Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Improper integrals with i

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Solve integrals

    [tex]\int^0_{-\infty}e^{(a-ik)x}dx[/tex]

    [tex]\int^{\infty}_{0}e^{-(a+ik)x}dx[/tex]


    2. Relevant equations

    [tex]\int e^x=e^x+C[/tex]



    3. The attempt at a solution

    My troble is with imaginary unit [tex]i[/tex]

    [tex]\int^0_{-\infty}e^{(a-ik)x}dx=\frac{e^{(a-ik)x}}{a-ik}|^0_{-\infty}[/tex]

    [tex]\int^{\infty}_{0}e^{-(a+ik)x}dx=\frac{e^{-(a+ik)x}}{-a-ik}|^{\infty}_0[/tex]

    I don't know how to get results because of [tex]i[/tex]?
     
  2. jcsd
  3. Oct 15, 2011 #2
    both integrals converge well if [itex]a > 0[/itex]. Just evaluate the limits and do some complex algebra.

    EDIT:
    Hint:
    [tex]
    e^{x + i y} = e^{x} \left(\cos{y} + i \sin{y}\right)
    [/tex]
     
  4. Oct 15, 2011 #3
    Ok. But here I have x in both terms. Can you show me idea with more details?
     
  5. Oct 15, 2011 #4
    Look at the modulus of the complex number:
    [tex]
    \left\vert e^{(a + i b) x}\right\vert = e^{a x}
    [/tex]
    When [itex]a < 0[/itex], what does this function tend to as [itex]x \rightarrow \infty[/itex]? How about when [itex]x \rightarrow -\infty[/itex]? How about when [itex]a > 0[/itex]?
     
  6. Oct 15, 2011 #5
    You tell me if I understand

    [tex]\int^0_{-\infty}e^{(a-ik)x}dx=\frac{e^{(a-ik)x}}{a-ik}|^0_{-\infty}=\frac{e^{ax}}{a-ik}|^0_{-\infty}=\frac{1}{a-ik}[/tex]

    [tex]\int^{\infty}_{0}e^{-(a+ik)x}dx=\frac{e^{-(a+ik)x}}{-a-ik}|^{\infty}_0=\frac{e^{-ax}}{-a-ik}|^{\infty}_0=\frac{1}{a+ik}[/tex]

    I still don't understand why I may do that :(
     
  7. Oct 15, 2011 #6
    The end results are correct, but the second step is not.
     
  8. Oct 15, 2011 #7
    Can you write second step?
     
  9. Oct 15, 2011 #8
    Sure, for the first integral:
    [tex]
    \frac{e^{0} - \lim_{x \rightarrow -\infty}{e^{(a - i k) x}}}{a - i k}
    [/tex]
     
  10. Oct 15, 2011 #9
    Ok. I don't know how to solve that limit. i is not bigger and is not less then zero. How to solve that?
     
  11. Oct 15, 2011 #10
    Do you know this equivalence

    [tex]
    \lim_{x \rightarrow x_0} f(x) = 0 \Leftrightarrow \lim_{x \rightarrow x_0} \vert f(x) \vert = 0
    [/tex]

    meaning that the limit of some function as [itex]x \rightarrow x_0[/itex] is zero if and only if the limit of the modulus of that function is also zero. It follows from the obvious identity:
    [tex]
    \vert f(x) - 0 \vert = \vert f(x) \vert = \vert \vert f(x) \vert \vert = \vert \vert f(x) \vert - 0 \vert
    [/tex]

    You should use this rule, as well as the fact that:
    [tex]
    \vert e^{(a - i k) x} \vert = e^{a x}
    [/tex]
    to show that that limit is zero!

    What is this limit:
    [tex]
    \lim_{x \rightarrow -\infty} {e^{a x}} = ?, \; a > 0
    [/tex]
     
  12. Oct 16, 2011 #11
    I don't know that equivalence. How can I show that this is equivalent?
     
  13. Oct 16, 2011 #12
    Doesn't matter. Just show that:
    [tex]
    \lim_{x \rightarrow -\infty}{\left e^{(a - i k) x} \vert} = 0
    [/tex]
    and use it to show that the limit in the numerator of the fraction is zero.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook