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!

Complex integral limit

  1. Jun 17, 2014 #1
    The problem statement, all variables and given/known data.

    Let ##\gamma_r:[0,\pi] \to \mathbb C## be given by ##\gamma_r(t)=re^{it}##. Prove that ##\lim_{r \to \infty} \int_{\gamma_r} \dfrac{e^{iz}}{z}dz=0##.

    The attempt at a solution.

    The only thing I could do was:
    ## \int_{\gamma_r} \dfrac{e^{iz}}{z}dz=\int_0^{\pi} \dfrac{e^{ire^{it}}ire^{it}}{re^{it}}dt##

    This last expression equals to ##\int_0^{\pi} \dfrac{e^{ire^{it}}ire^{it}}{re^{it}}dt=\int_0^{\pi} e^{ire^{it}}idt##. At this point I got stuck, this expression ## e^{ire^{it}}## looks horrible to me and I don't know what to do with it.

    The curve ##\gamma_r## is the upper half of the circle of radius ##r##, according to the statement as the radius of the circle tends to infinity, the value of the integral goes to zero. I would like to understand intuitively why this is the case and how the function ##\dfrac{e^{iz}}{z}## behaves.

    I would appreciate some help and suggestions. Thanks in advance.
     
  2. jcsd
  3. Jun 18, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    Use the bound [tex]
    \left|\int_\gamma f(z)\,dz\right| \leq L(\gamma) \sup_\gamma |f(z)|
    [/tex] where [itex]L(\gamma)[/itex] is the length of the curve [itex]\gamma[/itex].

    Note that [itex]e^{ire^{it}} = e^{ir\cos t}e^{-r\sin t}[/itex].
     
  4. Jun 18, 2014 #3
    Thanks, that inequality is the key to the solution. Now, working on the right member of the inequality

    ##L(\gamma) \sup_\gamma |f(z)|=\int_0^{\pi} |rie^{it}|dt\sup_\gamma|\dfrac{e^{ir\cos(t)}e^{-r\sin(t)}}{re^{it}}|##

    This last expression equals to ##\int_0^{\pi} |rie^{it}|dt\sup_\gamma|\dfrac{e^{ir\cos(t)}e^{-r\sin(t)}}{re^{it}}|=r\pi \sup_{\gamma}|\dfrac{e^{ir\cos(t)}e^{-r\sin(t)}}{re^{it}}|=r\pi \sup_{\gamma}|\dfrac{e^{-r\sin(t)}}{r}|##

    I have problems to find a bound for the expression ##r\pi\sup_\gamma|\dfrac{e^{-r\sin(t)}}{r}|##. I would like to get to an expression with an ##r## on the denominator, something like ##\dfrac{k}{r^c}## with ##k## constant and ##c\geq 1##, but I don't know how to get to that.
     
  5. Jun 19, 2014 #4
    The r cancels out, next the exponentional tends to zero as r tends to infinity, do you agree?
     
  6. Jun 19, 2014 #5

    pasmith

    User Avatar
    Homework Helper

    Unfortunately [itex]f(z) = e^{iz}/z[/itex] is a case where this bound doesn't give a sufficiently accurate bound for the integral. But one can instead use [tex]
    \left| \int_{t_1}^{t_2} f(Re^{it}) iRe^{it}\,dt \right| \leq R\int_{t_1}^{t_2} |f(Re^{it})|\,dt.
    [/tex] In this case you end up needing to find an estimate for [itex]\int_0^\pi e^{-R\sin t}\,dt[/itex]. Fortunately the symmetry about [itex]t = \pi/2[/itex] allows the use of Jordan's Inequality to find an easily integrable [itex]g_R: [0,\pi/2] \to \mathbb{R}[/itex] such that [itex]e^{-R\sin t} \leq g_R(t)[/itex] for all [itex]t \in [0,\pi/2][/itex].
     
  7. Jun 19, 2014 #6

    pasmith

    User Avatar
    Homework Helper

    The supremum of [itex]e^{-r\sin t}[/itex] on [itex]0 \leq t \leq \pi[/itex] (which must be calculated before taking the limit of r) is 1 (occuring at [itex]t = 0[/itex] and [itex]t = \pi[/itex]), so in fact there is no exponential.
     
  8. Jun 19, 2014 #7
    You're right, my bad.
     
  9. Jun 19, 2014 #8
    Hmmm, I couldn't go further:

    As you've written, it is true that ##|\int_0^{\pi}f(re^{it})rie^{it}dt|\leq r\int_0^{\pi} |f(re^{it})|dt##.

    The member from the right is ## r\int_0^{\pi} |f(re^{it})|dt=r\int_0^{\pi} |\dfrac{e^{ir\cos(t)}e^{-r\sin(t)}}{re^{it}}|dt##.

    Now, we are studying a limit for ##r \to \infty##, so lets analyze this last expression for ##r>1##:

    Using the fact that ##\dfrac{1}{r}<1## and that ##e^{-r\sin(t)}\leq 1## (given ##r>0## and ##t \in [0,\pi]##), we get

    ##r\int_0^{\pi} |\dfrac{e^{ir\cos(t)}e^{-r\sin(t)}}{re^{it}}|dt\leq r \int_0^{\pi}1=r\pi##

    Now, arriving to that expression doesn't help at all, I would get a sort of opposite conclusion: as ##r \to \infty##,##\space## ##\int_{\gamma_r}\dfrac{e^{iz}}{z}dz \to \infty##.

    What else can I do to get a useful bound for the integral's absolute value?
     
  10. Jun 19, 2014 #9

    pasmith

    User Avatar
    Homework Helper

    This can be further simplified: both [itex]e^{ir\cos t}[/itex] and [itex]e^{it}[/itex] are equal to 1. Thus [tex]
    r \int_0^\pi |f(re^{it})|\,dt = \int_0^\pi e^{-r \sin t}\,dt.
    [/tex]

    You have already demonstrated by using the crude "length of curve times supremum" method that the bound [itex]e^{-r\sin t} \leq 1[/itex] is not good enough to do the job. You need something which actually varies with [itex]r[/itex].

    Since [itex]e^{-r\sin t}[/itex] is symmetric about [itex]t = \pi/2[/itex] we have [tex]
    \int_0^\pi e^{-r\sin t}\,dt = 2\int_0^{\pi/2} e^{-r\sin t}\,dt.[/tex] If you could find an easily integrable [itex]g_r[/itex] such that [itex]g_r(t) \geq e^{-r\sin t}[/itex] for [itex]t \in [0,\pi/2][/itex] and [itex]\int_0^{\pi/2} g_r(t)\,dt \to 0[/itex] as [itex]r \to \infty[/itex] then it would follow that [tex]
    \int_0^\pi e^{-r\sin t}\,dt = 2\int_0^{\pi/2} e^{-r \sin t}\,dt \leq 2\int_0^{\pi/2} g_r(t)\,dt.[/tex] and thus [tex]
    0 \leq \lim_{r \to \infty} \int_0^\pi e^{-r\sin t}\,dt \leq 2\lim_{r \to \infty} \int_0^{\pi/2} g_r(t)\,dt = 0.[/tex]
    Try using Jordan's Inequality as I suggested.
     
  11. Jun 19, 2014 #10
    Yes, sorry, that suggestion was extremely helpful:

    We want to get a bound for
    ##2\int_0^{\frac{\pi}{2}} e^{-r\sin(t)}dt##.

    Since ##r\geq 0##, the function ##e^{-rx}## is decreasing for ##x\geq 0##. This means that if ##x_1<x_2 \implies e^{-rx_1}\geq e^{-rx_2}##.

    For ##t \in [0,\dfrac{\pi}{2}]##, ##\sin(t)\geq 0##, and using Jordan's inequality, we get ##\dfrac{2}{\pi}t\leq \sin(t)##. But this means ##e^{-r\sin(t)}\leq e^{-r\dfrac{2}{\pi}t}##.

    This last inequality implies ##2\int_0^{\frac{\pi}{2}} e^{-r\sin(t)}dt\leq 2\int_0^{\frac{\pi}{2}} e^{-r\frac{2}{\pi}t}dt##.

    The value of that integral is ##\dfrac{\pi}{r}(1-e^{-r})## and this expression tends to zero when ##r \to \infty##.

    Since by a chain of inequalities, we get ##0\leq |\int_{\gamma_r}\dfrac{e^{iz}}{z}dz|\leq \dfrac{\pi}{r}(1-e^{-r})##, it follows ##\int_{\gamma_r}\dfrac{e^{iz}}{z}dz \to 0## when ##r \to \infty##.

    pasmith, I really appreciate all the help and suggestions you gave me.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Complex integral limit
  1. Complex Limits (Replies: 3)

Loading...