Prime sine power

For $P_1 = 2$ , $\displaystyle I_2 = \int_0^\pi \sin^2 (2x) \ dx = \int_0^\pi \frac {1-2\cos (4x)}2 \ dx = \frac {x-8\sin (4x)}2 \bigg|_0^\pi = \frac \pi 2$

Besides $P_1 = 2$ , the other primes are odd. Then, we have:

$\begin{aligned} I_k & = \int_0^\pi \sin^k (kx) \ dx & \small {\color{#3D99F6}\text{Since the integrand is odd}} \\ & = \int_0^\frac \pi k \sin^k (kx) \ dx & \small {\color{#3D99F6}\text{Let }u=kx, \ du = k \ dx} \\ & = \frac 1k \int_0^\pi \sin^k u \ du & \small {\color{#3D99F6}\text{Since the integrand is symmetrical on }\frac \pi 2} \\ & = \frac 2k \int_0^\frac \pi 2 \sin^k u \cos^0 u \ du \\ & = \frac 1k B \left(\frac {k+1}2, \frac 12 \right) & \small {\color{#3D99F6} B(m,n) \text{ is beta function.}} \\ & = \frac {\Gamma \left(\frac {k+1}2 \right) \Gamma \left( \frac 12 \right)}{k \Gamma \left(\frac {k+2}2 \right)} & \small {\color{#3D99F6} \Gamma (n) \text{ is gamma function.}} \\ & = \frac {\left(\frac {k-1}2 \right)! \sqrt \pi}{k \frac {k!!}{2^{\frac {k+1}2}} \sqrt \pi} \\ & = \frac {2^{\frac {k+1}2}\left(\frac {k-1}2 \right)!}{k \cdot k!!} \end{aligned}$

Therefore, we have:

$\begin{aligned} v(6) & = \frac \pi 2 + \frac {2^2 \cdot 1!}{3 \cdot 3!!} + \frac {2^3 \cdot 2!}{5 \cdot 5!!} + \frac {2^4 \cdot 3!}{7 \cdot 7!!} + \frac {2^6 \cdot 5!}{11 \cdot 11!!} + \frac {2^7 \cdot 6!}{13 \cdot 13!!} \\ & = \frac \pi 2 + \frac 49 + \frac {16}{75} + \frac {32}{245} + \frac {512}{7623}+ \frac {2048}{39039} \\ & = \frac {22745812}{25050025} \end{aligned}$

$\implies q-p = 25050025 - 22745812 = \boxed{2304213}$