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Integral + Complex Conjugate

  1. Apr 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that the following = 0:
    [itex]\int_{-\infty}^{+\infty} \! i*(\overline{d/dx(sin(x)du/dx})*u \, \mathrm{d} x + \int_{-\infty}^{+\infty} \! \overline{u}*(d/dx(sin(x)du/dx) \, \mathrm{d} x[/itex] where [itex]\overline{u}[/itex] = complex conjugate of u and * is the dot product.

    2. Work so far
    My thoughts: [itex]\int_{-\infty}^{+\infty} \! i*(\overline{d/dx(sin(x)du/dx})*u \, \mathrm{d} x + \int_{-\infty}^{+\infty} \! \overline{u}*(d/dx(sin(x)du/dx) \, \mathrm{d} x[/itex]
    =
    [itex]\int_{-\infty}^{+\infty} \! -i*(d/dx(sin(x)du/dx)*u \, \mathrm{d} x + \int_{-\infty}^{+\infty} \! \overline{u}*(d/dx(sin(x)du/dx) \, \mathrm{d} x[/itex]

    But I don't even know if that's right.
     
  2. jcsd
  3. Apr 15, 2012 #2
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