I came across this equation (identity) (Eq. 4 in this paper):
$\int(-i d\psi/dx)^*\psi dx = \int \psi^*(-i d\psi/dx) dx + id(\psi^*\psi)/dx\mid_{-\infty}^{+\infty}$
I have trouble proving it. I tried to use integration by parts but could not reach there. How do we take complex conjugate (Hermitian Adjoint) of the differential operator that occurs in this equation and also of any general function.
Notice that \begin{align} i\frac{d(\psi^*\psi)}{dx} &=\frac{d\big[(-i\psi)^*\psi\big]}{dx} \\ &= \frac{d(-i\psi)^*}{dx}\psi + (-i\psi)^*\frac{d\psi}{dx} \\ &= \left(-i\frac{d\psi}{dx}\right)^*\psi + \psi^*\left(i\frac{d\psi}{dx}\right) \\ \end{align} Now subtract the second term on the right from both sides to get \begin{align} \psi^*\left(-i\frac{d\psi}{dx}\right)+i\frac{d(\psi^*\psi)}{dx} &= \left(-i\frac{d\psi}{dx}\right)^*\psi \end{align} and finally integrate both sides from $-\infty$ to $\infty$ to obtain (as Stan Liou pointed out in the comments) $$ \int_{-\infty}^\infty\psi^*\left(-i\frac{d\psi}{dx}\right)+i\psi^*\psi\Big|_{-\infty}^\infty = \int_{-\infty}^\infty\left(-i\frac{d\psi}{dx}\right)^*\psi $$ Notice that the boundary term you wrote in the identity has an erroneous derivative that goes away when you actually evaluate the integral and use the fundamental theorem of calculus.
