 Laplace transform of the integrated square of Brownian motion

The independent squared-normal series for the integrated square of Brownian motion gives the product ${\prod }_n(1+2\lambda /((n-1/2{)}^2{\pi }^2){)}^{-1/2}$. The half-integer cosine product reduces this to $(\cosh \sqrt{2\lambda }{)}^{-1/2}$. Expanding at zero gives mean $1/2$ and variance $1/3$.