<s>[INST] Solving a formal power series equation I want to find a function $f(x,y)$ which can satisfy the following equation,
$$\prod _{n=1} ^{\infty} \frac{1+x^n}{(1-x^{n/2}y^{n/2})(1-x^{n/2}y^{-n/2})} = \exp \left[ \sum _{n=1} ^\infty \frac{f(x^n,y^n)}{n(1-x^{2n})}\right]$$


*

*I would like to know how this is solved.


In a certain paper where I ran into this, it is claimed that the function is,
$$f(x,y) = \sqrt{x}(y + 1/y) + x(1+y^2 + 1/y) + x^{3/2}(y^3+1/y^3) + x^2(y^4+1/y^4) + \sum _{n=5}^\infty x^{n/2}(y^n + 1/y^n - y^{n-4} - 1/y^{n-4})$$
The paper doesn't state any proof or explanation for how this was obtained but perturbatively the above can be checked to be correct! 

Now I tried to do something obvious but it didn't work!
\begin{eqnarray} 
\prod _ {n=1} ^{\infty} \frac{ (1+x^n) }{1+x^n -x^{\frac{n}{2}} \left(y^{\frac{n}{2}}  + y^{-\frac{n}{2}}\right) } = \exp \left[ \sum _ {n=1} ^{\infty}   \frac{ I_{ST}(x^n,y^n) } {n (1-x^{2n}) } \right] \\ 
\Rightarrow \sum_{n=1}^{\infty} \left\{ \ln (1+x^n) - \ln(1-(\sqrt{xy})^n) - \ln\left(1- \left(\sqrt{\frac{x}{y}}\right)^n\right) \right\}  = \sum_{n=1}^\infty  \frac{I_{ST}(x^n,y^n)} {n(1-x^{2n})} 
\end{eqnarray} 
Now we expand the logarithms and we have, 
\begin{eqnarray} 
\sum _ {n=1} ^ {\infty} \left \{  \sum _{a=1}^{\infty} (-1)^{a+1} \frac{x^{na}}{a} + \sum_{b=1} ^{\infty} \frac{ (\sqrt{xy})^{nb} } {b} + \sum _{c=1}^{\infty} \frac{ \left(\sqrt{\frac{x}{y}}\right)^{nc} }{c}   \right \} =  \sum _{n=1} ^\infty  \frac{I_{ST}(x^n,y^n)} {n(1-x^{2n})} \\
\Rightarrow \sum _{a=1} ^{\infty} \frac{1}{a} \left\{  \sum _{n=1} ^{\infty} \left( (-1)^{a+1}x^{na} + (xy)^{\frac{na}{2}} + \left(\frac{x}{y}\right)^{\frac{na}{2}} \right)  \right\} =  \sum _{n=1} ^\infty  \frac{I_{ST}(x^n,y^n)} {n(1-x^{2n})}
\end{eqnarray} 
By matching the patterns on both sides one sees that one way this equality can hold is if, 
$$
\begin{eqnarray} 
I_{ST}(x,y) = (1-x^2) \sum _{n=1} ^{\infty} \left\{ x^n + (xy)^{\frac{n}{2}} + (\frac{x}{y})^{\frac{n}{2}}  \right\} \\
\Rightarrow I_{ST} (x,y) = (1-x^2) \left(-1 + \frac{1}{1-x} -1 + \frac{1}{1-\sqrt{xy}} - 1 + \frac{1}{1-\sqrt{\frac{x}{y}} } \right)
\end {eqnarray} 
$$
But this solution doesn't satisfy the original equation! 
 [/INST] What a goofy looking function.  Why do they need this anyway?  Taking the log of both sides, you should get:
$$ \log(1+x^n) - \log(1 - x^{n/2}y^{n/2}) - \log(1 - x^{n/2}y^{-n/2})  $$
Then
$$ \log ( 1 + x^n) = 1 - x^n + \frac{1}{2}x^{2n} - \frac{1}{3}x^{3n} + \dots $$
and also 
$$ \log (1 - x^{n/2}y^{n/2}) = 1 + (xy)^{n/2} + \frac{1}{2} (xy)^n + \dots $$
and 
$$ \log (1 - x^{n/2}y^{-n/2}) = 1 + (x/y)^{n/2} + \frac{1}{2} (x/y)^n + \dots $$
I suppose if you add from $n=1 \to \infty$ you will get the correct $f(x,y)$.
</s>