Hermitian Operator? Proof by complex eigenvalue Is the operator $\frac{1}{x} \frac{\partial }{\partial x}$ Hermitian?
I think it is because I found $f(x)=Ce^{\frac{1}{2}x^{2}\lambda}$ to be the set of eigenfuctions with ${\lambda}$ as the eigenvalue, and it works for ${\lambda} = i$.  I thought Hermitian operators couldn't have complex eigenvalues, so this made the operator not Hermitian, but I'm not sure that my "proof" is rigorous.
Edit: This is part of the questions of if the given operator can represent a physical quantity.  Its acting on all space, ie negative infinity to infinity.  I know the integration proof is the right way to determine if its Hermitian, but I'm not great at integrating, so if this way works as well, I'd like to know.
Thanks for the additional comments, I know I need practice integrating, but its been awhile since taking basic math classes, and I've forgotten alot.  The definition of a Hermitian operator given in my quantum textbook (where this problem is from) is  $$\int_{-\infty}^{\infty} g^{*} A f dx =  \int_{-\infty}^{\infty} f A^{*} g^{*} dx $$ where * is complex conjugate and f and g are well-behaved functions of x.
Setting A= $\frac{1}{x} \frac{\partial }{\partial x}$
I got the LHS =  $\int_{-\infty}^{\infty} -f(g^{*})'(1/x) + (1/x^{2})fg^{*} dx$
and the RHS = $\int_{-\infty}^{\infty} -f'g^{*}(1/x) + (1/x^{2})fg^{*} dx$
Since the first term in each of the integrals has a different function as a derivative, I'm assuming the integrals aren't equal, and the operator is not Herminitian.  Is that the correct way to go about proving it?  Thank you!
 A: Ignoring a lot of very important details here (the space the operator is acting on, for example, see @charlestoncrabb's comment), I will assume the operator $\frac{1}{x} \frac{\partial}{\partial x}$ is defined on some space with inner product
\begin{equation}
 \langle f,g \rangle = \int_{\Omega} f \bar{g} \,\text{d} x.
\end{equation}
Moreover, for the sake of argument, let's assume that $f$ and $g$ are functions of $x$ only, so we can write the operator in question as $A = \frac{1}{x} \frac{\text{d}}{\text{d} x}$.
Now, the first thing to check is whether $A$ is self-adjoint. That is, whether the following equality holds:
\begin{equation}
 \langle A f,g \rangle = \langle f,A g\rangle,
\end{equation}
i.e. whether
\begin{equation}
 \int_{\Omega} \frac{1}{x} f'(x) \bar{g}(x)\,\text{d} x \tag{1}
\end{equation}
is equal to
\begin{equation}
 \int_{\Omega} \frac{1}{x} f(x) \bar{g}'(x)\,\text{d} x. \tag{2}
\end{equation}
You can now use integration by parts to check whether $(1)$ can be rewritten to obtain $(2)$.
Addition based on edited question: To be short: stick to the integration, and if you're not great at it, this is a good reason to practice. If you're not going into issues concerning the space on which the operator acts (strictly speaking, without which the operator isn't even defined properly), it's hardly any use looking at functions which might be eigenfunctions of the operator in question. To give you an idea of the issues related to this: for $\lambda = a + b i$ with $a>0$, the functions $C e^{\frac{\lambda}{2}x^2}$ are unbounded as $x \to \pm \infty$, which you might consider undesirable (again, this is a question of a choice of function space!). If $a = 0$, the function would be bounded, but would not go to zero as $x \to \pm \infty$. Again, this may or may not be a problem, depending on any 'boundary conditions' you would like to impose: as the domain in question is all of $\mathbb{R}$, these 'boundary conditions' are conditions on the function as $x \to \pm \infty$. 
Addition based on second edit of question: You've made a mistake in the integration by parts leading to 'LHS $=$' and 'RHS$=$'. See eg. here for more info. The conclusion you obtain is correct, though: the operator cannot be Hermitian, even in the absense of proper definition of function spaces and what not.
