$\newcommand{\sech}{\operatorname{sech}}$ How to deduce from :
$$\int_0^{+∞} \sech^2(x)e^{-λx}\,dx=\int_{-∞}^0 \sech^2(x)e^{λx}\,dx = \frac \lambda 2 \left(\psi\left[\frac{λ+2r}4\right]-\psi\left( \frac \lambda 4\right)\right)-1,$$
to the result:
$$∫_{-∞}^{+∞} \sech^2(rV_i) λ/2e^{-λV_i}dV_i=(λ/r)[(λ/2r)(ψ[(λ+2r)/4r]-ψ(λ/4r)-1]$$
finally,we have the mathematical expectation :$E[\sech^2(rV_i)]$ is equal to right side of the equation,in other words,
$$E[\sech^2(rV_i)]=(λ/r)[(λ/2r)(ψ[(λ+2r)/4r]-ψ(λ/4r)-1].$$
where $\sech(x)$ is defined as secant of hyperbolic function, $ψ(x)$ is Euler psi function and $r, λ$ are two constants. Thank you for your help.