Question: is there a nice formula for $\int_{-\infty}^{+\infty} \frac{e^{-rx^2}}{e^x + e^{-x}} dx$ ? For a real parameter $r > 0$.

Maybe useful: consider the bilateral Laplace transform $$J(z) = \int_{-\infty}^{+\infty} \frac{e^{-rx^2}}{e^x + e^{-x}} e^{zx} dx $$ $J$ is an entire even function. We want to compute $J(0)$.

$$ \partial^n J(z) = \int_{-\infty}^{+\infty} \frac{e^{-rx^2}}{e^x + e^{-x}} x^n e^{zx} dx $$

$$ (e^{\partial} + e^{-\partial})J(z) = \int_{-\infty}^{+\infty} e^{-rx^2} e^{zx} dx $$

$$J(z-1) + J(z+1) = \sqrt{\frac{\pi}{r}} e^{\frac{z^2}{4r}} $$

Since $J(1)=J(-1)$, we can compute $J(1)$. The functional equation now provides a formula for $J(n)$ where $n$ is an odd integer.

Question: is there a nice formula for $\int_{-\infty}^{+\infty} \frac{e^{-rx^2}}{e^x + e^{-x}} dx$ ? For a real parameter $r > 0$.

Maybe useful: consider the bilateral Laplace transform $$J(z) = \int_{-\infty}^{+\infty} \frac{e^{-rx^2}}{e^x + e^{-x}} e^{zx} \, \, dx $$ $J$ is an entire even function. We want to compute $J(0)$.

$$ \partial^n J(z) = \int_{-\infty}^{+\infty} \frac{e^{-rx^2}}{e^x + e^{-x}} x^n e^{zx} \, dx $$

$$ (e^{\partial} + e^{-\partial})J(z) = \int_{-\infty}^{+\infty} e^{-rx^2} e^{zx} \, dx $$

$$J(z-1) + J(z+1) = \sqrt{\frac{\pi}{r}} e^{\frac{z^2}{4r}} $$

Since $J(1)=J(-1)$, we can compute $J(1)$. The functional equation now provides a formula for $J(n)$ where $n$ is an odd integer.