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Homework Help: Complex Analysis - Gaussian Function Integration

  1. Nov 15, 2013 #1
    I know that [itex]\displaystyle \int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}[/itex] (You calculate the square of the integral, combine both integrals, change variable to polar coordinates and you can finally integrate that with ease).
    But in this exercise I have the following statament:

    Let be [itex] P_{R} [/itex] the parallelogram define by the points [itex]R+Ri[/itex], [itex]R+1+Ri[/itex], [itex]-R-RI[/itex] and [itex]-R+1-Ri[/itex].
    Integrating the function [itex]f(z) = e^{-i \pi z^{2}} \tan( \pi z )[/itex] over [itex]P_{R}[/itex] and making [itex]R \longrightarrow \infty[/itex] prove that.
    [itex]\displaystyle \int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}[/itex]

    I don't have idea why that contour may help me, neither how can aid me the [itex]\tan( \pi z )[/itex].

    Any hints?
  2. jcsd
  3. Nov 15, 2013 #2
    Well, you should show us that you attempted the problem first. Then we can help you.
  4. Nov 15, 2013 #3
    I tried to compute the larger sides of the (the ones whose size goes to infinity) to see if the sum of their line integrals converges to zero, so the integral if set only by the other sides. But there's always only two poles of the function here, and I don't know how it can help me even if I decide to make their sides diverges (because the only segment of the real line inside this contour is always (0,1)).
    Integration by parts doesn't help me here because neither the Tan(z) integral nor the Tan(z) derivative are something that can help me with the integration.
  5. Nov 15, 2013 #4
    How about just brute-force run around it in a counter-clockwise direction and see what the integrals look like? Don't even worry about solving it yet. I'll set up the first one, the horizontal segment from -R to -R+1: Letting z=x+Ri, then

    [tex]\int_{C_1}e^{-i\pi z^2}\tan(\pi z)dz=\int_{-R}^{-R+1} e^{-i\pi(x+Ri)^2}\tan(\pi(x+Ri))dx[/tex]

    Ok, now do the next one, the diagonal line running from the point -R+1-Ri to R+1+Ri. How would you parameterize that line in terms of some parameter t such that z(t) tracks that line from start to finish?
    Last edited: Nov 15, 2013
  6. Nov 15, 2013 #5
    Edit: jackmell's got it.
  7. Nov 15, 2013 #6
    [itex]\displaystyle \int_{C} e^{-i \pi z^2} \tan (\pi z) dz =[/itex]

    [itex]\displaystyle \int_{R(-1-i))}^{R(-1-i)+1} e^{-i \pi t^2} \tan (\pi t) dt +
    \displaystyle \int_{R(-1-i)+1}^{R(1+i)+1} e^{-i \pi t^2} \tan (\pi t) dt +
    \displaystyle \int_{R(1+i)+1}^{R(1+i)} e^{-i \pi t^2} \tan (\pi t) dt +
    \displaystyle \int_{R(1+i)}^{R(-1-i)} e^{-i \pi t^2} \tan (\pi t) dt [/itex]

    [itex]\displaystyle \int_{-R}^{-R+1} e^{-i \pi (t+Ri)^2} \tan (\pi (t+Ri)) dt +
    \displaystyle \int_{R+1}^{R} e^{-i \pi (t-Ri)^2} \tan (\pi (t-Ri)) dt [/itex]
    [itex]\displaystyle \int_{-R}^{-R+1} e^{-i \pi (t+Ri)^2} \tan (\pi (t+Ri)) dt = - \displaystyle \int_{R}^{R+1} e^{-i \pi (-t'+Ri)^2} \tan (\pi (-t'+Ri)) dt' = [/itex]
    [itex]= \displaystyle \int_{R+1}^{R} e^{-i \pi (-t'+Ri)^2} \tan (\pi (-t'+Ri)) dt'= \displaystyle \int_{R+1}^{R} e^{-i \pi (t'-Ri)^2} \tan (\pi (-t'+Ri)) dt'[/itex]
    [itex]\displaystyle \int_{-R}^{-R+1} e^{-i \pi (t+Ri)^2} \tan (\pi (t+Ri)) dt +
    \displaystyle \int_{R+1}^{R} e^{-i \pi (t-Ri)^2} \tan (\pi (t-Ri)) dt =[/itex]
    [itex]\displaystyle \int_{R+1}^{R} e^{-i \pi (t-Ri)^2} (\tan (\pi (-t-Ri)) + \tan(\pi (-t+Ri)) dt = [/itex]
    I think if I make R be only naturals numbers both Tan will be this integral will be equal to zero.
    Anyway I still have no clue how this could help me.
  8. Nov 16, 2013 #7
    Firstly, we need a good plot to identify what and where we're integrating:


    So I can write:

    [tex]\mathop\int\limits_{\text{Red}} fdz+\mathop\int\limits_{\text{Blue}} fdz+\mathop\int\limits_{\text{Green}} fdz +\mathop\int\limits_{\text{Black}} fdz=2\pi i\;\text{Res}(f,1/2)[/tex]

    Now, the tan may turn out to be a problem. So why don't we convert everything to exponentials and see where that leads? Not sure though but sometimes you just have to try things. So:

    [tex]e^{-i\pi z^2}\tan(\pi z)=-ie^{-i\pi z^2}\left(\frac{e^{i\pi z}-e^{-i\pi z}}{e^{i\pi z}+e^{-i\pi z}}\right)[/tex]

    Attached Files:

    Last edited: Nov 16, 2013
  9. Nov 16, 2013 #8
    [itex]\displaystyle \int_{R}^{-R} e^{-((1+i)t)^{2} \pi} \tan ((1+i)\pi t) dt = \displaystyle -\int_{-R}^{R} e^{-((1+i)t)^{2} \pi} \tan ((1+i)\pi t) dt=[/itex]

    [itex]\displaystyle \int_{-R}^{0} e^{-((1+i)t)^{2} \pi} \tan ((1+i)\pi t) dt + \displaystyle \int_{0}^{R} e^{-((1+i)t)^{2} \pi} \tan ((1+i)\pi t) dt =[/itex]
    Let be t' = -t

    [itex]\displaystyle \int_{R}^{0} e^{-((1+i)t')^{2} \pi} \tan (-(1+i)\pi t') dt = [/itex]
    [itex]- \displaystyle \int_{0}^{R} e^{-((1+i)t')^{2} \pi} \tan (-(1+i)\pi t') dt[/itex]
    I know that in the real line, tan is a impair function so its integral in (-p,p) is zero, but I have to see if it works in the complex plane, at least in the line of 45º.
    I dropped the pi for the sake of simplicity.
    [itex]tan((1+i)t) = \frac{e^{(1+i)t}-e^{-(1+i)t}}{e^{(1+i)t}+e^{-(1+i)t}}[/itex]
    As [itex]cos((1+i)t)=cos(-(1+i)t)[/itex] we can add bot tans easily.

    [itex]tan((1+i)t) + tan(-(1+i)t)= \frac{e^{(1+i)t}-e^{-(1+i)t} + e^{-(1+i)t} - e^{(1+i)t}}{e^{(1+i)t}+e^{-(1+i)t}}=0[/itex]
    So, we have one null integral, two integrals whose sum is zero and the fourth integral.
    So the sum of the residues has to be equal to the fouth one:
    [itex]\displaystyle \int_{R(-1-i)+1}^{R(1+i)+1} e^{-i \pi t^2} \tan (\pi t) dt = \displaystyle \sum Res[/itex]

    What software did you use to do the plot jack?
    Last edited: Nov 16, 2013
  10. Nov 16, 2013 #9
    Dang it. Guess I'm stuck. Maybe I shouldn't reply unless I know how to do it. Ok, let me be clear on what I have so far:

    (1) I do not follow your work. Not to say it's wrong, but it is not clearly presented in my opinion.

    (2) This is what I have so far:

    (a) Proved the integral over the black contour is zero.

    (b) Showed the integral over the blue contour is:

    $$ \int_{\text{Blue}} e^{-i\pi z^2}\tan(\pi z)dz=\int_{-R+1}^{R+1} e^{-i\pi(x+(x-1)i)^2}\tan\left[\pi\{x+(x-1)\}\right](1+i)dx$$

    (c) Showed the integral over the red plus the green is:

    $$\int_{\text{Red}}fdz+\int_{\text{Green}} fdz=-\int_{R-1}^{R+1} e^{-i\pi(u+Ri)^2} \tan\left[\pi(u+Ri)\right] du$$

    So that:

    \oint e^{-i\pi z^2} \tan(\pi z)dz&=\int_{-R+1}^{R+1} e^{-i\pi[x+(x-1)i]^2}\tan[\pi\{x+(x-1)i\}](1+i)dx \\
    &-\int_{R-1}^{R+1} e^{-i\pi(u+Ri)^2}\tan[\pi(u+Ri)]du

    and I checked that quantity numerically with ##R=2## and it agreed with the residue calculation: ##2\pi i\text{Res}(f,1/2)## so I know it is correct.

    And I use Mathematica to plot the contour.
    Last edited: Nov 16, 2013
  11. Nov 17, 2013 #10
    . . . fine then. Had to resort to just looking it up. Here's one using

    $$\frac{e^{\pi iz^2}}{\sin(\pi z)}$$

    over the parallelogram ##\pm 1/2+te^{\pi i/4}##. I like the idea of use polar coordinates for the diagonal contours. Maybe that would make the tan problem easier to solve. But I think we should first go over the sine problem. That's pretty close to the tan problem. So here's my new suggestion:

    (1) Study and understand every detail of the sin problem in the reference:


    (2) Then see if we can apply what we learned in (1) above to the tan one. If not, then back to google.

    And no I don't want someone posting a reference to the tan problem. Not yet anyway. :)
    Last edited: Nov 17, 2013
  12. Nov 17, 2013 #11


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    Homework Helper

    ah, that's a nice link. Yeah, it might be best to try the sin method first. It looks very similar to the tan method, so you might be able to work out the tan method once you've tried the sin method. The tan method looks more difficult for two reasons: 1) there is a pole lying on the contour integral 2) there is another pole inside the region. But the sin method looks nicer because there is simply one pole inside the region, and no pole lying on the contour.

    edit: ah, the two methods are not as similar as I thought. The tan method has an ##e^{-i\pi z^2}## while the sin method has an ##e^{i\pi z^2}##... maybe the two methods are still related though. Also I've not tried to do the sin method or the tan method yet, so don't take what I'm saying too seriously.

    edit again: what the heck am I talking about?! The tan method doesn't have a pole lying on the contour integral, sorry, I saw something that was not there because I tried to think about it too quickly. Alright, so maybe the tan method is not much more difficult than the sin method, but anyway, you have a reference for the sin method, so at least you can practice and compare with that one.
    Last edited: Nov 17, 2013
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