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Integration of sinx/x

  1. Sep 24, 2011 #1
    Wherever I look on the internet it says that sinx/x cannot be integrated by elementary techniques, but it seems that there is a method using integration by parts of the quotient rule. However, when I compute definite integrals with this, the answer that my calculator returns is different than the answer I get with the definite integral. Can anyone tell me where my problem is or if my method does actually work.

    [tex]\frac{d}{dx}\frac{f(x)}{g(x)} = \frac{f'(x)g(x)-g'(x)f(x)}{g^2(x)}[/tex]
    [tex]\frac{f(x)}{g(x)}=\int \frac{f'(x)g(x)}{g^2(x)}\,dx-\int \frac{g'(x)f(x)}{g^2(x)}\,dx[/tex]
    [tex]\int \frac{g'(x)f(x)}{g^2(x)}\,dx=\int \frac{f'(x)}{g(x)}\,dx-\frac{f(x)}{g(x)}\,dx[/tex]
    [tex]\int \frac{sin(x)}{x}\,dx=\int \frac{xsin(x)}{x^2}\,dx=\frac{1}{2}\int \frac{2xsin(x)}{x^2}\,dx[/tex]
    [tex]g(x)=x^2[/tex]
    [tex]f(x)=sin(x)[/tex]
    [tex]g'(x)=2x[/tex]
    [tex]f'(x)=cos(x)[/tex]
    [tex]\int \frac{g'(x)f(x)}{g^2(x)}\,dx=\int \frac{f'(x)}{g(x)}\,dx-\frac{f(x)}{g(x)}\,dx[/tex]
    [tex]\frac{1}{2}\int \frac{2xsin(x)}{x^2}\,dx=\frac{1}{2}\int \frac{cos(x)}{x}\,dx-\frac{sin(x)}{x}\,dx[/tex]
    [tex]\int \frac{cos(x)}{x}\,dx=\frac{1}{2}\int \frac{2xcos(x)}{x^2}\,dx[/tex]
    [tex]\frac{1}{2}\int \frac{2xcos(x)}{x^2}\,dx=\frac{1}{2}\int \frac{-sin(x)}{x}\,dx-\frac{1}{2}\frac{cos(x)}{x}\,dx[/tex]
    [tex]\frac{1}{2}\int \frac{2xcos(x)}{x^2}\,dx=\frac{1}{2}\int \frac{-sin(x)}{x}\,dx-\frac{1}{2}\frac{cos(x)}{x}\,dx[/tex]
    [tex]\int \frac{sin(x)}{x}\,dx=\frac{1}{2}\left(\frac{1}{2} \int \frac{-sin(x)}{x} \,dx - \frac{1}{2}\frac{cos(x)}{x}\right)-\frac{1}{2}\frac{sin(x)}{x}[/tex]
    [tex]\frac{5}{4}\int \frac{sin(x)}{x}\,dx=-\frac{1}{4}\frac{cos(x)}{x}-\frac{1}{2}\frac{sin(x)}{x}[/tex]
    [tex]\int \frac{sin(x)}{x}\,dx=-\frac{1}{5}\frac{cos(x)}{x}-\frac{2}{5}\frac{sin(x)}{x}[/tex]
     
  2. jcsd
  3. Sep 24, 2011 #2

    rude man

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    I got a bit of a headache following you since there are extra dx's everywhere & at least one is missing ....

    But, could you check your result against a power series expansion of
    sin(x)/x = 1 - x^2/3! + x^4/5! - ....

    Integral sin(x)/x = x - x^3/3(3!) + x^5/5(5!) - ...

    and compare that with the series for your answer?
     
  4. Sep 24, 2011 #3

    gb7nash

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    I haven't looked at your method yet, but there's an easy way to tell if your method works. Take the derivative of your answer and see if you get sin(x)/x
     
  5. Sep 24, 2011 #4

    rude man

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    good point gb nash! Somebody please do it - I gotta go hike.
     
  6. Sep 24, 2011 #5
    your mistake is that you took [itex]g(x) = x^{2}[/itex], whereas, according to the formula you derived the denominator is [itex]g^{2}(x)[/itex] and should be equal to [itex]x^{2}[/itex], thus [itex]g(x) = x[/itex]. But, then, [itex]g'(x) = 1[/itex] and not [itex]2 x[/itex] as you had assumed.
     
  7. Sep 24, 2011 #6
    Thanks. I guess that means the method is totally trashed by that explains why it wasn't working.
     
  8. Sep 24, 2011 #7
    Yes.

    Alternatively, you could have used the ordinary integration by parts:

    [tex]
    \int{\frac{\sin{x}}{x} \, dx} = -\int{\frac{1}{x} \, d(\cos{x})} = -\frac{\cos{x}}{x} - \int{\frac{\cos{x}}{x^{2}} \, dx} = -\frac{\cos{x}}{x} - \int{\frac{1}{x^{2}} \, d(\sin{x})} = -\frac{\cos{x}}{x} - \frac{\sin{x}}{x^{2}} - 2 \int{\frac{\sin{x}}{x^{3}} \, dx}
    [/tex]

    You may notice a recursion for the integrals:
    [tex]
    I_{n} \equiv \int{\frac{\sin{x}}{x^{2 n + 1}} \, dx}
    [/tex]

    By similar double integration by parts, you get:
    [tex]
    I_{n} = -\frac{\cos{x}}{x^{2 n + 1}} - \frac{(2 n + 1) \sin{x}}{x^{2 n + 2}} - (2 n + 1) (2 n + 2) I_{n + 1}
    [/tex]

    As you can see, this procedure goes on indefinitely and you do not get a closed form.
     
  9. Sep 25, 2011 #8
    Ohhh. That's genius. Thanks.
     
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