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Miscellaneous Definite Integrals

  1. Jun 6, 2012 #1
    1. The problem statement, all variables and given/known data

    show that

    [itex]\int^{∞}_{0}\frac{sin^{2}x}{x^{2}}dx= \frac{\pi}{2}[/itex]


    2. Relevant equations

    consider

    [itex]\oint_{C}\frac{1-e^{i2z}}{z^{2}}dz[/itex]

    where C is a semi circle of radius R, about 0,0 with an indent (another semi circle) excluding 0,0.

    3. The attempt at a solution


    curve splits into
    C1, line segment from -R to -ε
    C2, semi circle z=εe θ goes from ∏ to 0
    C3, line segment from ε to R
    C4, semi circle z=Re θ goes from 0 to ∏

    function is holomorphic in/on C. so integral =0

    for C4, applying limit R > infinity integral = 0

    C3. really stuck here

    I sub in the value of z

    to get

    [itex]\int^{0}_{∏}\frac{1-e^{i2e^{iθ}}}{ε^{2}e^{2iθ}}εe^{iθ}[/itex]


    but I can't take the limit ε -> 0 of this since theres an ε on the denominator
     
  2. jcsd
  3. Jun 6, 2012 #2

    vela

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    You forgot an ε in the exponent and a factor of i.
    $$\int^0_\pi \frac{1-e^{i2\varepsilon e^{i\theta}}}{\varepsilon^2 e^{2i\theta}} i\varepsilon e^{i\theta}\,d\theta$$
     
  4. Jun 6, 2012 #3
    Thank you. I was half awake when I wrote that.

    I just don't know how to get that ε out from the denominator, I cannot take the limit otherwise.
     
  5. Jun 6, 2012 #4

    vela

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    Try expanding the exponential and the numerator as a series.
     
  6. Jun 6, 2012 #5
    The teacher gave me a hint, use L'hopitals Rule, which isn't clear to be possible since i forgot to add in the limit
     
  7. Jun 6, 2012 #6

    vela

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    I'm not sure what you mean by that. In any case, either way should work. The numerator is first-order in ε. Along with the factor of ε from dz, it cancels the ε2 in the denominator, so you will get a finite result.
     
  8. Jun 7, 2012 #7
    I mean, differentiating the top and bottom before taking the limit.

    as in a taylor series?
     
  9. Jun 7, 2012 #8

    vela

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    I still have no clue what you mean.
     
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