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Homework Help: Contour integration (related to deformation of path)

  1. Oct 19, 2013 #1
    1. The problem statement, all variables and given/known data
    Use the principle of deformation of path to deduce
    [itex]\int_0^\infty t^n \textbf{cos}(bt) e^{-at}dt=\frac{n!}{e^{n+1}}\textbf{cos}((n+1)\phi)[/itex] and [itex]\int_0^\infty t^n \textbf{sin}(bt) e^{-at}dt=\frac{n!}{e^{n+1}}\textbf{sin}((n+1)\phi)[/itex]
    where [itex] a>0, b>0, c=\sqrt{a^2+b^2},[/itex] and [itex]\phi=\textbf{tan}^{-1}(\frac{b}{a}) for 0\leq\phi<\frac{\pi}{2}[/itex]

    It also gives the hint to solve this problem.
    (a) Consider the integral of [itex]f(z)=z^n e^{-z}[/itex] along three directed smooth curves:
    (i)[itex]\textbf{Im} z=0[/itex], (ii)[itex]z=Re^{i\theta}[/itex], where [itex]0\leq\theta\leq\phi[/itex], (iii)[itex]z=ce^{i\phi}t[/itex], where [itex]t[/itex] goes from [itex]0[/itex] to [itex]\infty[/itex]
    (b) Find the bound for modulus of the integral on (ii).
    Use the inequality [itex]\textbf{cos}\theta\leq1-\frac{2\theta}{\pi}[/itex] for [itex]0\leq\theta <\frac{\pi}{2}[/itex]

    2. Relevant equations
    Cauchy's Integral Formula and Cauchy's Integral Theorem (I have only learnt these two in the topic of Contour integration)

    3. The attempt at a solution
    I know the question should be solved by comparing real and imaginary part, but I don't know how to evaluate the integrals and hence I follow the hint.
    In the previous part, I have shown that [itex]\int_0^\infty x^n e^{-x} dx=n![/itex]

    Hence, I can also solved the (a)(i) of the hint:
    [itex]\int_{Im z=0} f(z)dz=2\int_0^\infty x^n e^{-x} dx=2n![/itex]

    But I don't know how to solve (ii) and also (iii) by deformation of path...
    [itex]\int_{z=Re^{i\theta}}z^n e^{-z}dz=\int_0^\phi R^n e^{in\theta-Re^{i\theta}}Rie^{i\theta}d\theta \\=R^{n+1}i\int_0^\phi e^{i(n+1)\theta-Re^{i\theta}}d\theta[/itex]
    Then I do not know what should I do in the next step...
    It has the similar case in (iii)

    Can anyone help me? Thank you.
  2. jcsd
  3. Oct 19, 2013 #2
    How about I ask you this: See that (closed) contour down there? Now suppose I wish to integrate a function over that contour that is analytic inside and on the contour. Since the function is analytic, the integral from 1 to 10 over the red contour should be the same as if I start at the origin, integrate over the blue, then integrate down over the green to the point (10,0) right? Integrating from 0 to 10 is independent of how I get to the end point. So I could write:

    [tex]\mathop\int\limits_{\text{red}}= \mathop\int\limits_{\text{blue}}+\mathop\int\limits_{\text{green}}[/tex]

    as long as you're careful to keep straight, the path directions of the integrals. However, if I just integrate over all of them in a counterclock-wise direction, then by Cauchy's Theoerm, since the function is analytic,

    [tex]\mathop\int\limits_{\text{red}}+ \mathop\int\limits_{\text{blue}}+\mathop\int\limits_{\text{green}}=0[/tex]

    Well, that's the three integration paths you have in i, ii, and iii above. Now, I haven't worked it out but I'd start by just plugging in all those integrals into these two formulas and try to just muscle-through all the algebra to see what happens.

    Attached Files:

  4. Oct 19, 2013 #3
    I get what you mean.
    But the path in ii does not at infinity. Or I can just change the R to infinity by having limit?
    Last edited: Oct 19, 2013
  5. Oct 19, 2013 #4
    Yes, we let R go to infinity.
  6. Oct 20, 2013 #5
    I found I typed something wrong (the inequality) and it should be the following:
    After plugging in and having [itex]\int_{\text{Im}z=0}z^ne^{-z}dz=2n![/itex]
    I got
    [itex]\int_{z=Re^{i\theta}}z^n e^{-z}dz=\int_{ce^{i\phi}t}z^n e^{-z}dz-2n![/itex] with [itex]R\rightarrow \infty[/itex]
    How can I use the inequality to get the bound for modulus of it?
    Besides, I found the integral iii is used to deduce the two required integrals and [itex]\int_{z=Re^{i\theta}}z^n e^{-z}dz=-n![/itex] should be true in order to deduce the two intregrals
    But I do not know how to get this result...
  7. Oct 20, 2013 #6
    Ok, let's take it slow. First off, that bounds is a mistake and would lead us in the wrong direction. I mean just plot [itex]\cos(t)[/itex] and [itex]1-2t/\pi[/itex] in the range from 0 to pi/2. What you get? I'll tell you. Cos(t) is always above that line except at the end points and that's good because we want to know the bounds for the integral:

    [tex]\int_0^{t} z^n e^{-z}dz[/tex]

    over the circular arc [itex]z=Re^{it}[/itex] for t in the range of 0 to pi/2. So plug in the substitution [itex]z->Re^{it}[/itex] in that expression and compute an upper limit on the value of the integral as a function of R. The first term is easy:

    [tex]|(Re^{it})^n|\leq R^n[/tex]

    How about the second one? Well,


    and therefore:

    [tex]|e^{-Re^{it}}|\leq |e^{-R\cos(t)}|[/tex]

    Now, we have just shown by plotting the functions that [itex]\cos(t)\geq 1-2t/\pi[/itex] in the interval [itex][0,\pi/2][/itex]. Ok, if that is the case, then what can we say about the upper bound on the expression:


    in that interval and therefore, what can we say about the upper bound on the integral:

    [tex]\lim_{R\to\infty}\int_0^t z^n e^{-z} dt,\quad z=Re^{it}, 0\leq t\leq \pi/2[/tex]
  8. Oct 20, 2013 #7
    I want to ask a few questions.
    Why [itex]|(Re^{it})^n|\leq R^n[/itex], [itex]|e^{-Re^{it}}|\leq |e^{-R\cos(t)}|[/itex] are ≤ but not =?
    And I do like this:
    [tex]\lim_{R\to\infty}|\int_{z=Re^{i\theta}} z^n e^{-z} dz|,\quad 0\leq \theta \leq \phi\\
    =\lim_{R\to\infty}|\int_0^\phi (Re^{i\theta})^n e^{-(Re^{i\theta})} Rie^{i\theta}d\theta|\\
    \leq \lim_{R\to\infty}|\int_0^\phi R^n e^{-R(1-\frac{2\theta}{\pi})} Rie^{i\theta}d\theta|\\
    =\lim_{R\to\infty}|iR^{n+1}\int_0^\phi e^{-R(1-\frac{2\theta}{\pi})} e^{i\theta}d\theta|\\
    \leq \lim_{R\to\infty}|\frac{iR^{n+1}}{e^R} \int_0^\phi e^{\frac{2\theta}{\pi}}d\theta|\\
    =\lim_{R\to\infty}|\frac{iR^{n+1}}{e^R}\frac{\pi}{\theta} [e^{\frac{2\theta}{\pi}}]_0^\phi|\\
    =\lim_{R\to\infty}|\frac{iR^{n+1}}{e^R}\frac{\pi}{\theta} [e^{\frac{2\theta}{\pi}}]_0^\phi|\\
    =0\quad \because \lim_{R\to\infty}\frac{iR^{n+1}}{e^R}=0
    Is it right?
  9. Oct 20, 2013 #8
    No. You have:

    [tex]\int_0^{\phi} e^{-R(1-2t/\pi)}dt,\quad 0<\phi<\pi/2[/tex]

    What's that? Then, if [itex]0<\phi<\pi/2[/itex], what's

    [tex]\lim_{R\to\infty}\left\{\int_0^{\phi} e^{-R(1-2t/\pi)}dt\right\}[/tex]
  10. Oct 20, 2013 #9
    [tex]\int_0^{\phi} e^{-R(1-2t/\pi)}dt,\quad 0<\phi<\pi/2\\
    =\frac{\pi}{2}e^{-R} [e^{\frac{2t}{\pi}}]_0^\phi|\\
    [tex]\lim_{R\to\infty}\left\{\int_0^{\phi} e^{-R(1-2t/\pi)}dt\right\}\\
    Is it right?
  11. Oct 20, 2013 #10
    That's still not right equalP. What is:

    [tex]\int e^{-R} e^{(\frac{2R}{\pi})t} dt[/tex]
  12. Oct 20, 2013 #11
    O, I found my mistake...
    I forgot the R...

    [tex]\int_0^{\phi} e^{-R(1-2t/\pi)}dt,\quad 0<\phi<\pi/2\\
    =\frac{\pi}{2R}e^{-R} [e^{\frac{2Rt}{\pi}}]_0^\phi\\
    [tex]\lim_{R\to\infty}\left\{\int_0^{\phi} e^{-R(1-2t/\pi)}dt\right\}\\
    Is it right?
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