Using LJ potential with N=2

Z = Ω \int_{1.00000000000000e-5}^{L} \int_{1.00000000000000e-5}^{L} \exp ((- β V (\sqrt{{r 12 x}^{2} + {r 12 y}^{2}}))) 𝑑 r 12 x 𝑑 r 12 y

Specialize to the LJ potential

Z = Ω \int_{1.00000000000000e-5}^{L} \int_{1.00000000000000e-5}^{L} \exp ((- β (\frac{4}{{({r 12 x}^{2} + {r 12 y}^{2})}^{6}} - \frac{4}{{({r 12 x}^{2} + {r 12 y}^{2})}^{3}}))) 𝑑 r 12 x 𝑑 r 12 y

Insert value for box size

Z = Ω \int_{1.00000000000000e-5}^{1.00000000000000} \int_{1.00000000000000e-5}^{1.00000000000000} \exp ((- β (\frac{4}{{({r 12 x}^{2} + {r 12 y}^{2})}^{6}} - \frac{4}{{({r 12 x}^{2} + {r 12 y}^{2})}^{3}}))) 𝑑 r 12 x 𝑑 r 12 y

Insert value for box volume

Z = \int_{1.00000000000000e-5}^{1.00000000000000} \int_{1.00000000000000e-5}^{1.00000000000000} \exp ((- β (\frac{4}{{({r 12 x}^{2} + {r 12 y}^{2})}^{6}} - \frac{4}{{({r 12 x}^{2} + {r 12 y}^{2})}^{3}}))) 𝑑 r 12 x 𝑑 r 12 y

Insert value for temperature

Z = \int_{1.00000000000000e-5}^{1.00000000000000} \int_{1.00000000000000e-5}^{1.00000000000000} \exp ((- \frac{4.00000000000000}{{({r 12 x}^{2} + {r 12 y}^{2})}^{6}} + \frac{4.00000000000000}{{({r 12 x}^{2} + {r 12 y}^{2})}^{3}})) 𝑑 r 12 x 𝑑 r 12 y