Integral of a real function multiplied by an imaginary function.

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SUMMARY

The integral of a real function multiplied by an imaginary function, represented as \int_{-\infty}^{\infty} a(k)^{*} i \dfrac{d\,a(k)}{dk} dk, evaluates to zero when a(k) is constrained to be real. This is due to the orthogonality of the real and imaginary components, leading to the conclusion that the integral of their product over the entire real line results in cancellation. The text specifies that the integral evaluates to \frac{i}{2}(a^2(\infty) - a^2(-\infty)), reinforcing the notion that the endpoints may contribute zero, further supporting the integral's value of zero.

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  • Understanding of real and imaginary functions in mathematics
  • Familiarity with calculus, particularly integration techniques
  • Knowledge of quantum mechanics principles
  • Basic concepts of orthogonality in function spaces
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  • Learn about the implications of complex functions in quantum mechanics
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cnelson
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So I'm reviewing some mathematics for quantum mechanics and this came equation came up
[itex]\int_{-\infty}^{\infty} a \left( k \right)^{*} i \dfrac{d\,a\left(k\right)}{dk}dk[/itex].

If [itex]a \left( k \right)[/itex] is constrained to be real then this integral is zero or so the text says. Why is this the case? Is it because this is the summation of two orthogonal functions so the integral must be zero. If so how what would be the first steps to proving this to myself?
 
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Does the text say the integral is real? The integral is i{a2(∞) -a2(-∞)}/2

I don't know anything about a, but its values at the end points may be 0.
 
The text just says that [itex]a \left(k\right)[/itex] is real. So its derivative is also real. But the derivative is multiplied by [itex]i[/itex] making it purely imaginary. So some how when these are multiplied and then integrated the integral is zero which I don't understand. Let me know if more clarification is needed.
 

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