Problem integrating complex function

[Frank Einstein]

Homework Statement

Hello, I have been tasked with the next problem, I have to prove that the next two integrals are complex numbers; but I have no idea of how to attack this problem.

Homework Equations

∫dx f^*(x) x (-ih) (∂/∂x) f(x) integrating between -∞ and ∞
∫dx f^*(x) (-ih) (∂/∂x) (x f(x)) integrating between -∞ and ∞
Where h is a constant, i = √-1 and f^* the complex conjugate of the function f

The Attempt at a Solution

Well, the only thing I can think of for solving is a direct integration by parts, using u =f^* x and dv= (∂/∂x) f(x) dx for the first integral, with du=f^* and v= ∫(∂/∂x) f(x)dx=f(x). But then, I find that
-ih∫u dv= uv -∫v du = -ih( [f f^* x]_-∞∞+∫f f^*dx).
f f^*=1, so I find myself with -ih[(∞) (∞-∞)]

All help is appreciated.

 

[Matterwave]

You shouldn't have $f^*(x)f(x)=1$ (where did the x dependence go?) instead, the normalization condition is that $\int_{-\infty}^\infty f^*(x) f(x) dx = 1$.

 

[Frank Einstein]

That was a bad mistake I made; even worse for the fact that f is actually a density of probability. Thanks for pointing this error.
Then I get rid of the ∞-∞, but I still have [f f^* x]_-∞ ^∞, which result in a complex but infinite result

 

[Matterwave]

That was a bad mistake I made; even worse for the fact that f is actually a density of probability. Thanks for pointing this error.
Then I get rid of the ∞-∞, but I still have [f f^* x]_-∞ ^∞, which result in a complex but infinite result

Why is it infinite? What conditions must $f(x)$ fulfill at the boundaries if we are to be able to normalize them? Why do you think $f^*(x)f(x)$ will diverge at both limits?

 

[Ray Vickson]

That was a bad mistake I made; even worse for the fact that f is actually a density of probability. Thanks for pointing this error.
Then I get rid of the ∞-∞, but I still have [f f^* x]_-∞ ^∞, which result in a complex but infinite result

No: f is NOT a probability density. In general, f has complex values, so cannot possibly be a probability density. However, $|f|^2 = f^*f$ is a probability density.

Also, you did not perform integration by parts correctly. You wrote
$$u = f^*(x) x, \;\; dv = (\partial/\partial x) f(x) \, dx \\<br /> \Longrightarrow du = f^* \; \text{ and }\; v = f.$$
The $v = f$ part is OK, but $du = f^*$ is wrong, and does not even make sense, because a $dx$ is missing. Anyway, you need to use the product rule, which you have not done, so it would be wrong even if you supplied the missing $dx$.

 

[Frank Einstein]

You are right. Posting at late night is a bad idea.
Today I have remade my calculations and I think I have arrived to coherent results.
If I make the integration by parts with u = x f^* and dv= (∂/∂x)(f(x)) then du = (f* + x (∂f^*/∂x)) dx
Taking into account that [f f^* x]_-∞ ^∞ is 0 because f goes way faster to 0 than x to ∞ and that ∫f f^* dx is 1 I find that the original inegral is it's own complex conjugate plus -ih, what means that the operator is not hermitian and therefore, the solution is not real.

I will use the same procedure for the second integral.