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Double Integral of Sin(x)/x

  1. Dec 10, 2012 #1
    Evaluate the integral:

    [tex]\int_0^\pi \int_x^\pi \frac{sin(y)}{y} [/tex]


    Look, I've been at this problem for near an hour and a half. I've tried by parts, but I just get stuck in a loop. And I can't think of any way to do this. I've been reading things about taylor expanding it in order to integrate, but that is not something we covered in this course, and is not how it will be expected to be integrated on the test.

    Is there a way, with the use of some algebra, where we could utilize a trig identity?
     
  2. jcsd
  3. Dec 10, 2012 #2

    LCKurtz

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    You forgot the ##dydx## in your integral, which is important. You need to draw the region and reverse the order of integration.
     
  4. Dec 10, 2012 #3
    ahhhhh. I thought that, but was not quite sure.

    Do you know of a mathematical approach to reversing the orders, rather than the graphical method?

    Or at least, could you walk me through the process of drawing it out and picking my new constraints? I haven't had much practice with it, and my professor does a terrible job and explaining anything.
     
  5. Dec 10, 2012 #4

    lurflurf

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    define the function

    [tex]\text{Si}(x)=\int_0^x \dfrac{\sin(t)}{t} \mathop{\text{dt}}[/tex]

    Which is a well known and tabulated function called sine integral.

    Then we have

    [tex]\int_0^\pi \int_x^\pi \dfrac{\sin(y)}{y} \mathop{\text{dy}} \mathop{\text{dx}}=\int_0^\pi (\text{Si}(\pi)-\text{Si}(x)) \mathop{\text{dx}}=\pi \text{Si}(\pi)-\int_0^\pi \text{Si}(x) \mathop{\text{dx}}[/tex]

    then integrate by parts
     
  6. Dec 10, 2012 #5

    Dick

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    The 'graphical method' is the method. Try and draw the domain of integration. It's just a triangle that is the upper half of a square. If you want to get the answer without using the Si function, integrate over the whole square and then subtract the integral over the lower half of the square. Both are pretty easy.
     
  7. Dec 10, 2012 #6

    LCKurtz

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    Your inside limits are ##y=x## and ##y=\pi##. Nothing tricky about drawing those two lines and then, according to the outer limits, taking the part where ##0\le x \le \pi##. You should always draw the region and use the picture to determine the limits when reversing the order of integration.
     
  8. Dec 11, 2012 #7
    Ahhhh. The reason I've always been hesitant to draw it out, because I was under the impression that I would have had to draw out the function itself and compare it with the line of its constraints. And I have no idea how to draw practically any of the functions that we're assigned to integrate.

    I appreciate all the help! It seems so easy now, like I've been stressing over the simple things...
     
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