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!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Improper integrals with i
  1. Improper integrals (Replies: 2)

  2. Improper Integral (Replies: 7)

Loading...