Let: $f(x,a,l)=\prod _{k=a}^l\sin \left(\frac{\pi x}{k}\right)$ and
$f(x,k)=\sin \left(\frac{\pi x}{k}\right)$
I came up with this equation to find the period $T(f(x,a,l))=2\,{{\pi }^{l-a+1} \left( \prod _{k=a}^{l} \left( \sqrt{-{\frac {{ \frac {\rm d}{{\rm d}x}}f(x,k)}{\int \!f(x,k)\,{\rm d}x}}} \right) \right) ^{-1 } \left( \prod _{k=a}^{l}{\it ifactor} \left( {\frac {k}{(op(1,{\it ifactor} \left( k \right) )}} \right) \right) ^{-1}}$
Note: ${op(1,{\it ifactor} \left( k \right)) }$=the smallest prime factorization of k
and
${{\it ifactor} \left( k \right) }$, returns all factors of k
*Now I am looking for an equation correlating the period of $f(x,a,l)$, and its zeros; such that when integrated from one point to another it returns the amount of zeroes between those points. ie. ${\int_{q}^p \!W(T(f(x,a,l)),f(x,a,l))\,{\rm d}x}=$ count {${f(x,a,l)=0}$},q..p
Can someone please help, or provide guidance?*
Simplification:
$2\,{\pi \left( \prod _{k=2}^{l} \left( \sqrt {-{\frac {{\frac { \partial }{\partial x}}f \left( x,k \right) }{\int \!f \left( x,k \right) \,{\rm d}x}}} \right) \right) ^{-1}}$
$f \left( x,k \right) =\sin \left( {\frac {\pi \,x}{k}} \right) $
${\frac {\partial }{\partial x}}f \left( x,k \right) ={\frac {\pi }{k} \cos \left( {\frac {\pi \,x}{k}} \right) } $
$-\int \!f \left( x,k \right) \,{\rm d}x={\frac {k}{\pi }\cos \left( { \frac {\pi \,x}{k}} \right) } $
${\frac {{\frac {\pi }{k} \cos \left( {\frac {\pi \,x}{k}} \right) }}{{\frac {k}{\pi }\cos \left( { \frac {\pi \,x}{k}} \right) }}}={\frac {{\pi }^{2}}{{k}^{2}}}$
$2\,{\pi \left( \prod _{k=2}^{l} \left( \sqrt {-{\frac {{\frac { \partial }{\partial x}}f \left( x,k \right) }{\int \!f \left( x,k \right) \,{\rm d}x}}} \right) \right) ^{-1}}=2\,{\frac {\pi }{\prod _{k=a}^{l}\pi \, \sqrt{{k}^{-2}}}}=2\,{\frac { \pi \,{\pi }^{a} \sqrt{ \left( \Gamma \left( a \right) \right) ^{-2} }}{{\pi }^{l-a} \sqrt{ \left( \Gamma \left( l+1 \right) \right) ^{-2 }}}}$
$T(a,l)={\frac {2\,{\frac { \pi \,{\pi }^{a} \sqrt{ \left( \Gamma \left( a \right) \right) ^{-2} }}{{\pi }^{l-a} \sqrt{ \left( \Gamma \left( l+1 \right) \right) ^{-2 }}}}}{\left(\prod _{k=a}^{l}{\it ifactor} \left( {\frac {k}{(op(1,{\it ifactor} \left( k \right) )}} \right) \right)}} $
Through cancellation of ${\pi }->$
${2\,{\frac {\sqrt { \left( \Gamma \left( a \right) \right) ^{-2}}}{ \sqrt { \left( \Gamma \left( 1+l \right) \right) ^{-2}}{\left(\prod _{k=a}^{l}{\it ifactor} \left( {\frac {k}{(op(1,{\it ifactor} \left( k \right) )}} \right) \right)}}}={\frac {2\Gamma \left( 1+l \right) }{\Gamma \left( a \right)2(PF\left({(k)}\right)}}}$
*A 2 can come out due to a 'strange factorization of 4' and implying 4 lies between a and l
$={\frac {l!}{ \left( a-1 \right) !\,\left(PF({k})\right)}}$
*PF is Strange Prime Factorization
PF(4)=2
PF(Prime)=1
${op(2,{\it ifactor} \left( k \right)) }={\frac {k }{{op(1,{\it ifactor} \left( k \right)) }}}=$Largest factor of K
$PF(k) = \prod\cases{ 2 & \text{if } k = 4\cr 1 & \text{if } k\in {Primes} \cr 2*\prod _{k=a}^l{op(2,{\it ifactor} \left( k \right)) } & \text{if } k\not\in {Primes} }$
Practical solution $PF(k) = \prod\cases{ 2 & \text{if } k = 4\cr 1 & \text{if } 0< \left| \prod _{u=2}^{k-1}-i/2 \left( {{\rm e}^{{\frac {ik\pi }{u}}} }-{{\rm e}^{{\frac {-ik\pi }{u}}}} \right) uk \right| \cr 2*\prod _{k=a}^l{op(2,{\it ifactor} \left( k \right)) } & \text{if } \left| \prod _{u=2}^{k-1}-i/2 \left( {{\rm e}^{{\frac {ik\pi }{u}}} }-{{\rm e}^{{\frac {-ik\pi }{u}}}} \right) uk \right| = 0 }$
$T(a,l)={ {\frac {2*l!}{ \left( a-1 \right) !{\prod\cases{ 2 & \text{if } k = 4\cr 1 & \text{if } 0< \left| \prod _{u=2}^{k-1}-i/2 \left( {{\rm e}^{{\frac {ik\pi }{u}}} }-{{\rm e}^{{\frac {-ik\pi }{u}}}} \right) uk \right| \cr 2*\prod _{k=a}^l{op(2,{\it ifactor} \left( k \right)) } & \text{if } \left| \prod _{u=2}^{k-1}-i/2 \left( {{\rm e}^{{\frac {ik\pi }{u}}} }-{{\rm e}^{{\frac {-ik\pi }{u}}}} \right) uk \right| = 0}} }}} $