Here are some preliminaries. Let's start with a generic issue:

For given real exponents $1\le p_1\le \dots\le p_r$ and $q_1,\dots,q_r$, how to write the second derivative of the function $$F(x):=\log\prod_{j=1}^r (1-x^{p_j})^{-q_j}=-\sum_{j=1}^r q_j \log(1-x^{p_j }),\qquad x\in(0,1) $$ without opening the Pandora's box of derivatives of quotients?

Notice that the above $F$ has no terms $1+x^p$, yet of course we may write any of them as ${1-x^{2p}\over 1-x^p}$, so that this formulation actually includes your problem. Precisely, in your situation, $r=7$, and $p:=(2n,2n+1,2n+2,4n-1,4n,4n+2,8n-2)$ with $q:=(+1,-2,+1,+1,-1,+1,-1)$ -just for fun we may think these data as describing $7$ point charges $q_j$ located at positions $p_j$. (It will appear that these particular exponents $(p,q)$ from your function have a remarkable combinatorial property).

To describe more clearly the dependence from $(p,q)$ it is convenient to introduce, for $s>0$, the function $$\varphi(s):=-\log(1-e^{-s})\ .$$

Then, given the data $(p,q)$, consider the distribution $m\in\mathcal{D'}(\mathbb{R}_+)$ defined by $f\mapsto\langle m,f\rangle:= \sum_{j=1}^r q_j f(p_j)$, just a linear combination of evaluations at the points $p_j$: $$m:=\sum_{j=1}^r q_j\delta_{p_j}.$$ Introducing a parameter $t>0$, we then have a function $\Phi=\Phi_m$ defined by the pairing w.r.to the variable $s\in\mathbb{R}_+$: $$\Phi(t) :=\big \langle m, \varphi(ts) \big\rangle_s =-\sum_{j=1}^r q_j \log(1-e^{-p_jt}),$$ so that for $0<x<1$ the function $F=F_m$ writes $$F(x)=\Phi(-\log x).$$ Since $F''(x)={1\over x^2}\big(\Phi''+\Phi')(-\log x),$ we are interested in $\Phi''(t)+\Phi'(t)$, that is $$(\Phi''+\Phi')(t)=\big \langle m, (\partial_t^2+\partial_{t})\varphi(ts)\big\rangle_s$$

Integrating by parts $$=-\big \langle \chi_{\mathbb{R}_+}*m, \ \partial_s(\partial_t^2+\partial_{t})\varphi(ts)\big\rangle_s=\big\langle m_1, \kappa(t,s)\big\rangle_s ,$$

with $$\kappa(t,s):=-\partial_s(\partial_t^2+\partial_{t})\varphi(ts)=-\partial_s\big(s^2\ddot\varphi(ts)+s\dot\varphi(st)\big)$$ $$=-ts^2 \dddot\varphi(ts)-(2s+ts)\ddot\varphi(ts)-\dot\varphi(ts)= $$ $$={\frac { \left( t{s}^{2} -ts-2\,s+1 \right) {{\rm e}^{2\,ts}}+ \left( { s}^{2}t+ts+2\,s-2 \right) {{\rm e}^{ts}}+1}{ \left( {{\rm e}^{ts}}-1 \right) ^{3}}} $$ and $$m_1:=\chi_{\mathbb{R}_+}*m=\sum_{j=1}^rq_j (\chi_{\mathbb{R}_+}*\delta_{p_j})=\sum_{j=1}^rq_j \chi_{[p_j,+\infty)} =\sum_{j=1}^{r}\left(\sum_{i=1}^j q_i\right) \chi_{[p_j,p_{j+1})} $$ (where we put $p_{r+1}:=+\infty$). In the case of our $7$ point charges, this is $$m_1:=\chi_{[2n,2n+1]} { -} \chi_{[2n+1,2n+2]}+\chi_{[4n-1,4n]} +\chi_{[4n+2,8n-2]}. $$ Summarizing, we have the representation for the second derivative, for $0<x<1$ and $t:=-\log x$:

$$F''(x)={1\over x^2}\int_0^{+\infty}\kappa(t,s)m_1(s)ds.$$

Of course, one would be happy to have both $m_1$ and $\kappa$ positive; neither is, but positivity is hidden there.

Here are some preliminaries. Let's start with a generic issue:

For given real exponents $1\le p_1\le \dots\le p_r$ and $q_1,\dots,q_r$, how to write the second derivative of the function $$F(x):=\log\prod_{j=1}^r (1-x^{p_j})^{-q_j}=-\sum_{j=1}^r q_j \log(1-x^{p_j }),\qquad x\in(0,1) $$ without opening the Pandora's box of derivatives of quotients?

Notice that the above $F$ has no terms $1+x^p$, yet of course we may write any of them as ${1-x^{2p}\over 1-x^p}$, so that this formulation actually includes your problem. Precisely, in your situation, $r=7$, and $p:=(2n,2n+1,2n+2,4n-1,4n,4n+2,8n-2)$ with $q:=(+1,-2,+1,+1,-1,+1,-1)$ -just for fun we may think these data as describing $7$ point charges $q_j$ located at positions $p_j$.

To describe more clearly the dependence from $(p,q)$ it is convenient to introduce, for $s>0$, the function $$\varphi(s):=-\log(1-e^{-s})\ .$$

Then, given the data $(p,q)$, consider the distribution $m\in\mathcal{D'}(\mathbb{R}_+)$ defined by $f\mapsto\langle m,f\rangle:= \sum_{j=1}^r q_j f(p_j)$, just a linear combination of evaluations at the points $p_j$: $$m:=\sum_{j=1}^r q_j\delta_{p_j}.$$ Introducing a parameter $t>0$, we then have a function $\Phi=\Phi_m$ defined by the pairing w.r.to the variable $s\in\mathbb{R}_+$: $$\Phi(t) :=\big \langle m, \varphi(ts) \big\rangle_s =-\sum_{j=1}^r q_j \log(1-e^{-p_jt}),$$ so that for $0<x<1$ the function $F=F_m$ writes $$F(x)=\Phi(-\log x).$$ Since $F''(x)={1\over x^2}\big(\Phi''+\Phi')(-\log x),$ we are interested in $\Phi''(t)+\Phi'(t)$, that is $$(\Phi''+\Phi')(t)=\big \langle m, (\partial_t^2+\partial_{t})\varphi(ts)\big\rangle_s$$

Integrating by parts $$=-\big \langle \chi_{\mathbb{R}_+}*m, \ \partial_s(\partial_t^2+\partial_{t})\varphi(ts)\big\rangle_s=\big\langle m_1, \kappa(t,s)\big\rangle_s ,$$

with $$\kappa(t,s):=-\partial_s(\partial_t^2+\partial_{t})\varphi(ts)=-\partial_s\big(s^2\ddot\varphi(ts)+s\dot\varphi(st)\big)$$ $$=-ts^2 \dddot\varphi(ts)-(2s+ts)\ddot\varphi(ts)-\dot\varphi(ts)= $$ $$={\frac { \left( t{s}^{2} -ts-2\,s+1 \right) {{\rm e}^{2\,ts}}+ \left( { s}^{2}t+ts+2\,s-2 \right) {{\rm e}^{ts}}+1}{ \left( {{\rm e}^{ts}}-1 \right) ^{3}}} $$ and $$m_1:=\chi_{\mathbb{R}_+}*m=\sum_{j=1}^rq_j (\chi_{\mathbb{R}_+}*\delta_{p_j})=\sum_{j=1}^rq_j \chi_{[p_j,+\infty)} =\sum_{j=1}^{r}\left(\sum_{i=1}^j q_i\right) \chi_{[p_j,p_{j+1})} $$ (where we put $p_{r+1}:=+\infty$). In the case of our $7$ point charges, this is $$m_1:=\chi_{[2n,2n+1]} { -} \chi_{[2n+1,2n+2]}+\chi_{[4n-1,4n]} +\chi_{[4n+2,8n-2]}. $$ Summarizing, we have the representation for the second derivative, for $0<x<1$ and $t:=-\log x$:

$$F''(x)={1\over x^2}\int_0^{+\infty}\kappa(t,s)m_1(s)ds.$$

Of course, one would be happy to have both $m_1$ and $\kappa$ positive. The idea then would be changing the representation using an identity $$\langle \mu,\kappa\rangle=\langle {^{t}L}^{-1}\mu, L\kappa\rangle,$$ choosing a suitable invertible operator $L$.

This gave me some hope to get a quick answer based on an integral formula for $F''$. For instance, the choice $L:=I-H$ with $$Hf(t):={1\over2}f({t\over2})$$ makes $^{t}L^{-1}\mu$ positive, but, unfortunately, not everywhere positive $L\kappa$.

Here are some preliminaries. Let's start with a generic issue:

For given real exponents $1\le p_1\le \dots\le p_r$ and $q_1,\dots,q_r$, how to write the second derivative of the function $$F(x):=\log\prod_{j=1}^r (1-x^{p_j})^{-q_j}=-\sum_{j=1}^r q_j \log(1-x^{p_j }),\qquad x\in(0,1) $$ without opening the Pandora's box of derivatives of quotients?

Notice that the above $F$ has no terms $1+x^p$, yet of course we may write any of them as ${1-x^{2p}\over 1-x^p}$, so that this formulation actually includes your problem. Precisely, in your situation, $r=7$, and $p:=(2n,2n+1,2n+2,4n-1,4n,4n+2,8n-2)$ with $q:=(+1,-2,+1,+1,-1,+1,-1)$ -just for fun we may think these data as describing $7$ point charges $q_j$ located at points $p_j$. (It will appear that these particular exponents $(p,q)$ from your function have a remarkable combinatorial property).

To describe more clearly the dependence from $(p,q)$ it is convenient to introduce, for $s>0$, the function $$\varphi(s):=-\log(1-e^{-s})\ .$$

Then, given the data $(p,q)$, consider the distribution $m\in\mathcal{D'}(\mathbb{R}_+)$ defined by $f\mapsto\langle m,f\rangle:= \sum_{j=1}^r q_j f(p_j)$, that is

$$m:=\sum_{j=1}^r q_j\delta_{p_j}.$$ Introducing a parameter $t>0$, we then have a function $\Phi=\Phi_m$ defined by the pairing w.r.to the variable $s\in\mathbb{R}_+$: $$\Phi(t) :=\big \langle m, \varphi(ts) \big\rangle_s =-\sum_{j=1}^r q_j \log(1-e^{-p_jt}),$$ so that for $0<x<1$ the function $F=F_m$ writes $$F(x)=\Phi(-\log x).$$ Since $F''(x)={1\over x^2}\big(\Phi''+\Phi')(-\log x),$ we are interested in $\Phi''(t)+\Phi'(t)$, that is $$(\Phi''+\Phi')(t)=\big \langle m, (\partial_t^2+\partial_{t})\varphi(st)\big\rangle_s$$

Integrating by parts $$=-\big \langle \chi_{\mathbb{R}_+}*m, \ \partial_s(\partial_t^2+\partial_{t})\varphi(st)\big\rangle_s=\big\langle m_1, \kappa(t,s)\big\rangle_s ,$$

with $$\kappa(t,s):=-\partial_s(\partial_t^2+\partial_{t})\varphi(st)=-\partial_s\big(s^2\ddot\varphi(st)+s\dot\varphi(st)\big)$$ $$=-s^2t\dddot\varphi(st)-(2s+st)\ddot\varphi(st)-\dot\varphi(st)= $$ $$={\frac { \left( {s}^{2}t-ts-2\,s+1 \right) {{\rm e}^{2\,ts}}+ \left( { s}^{2}t+ts+2\,s-2 \right) {{\rm e}^{ts}}+1}{ \left( {{\rm e}^{ts}}-1 \right) ^{3}}} $$ and $$m_1:=\chi_{\mathbb{R}_+}*m=\sum_{j=1}^rq_j (\chi_{\mathbb{R}_+}*\delta_{p_j})=\sum_{j=1}^rq_j \chi_{[p_j,+\infty)} =\sum_{j=1}^{r}\left(\sum_{i=1}^j q_i\right) \chi_{[p_j,p_{j+1})} $$ (where we put $p_{r+1}:=+\infty$). In the case of our $7$ point charges, this is $$m_1:=\chi_{[2n,2n+1]} { -} \chi_{[2n+1,2n+2]}+\chi_{[4n-1,4n]} +\chi_{[4n+2,8n-2]}. $$ Summarizing, we have the representation for the second derivative, for $0<x<1$ and $t:=-\log x$:

$$F''(x)={1\over x^2}\int_0^{+\infty}\kappa(t,s)m_1(s)ds.$$

Of course, one would be happy to have both $m_1$ and $\kappa$ positive; neither is, but positivity is hidden there.

Here are some preliminaries. Let's start with a generic issue:

For given real exponents $1\le p_1\le \dots\le p_r$ and $q_1,\dots,q_r$, how to write the second derivative of the function $$F(x):=\log\prod_{j=1}^r (1-x^{p_j})^{-q_j}=-\sum_{j=1}^r q_j \log(1-x^{p_j }),\qquad x\in(0,1) $$ without opening the Pandora's box of derivatives of quotients?

Notice that the above $F$ has no terms $1+x^p$, yet of course we may write any of them as ${1-x^{2p}\over 1-x^p}$, so that this formulation actually includes your problem. Precisely, in your situation, $r=7$, and $p:=(2n,2n+1,2n+2,4n-1,4n,4n+2,8n-2)$ with $q:=(+1,-2,+1,+1,-1,+1,-1)$ -just for fun we may think these data as describing $7$ point charges $q_j$ located at positions $p_j$. (It will appear that these particular exponents $(p,q)$ from your function have a remarkable combinatorial property).

To describe more clearly the dependence from $(p,q)$ it is convenient to introduce, for $s>0$, the function $$\varphi(s):=-\log(1-e^{-s})\ .$$

Then, given the data $(p,q)$, consider the distribution $m\in\mathcal{D'}(\mathbb{R}_+)$ defined by $f\mapsto\langle m,f\rangle:= \sum_{j=1}^r q_j f(p_j)$, just a linear combination of evaluations at the points $p_j$: $$m:=\sum_{j=1}^r q_j\delta_{p_j}.$$ Introducing a parameter $t>0$, we then have a function $\Phi=\Phi_m$ defined by the pairing w.r.to the variable $s\in\mathbb{R}_+$: $$\Phi(t) :=\big \langle m, \varphi(ts) \big\rangle_s =-\sum_{j=1}^r q_j \log(1-e^{-p_jt}),$$ so that for $0<x<1$ the function $F=F_m$ writes $$F(x)=\Phi(-\log x).$$ Since $F''(x)={1\over x^2}\big(\Phi''+\Phi')(-\log x),$ we are interested in $\Phi''(t)+\Phi'(t)$, that is $$(\Phi''+\Phi')(t)=\big \langle m, (\partial_t^2+\partial_{t})\varphi(ts)\big\rangle_s$$

Integrating by parts $$=-\big \langle \chi_{\mathbb{R}_+}*m, \ \partial_s(\partial_t^2+\partial_{t})\varphi(ts)\big\rangle_s=\big\langle m_1, \kappa(t,s)\big\rangle_s ,$$

with $$\kappa(t,s):=-\partial_s(\partial_t^2+\partial_{t})\varphi(ts)=-\partial_s\big(s^2\ddot\varphi(ts)+s\dot\varphi(st)\big)$$ $$=-ts^2 \dddot\varphi(ts)-(2s+ts)\ddot\varphi(ts)-\dot\varphi(ts)= $$ $$={\frac { \left( t{s}^{2} -ts-2\,s+1 \right) {{\rm e}^{2\,ts}}+ \left( { s}^{2}t+ts+2\,s-2 \right) {{\rm e}^{ts}}+1}{ \left( {{\rm e}^{ts}}-1 \right) ^{3}}} $$ and $$m_1:=\chi_{\mathbb{R}_+}*m=\sum_{j=1}^rq_j (\chi_{\mathbb{R}_+}*\delta_{p_j})=\sum_{j=1}^rq_j \chi_{[p_j,+\infty)} =\sum_{j=1}^{r}\left(\sum_{i=1}^j q_i\right) \chi_{[p_j,p_{j+1})} $$ (where we put $p_{r+1}:=+\infty$). In the case of our $7$ point charges, this is $$m_1:=\chi_{[2n,2n+1]} { -} \chi_{[2n+1,2n+2]}+\chi_{[4n-1,4n]} +\chi_{[4n+2,8n-2]}. $$ Summarizing, we have the representation for the second derivative, for $0<x<1$ and $t:=-\log x$:

$$F''(x)={1\over x^2}\int_0^{+\infty}\kappa(t,s)m_1(s)ds.$$

Of course, one would be happy to have both $m_1$ and $\kappa$ positive; neither is, but positivity is hidden there.