Thermal trace of an interval image kernel ($Z={\int }_0^L{K}_D(q,q;\beta )dq$)

For a particle with mass $m$ on $(0,L)$, put $K_0(x;\beta )=\sqrt{m/(2\pi {\hslash }^2\beta )}{e}^{-m{x}^2/(2{\hslash }^2\beta )}$. The method of images gives $K_D(q_f,q_i;\beta )={\sum }_r[K_0(q_f-q_i+2rL;\beta )-K_0(q_f+q_i+2rL;\beta )]$. The diagonal integral isThe reflected intervals tile the real line, giving the subtraction $1/2$. The Poisson summation formula converts this to ${\sum }_{n\ge 1}{e}^{-\beta {\hslash }^2{\pi }^2{n}^2/(2m{L}^2)}$, proving equality between the image and energy eigenstate calculations.