1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problem integrating complex function

  1. Nov 26, 2014 #1
    1. The problem statement, all variables and given/known data
    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.
    2. Relevant 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

    3. 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.
  2. jcsd
  3. Nov 26, 2014 #2


    User Avatar
    Science Advisor
    Gold Member

    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##.
  4. Nov 26, 2014 #3
    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
  5. Nov 26, 2014 #4


    User Avatar
    Science Advisor
    Gold Member

    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?
  6. Nov 27, 2014 #5

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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
    [tex] u = f^*(x) x, \;\; dv = (\partial/\partial x) f(x) \, dx \\
    \Longrightarrow du = f^* \; \text{ and }\; v = f.[/tex]
    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##.
  7. Nov 27, 2014 #6
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Problem integrating complex function