Is this differential operator Hermitian? The operator is
$$\hat{A} = -i \left(x \frac{d}{dx} + \frac{1}{2} \right).$$
Is it true that 
$$\langle \hat{A} \psi_1(x)|\psi_2(x)\rangle = \langle \psi_1(x)|\hat{A}\psi_2(x)\rangle\ ?$$
Here, $\langle\ldots|\ldots\rangle$ is a scalar product defined as
$$\langle\psi_1(x)|\psi_2(x)\rangle = \int_{\Omega} \psi_1(x) \psi_2^*(x) \ dx.$$
 A: Here is a different route.  Consider the operators $\hat{x}$ and $\hat{p}$ where
$$\hat{x}\psi(x)=x\psi(x)$$
and
$$\hat{p}\psi(x)=-i\psi'(x).$$
Show that $\hat{x}$ and $\hat{p}$ are Hermitian operators.  Also, show that $[\hat{x},\hat{p}]=\hat{x}\hat{p}-\hat{p}\hat{x}=i$.
It can be proven that $\hat{P}\hat{Q}+\hat{Q}\hat{P}$ is a Hermitian operator for any Hermitian operators $\hat{P}$ and $\hat{Q}$.  The same situation applies to $\hat{x}$ and $\hat{p}$.  We have
$\hat{x}\hat{p}+\hat{p}\hat{x}$
is Hermitian, but
$$\hat{x}\hat{p}+\hat{p}\hat{x}=2\hat{x}\hat{p}-[\hat{x},\hat{p}]=2\hat{x}\hat{p}-i=2\hat{A}.$$
So, $2\hat{A}$ is Hermitian, and so $\hat{A}$ is Hermitian, since $2$ is a real number.
A: The short answer is: Yes it is. You can see this simply by doing an integration by parts. Let us leave out the $-i$ and show that $x \frac{d}{dx} + \frac{1}{2}$ is antisymmetric instead.
\begin{equation}
\int_{\Omega} \left( \left( x \frac{d}{dx} + \frac{1}{2} \right) \psi_1 \right) \overline{\psi_2} \: dx= -\int_{\Omega} \left( x \frac{d}{dx} \overline{\psi_2} \right) \psi_1 + \psi_1 \overline{\psi_2} \: dx + \frac{1}{2} \int_{\Omega} \psi_1 \overline{\psi_2} \: dx
\end{equation}
By integration by parts and since $\frac{d}{dx} \left( x \overline{\psi_2} \right) = x \frac{d}{dx} \overline{\psi_2} + \overline{\psi_2}$. Clearly this equals $- \int_{\Omega} \overline{\left( \left( x \frac{d}{dx} + \frac{1}{2} \right) \psi_2 \right)} \psi_1 \: dx$. Thus your operator is indeed symmetric (since multiplication by $i$ turns antisymmetric into symmetric operators and vice versa).
More importantly perhaps, you should note that this is an unbounded differential operator. Although you did not state it explicitly, you are probably thinking about it as acting on $L^2(\Omega)$, where it is densely defined. There the natural scalar product is the one you gave. If you choose the Sobolev space $H^1$ as its domain, the operator even becomes self-adjoint, a much stronger property of unbounded operators than just being symmetric.
You can learn more about the difference between symmetry and self-adjointness of unbounded operators for instance in the book of Teschl.
