}\right) \end{aligned}$$
and
$$\begin{aligned} \alpha _1(s)&= \frac{(8 i)^s \Gamma \left( \frac{s+1}{2}\right) \Gamma \left( \epsilon +\frac{s}{2}+\frac{1}{2}\right) }{\Gamma \left( \epsilon +\frac{1}{2}\right) \Gamma \left( \frac{1-s}{2}\right) \Gamma \left( \frac{s}{2}+1\right) ^3}, \\ \alpha _2(s)&= \frac{i^{s+1} 2^{3 s-2} \Gamma \left( \frac{s}{2}\right) \Gamma \left( \epsilon +\frac{s}{2}\right) }{\Gamma \left( \epsilon +\frac{1}{2}\right) \Gamma \left( -\frac{s}{2}\right) \Gamma \left( \frac{s+1}{2}\right) ^3},\\ {} \alpha _3(s)&=-\frac{i \Gamma (\epsilon ) \Gamma \left( -\frac{s}{2}-\frac{1}{2}\right) }{8 \pi ^{3/2} \Gamma \left( \epsilon +\frac{1}{2}\right) \Gamma \left( \frac{1-s}{2}\right) },\\ \alpha _4(s)&= -\frac{i (-1)^{-\epsilon } 8^{-2\epsilon -1} \Gamma (-\epsilon ) \Gamma \left( -\epsilon -\frac{s}{2}-\frac{1}{2}\right) \Gamma \left( -\epsilon -\frac{s}{2}\right) }{\sqrt{\pi } \Gamma \left( \frac{1}{2}-\epsilon \right) ^3 \Gamma \left( \frac{1-s}{2}\right) \Gamma \left( -\frac{s}{2}\right) }. \end{aligned}$$
We expand \(\alpha _3(s)\) and \(\alpha _4(s)\) in powers of \(\epsilon \). Note that \(\Gamma (\epsilon ) = \Gamma (\epsilon +1)/ \epsilon \), so that for s fixed
$$\begin{aligned} \alpha _3(s) = \frac{1}{\epsilon } \cdot \frac{i}{4 \pi ^2 (1+s)} + \mathcal {O}(1) \end{aligned}$$
and
$$\begin{aligned} \alpha _4(s) = \frac{1}{\epsilon } \cdot \frac{-i}{4 \pi ^2 (1+s)} + \mathcal {O}(1), \end{aligned}$$
hence
$$\begin{aligned} \lim _{\epsilon \rightarrow 0}&\left( \alpha _3(s) Y_3(k) + \alpha _4(s) Y_4(k) \right) =\lim _{\epsilon \rightarrow 0} (\alpha _3(s) \\&+ \alpha _4(s)) Y_3(k) - \lim _{\epsilon \rightarrow 0} \alpha _4(s) \epsilon \left( \frac{Y_3(k) - Y_4(k)}{\epsilon } \right) . \end{aligned}$$
By L’Hôpital’s rule,
$$\begin{aligned} \lim _{\epsilon \rightarrow 0}&(\alpha _3(s) + \alpha _4(s)) = \frac{\mathrm {d}}{\mathrm {d}\epsilon } \epsilon (\alpha _3(s) + \alpha _4(s)) |_{\epsilon =0}\\ =&-\frac{1}{4 \pi (s+1)} + i\frac{1}{4 \pi ^2(s+1)} \left( \psi (1) \right. \\&\left. + \psi \left( \frac{-s}{2}\right) +\psi \left( \frac{-s-1}{2} \right) - 3 \psi \left( \frac{1}{2} \right) + \log (256) \right) , \end{aligned}$$
where \(\psi \) is the digamma function. Furthermore, notice that
$$\begin{aligned}&G_{4,4}^{2,4} \left( \frac{2+s}{2}, \frac{2+s}{2}, \frac{2+s}{2}, \frac{2+s}{2};\frac{1+s}{2}, \frac{1+s}{2}, 0 ,\frac{1}{2} ;\frac{k^2}{64} \right) \\&= \frac{\mathrm {d}}{\mathrm {d}\epsilon } \left( - \frac{\Gamma (1-\epsilon ) \Gamma (\frac{1}{2} + \epsilon )^4}{8^{1+s+2 \epsilon }\Gamma (\frac{3+s}{2} + \epsilon ) \Gamma (1 + \frac{s}{2} + \epsilon )} Y_4(k)+\frac{\Gamma (1+\epsilon ) \Gamma (\frac{1}{2})^4}{8^{1+s}\Gamma (\frac{3+s}{2}) \Gamma (1+\frac{s}{2})} Y_3(k)\right) \Biggr |_{\epsilon = 0}\\&= \frac{\pi ^2}{8^{1+s}\Gamma (\frac{3+s}{2}) \Gamma (1+\frac{s}{2})} \left( \frac{\mathrm {d}}{\mathrm {d}\epsilon } (Y_3(k) - Y_4(k))\Bigr |_{\epsilon =0} - C Y_3(k) \Bigr |_{\epsilon = 0} \right) , \end{aligned}$$
where
$$\begin{aligned} C=-2\psi (1) + 4 \psi \left( \frac{1}{2}\right) - \psi \left( 1+\frac{s}{2}\right) - \psi \left( \frac{3+s}{2}\right) - \log (64). \end{aligned}$$
Thus we can write
$$\begin{aligned} \lim _{\epsilon \rightarrow 0}&\left( \alpha _3(s) Y_3(k) + \alpha _4(s) Y_4(k) \right) \\ {}&= \left( \frac{-1}{4 \pi (s+1)} + i\frac{ \cot (\pi s)}{2 \pi (s+1)} \right) |k|^{1+s} {}_4 F_{3} \left( \frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1+\frac{s}{2},\frac{3+s}{2};\frac{k^2}{64}\right) \\&\quad + i \frac{4^{s} \Gamma (1+s)}{\pi ^{7/2}} G_{4,4}^{2,4} \left( \frac{2+s}{2}, \frac{2+s}{2}, \frac{2+s}{2}, \frac{2+s}{2};\frac{1+s}{2}, \frac{1+s}{2}, 0 ,\frac{1}{2} ;\frac{k^2}{64} \right) . \end{aligned}$$
Finally, using (22) we arrive at the result. \(\square \)
We can alternatively represent the Meijer G-function in (8) as the following triple integral.
Proposition 4.15
For \({\text {Re}}(s)>-1\) and \(0< |k| < 8\),
$$\begin{aligned}&G^{2,4}_{4,4} \left( \frac{2+s}{2},\frac{2+s}{2},\frac{2+s}{2},\frac{2+s}{2};\frac{1+s}{2}, \frac{1+s}{2},0,\frac{1}{2}; \frac{k^2}{64}\right) \\&\quad = \frac{ \sqrt{\pi }|k|^{1+s}}{\Gamma (1 +s) 2^{3+2s}}\int _{[0,1]^3} \frac{ (1-x_2)^{\frac{s}{2}} (1-x_3)^{\frac{s-1}{2}} \, \mathrm {d}x_1 \, \mathrm {d}x_2 \, \mathrm {d}x_3}{\sqrt{x_1 x_2 x_3 (1-x_1)(1-x_1 + \frac{k^2}{64}x_1x_2x_3)}}. \end{aligned}$$
Proof
First, for all \(z \in \mathbb {C}\) we have
$$\begin{aligned}&G^{2,4}_{4,4} \left( \frac{2+s}{2},\frac{2+s}{2},\frac{2+s}{2},\frac{2+s}{2};\frac{1+s}{2}, \frac{1+s}{2},0,\frac{1}{2}; z \right) \\&= z^{\frac{1+s}{2}} G^{2,4}_{4,4} \left( \frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};0, 0,-\frac{1+s}{2},-\frac{s}{2}; z\right) . \end{aligned}$$
Applying Nesterenko’s theorem [17, Proposition 1] with \(a_0 = \dots = a_3 = \frac{1}{2}\), \(b_1 = 1\), \(b_2 =1 + \frac{s}{2}\) and \(b_3 = \frac{3+s}{2}\) to the right hand side gives
$$\begin{aligned} G^{2,4}_{4,4}&\left( \frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};0, 0,-\frac{1+s}{2},-\frac{s}{2}; z\right) \\&= \frac{2^s \sqrt{\pi }}{\Gamma (1+s)}\int _{[0,1]^3} \frac{ (1-x_2)^{\frac{s-1}{2}} (1-x_3)^{\frac{s}{2}} \mathrm {d}x_1 \mathrm {d}x_2 \mathrm {d}x_3}{\sqrt{x_1 x_2 x_3 (1-x_1)(1-x_1 + \frac{k^2}{64}x_1x_2x_3)}}. \end{aligned}$$
\(\square \)
As a consequence of Theorem 1.7 we can compute the Mahler measure of the polynomial \(k + (x+x^{-1})(y+y^{-1})(z+z^{-1})\) for \(|k|<8\).
Corollary 4.16
The Mahler measure of \(k + (x+x^{-1})(y+y^{-1})(z+z^{-1})\) for \(|k|<8\) is given by
$$\begin{aligned} \frac{1}{2 \pi ^{5/2}} G_{4,4}^{2,4} \left( 1, 1, 1, 1;\frac{1}{2}, \frac{1}{2}, 0 ,\frac{1}{2} ;\frac{k^2}{64} \right) . \end{aligned}$$
(23)
Proof
We expand Eq. (8) in s. Note that
$$\begin{aligned}&\frac{\tan (\frac{\pi s}{2})^2}{4 \pi (1+s)} = \mathcal {O}(s^2),\\&\quad {}_4 F_{3} \left( \frac{-s}{2},\frac{-s}{2},\frac{-s}{2},\frac{-s}{2};\frac{1-s}{2},\frac{1-s}{2},\frac{1}{2};\frac{k^2}{64}\right) = \mathcal {O}(s^2), \end{aligned}$$
and
$$\begin{aligned} \frac{4^s \tan (\frac{\pi s}{2}) \Gamma (s+1)}{\pi ^{7/2}}&=\frac{1}{2 \pi ^{5/2}}s+\mathcal {O}(s^2). \end{aligned}$$
Hence,
$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}s}W_3(k;s) |_{s = 0}&= \frac{1}{2 \pi ^{5/2}} G_{4,4}^{2,4} \left( 1, 1, 1, 1;\frac{1}{2}, \frac{1}{2}, 0 ,\frac{1}{2} ;\frac{k^2}{64} \right) . \end{aligned}$$
\(\square \)
We can further write this Mahler measure for \(0<|k|<8\) as the triple integral (compare with equation (21)):
$$\begin{aligned}&{\text {m}}(k + (x+x^{-1})(y+y^{-1})(z+z^{-1})) \nonumber \\&\qquad =\frac{|k|}{16 \pi ^2} \int _{[0,1]^3} \frac{\mathrm {d}x_1 \mathrm {d}x_2 \mathrm {d}x_3}{\sqrt{x_1 x_2 x_3(1-x_1)(1-x_2)(1-x_1 + \frac{k^2}{64} x_1x_2x_3)}}. \end{aligned}$$