By using integration by parts, I need to show for $$A = \frac{\mathrm d}{\mathrm dx} + \tanh x, \qquad A^{\dagger} = - \frac{\mathrm d}{\mathrm dx} + \tanh x,$$ that
$$\int_{-\infty}^{\infty}\psi^* A^{\dagger}A\psi\;\mathrm dx = \int_{-\infty}^{\infty}(A\psi)^*(A\psi)\;\mathrm dx $$
Where $\psi$ is a normalized wavefunction. I thought that this was the definition of the Hermitian conjugate of an operator $A$, but the problem asks for me to use integration by parts. I don't really see where the result is going to come from here, surely it is not messy calculation, since it is true in general, right?
I started by by integrating by parts, noting that the surface terms must vanish since $\psi$ is normalized, but after that I just get into a big mess...
I asked this question of Physics.SE and didn't get an answer. I then asked in the Maths meta if it would be OK for me to post the question here and I was told that it would be fine to do so, so here it is. Thank you.
