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Integral 1/(x^4+1)

  1. Sep 18, 2011 #1
    I know this is not a school forum...

    But I think this problem has been solved multiple times all over the internett, but now I am unable to find a complete soultion for this problem.

    1/(x^4+1)

    I think it also have been solved multiple times on this site.

    Could anyone help me find a complete solution? (Link to thread here)

    The basic outline I remember from reading about it before is

    1. Rationalize the denominator

    2. Split the integral into two mean integrals

    3. Divide top and bottom by x^2

    4. Factor

    5. Substitution

    But as I said, I am unable to find it now.
     
  2. jcsd
  3. Sep 18, 2011 #2
    you can use trig substitution 1 + tan^2(x) = sec^2(x)

    so x = sqrt(tan(x))

    dx = ?

    then use substitution again
     
  4. Sep 18, 2011 #3
    Yes draw your triangle and proceed with a trig substitution
     
  5. Sep 18, 2011 #4
    Gargl... I tried to submt this into the non homework section but alas, I was redirected here...

    For once this is not a problem any teacher, except the sadistic ones would give to any student.

    The only time you will see this integral is in the course Complex Analysis, and then you will evaluate this using contour integration. Simply this is just a mean proble, and I wanted to look at the solution. Not trying to solve it myself...

    Trig substitution will not work here. Since you will end up with [itex]\frac{1}{\sqrt{\tan(x)}}[/itex]... Which is just more complicated...

    My work so far. I do not really want to use trig integrals. And I belive atleast one of these integrals below can be solved without trig. (I want to avoid trig substitutions because of the mean / near impossible back-substitutions)

    Does this look correct?

    [tex]
    \begin{array}{l}
    I = \int {\frac{1}{{{x^4} + 1}}dx} \\
    I = \int {\frac{1}{{\left( {{x^2} - \sqrt 2 x + 1} \right)\left( {{x^2} + \sqrt 2 x + 1} \right)}}dx} \\
    I = \frac{1}{{2\sqrt 2 }}\int {\frac{{\sqrt 2 - x}}{{{x^2} - \sqrt 2 x + 1}} + \frac{{x + \sqrt 2 }}{{{x^2} + \sqrt 2 x + 1}}dx} \\
    I = \frac{1}{{2\sqrt 2 }}\int {\frac{{\sqrt 2 - x}}{{{{\left( {x - \frac{{\sqrt 2 }}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2}}} + \frac{{x + \sqrt 2 }}{{{{\left( {x + \frac{{\sqrt 2 }}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2}}}dx} \\
    \end{array}
    [/tex]

    any tips how the easiest way to proceed is / any threads dealing with this integral?
     
  6. Sep 18, 2011 #5
    [tex]
    z^{4} + 1 = 0
    [/tex]
    has 4 complex roots:
    [tex]
    z_{1} = e^{i \frac{\pi}{4}} = \frac{1 + i}{\sqrt{2}}
    [/tex]

    [tex]
    z_{2} = e^{i \frac{3 \pi}{4}} = \frac{-1 + i}{\sqrt{2}}
    [/tex]

    [tex]
    z_{3} = e^{i \frac{5 \pi}{4}} = \frac{-1 - i}{\sqrt{2}}
    [/tex]

    [tex]
    z_{4} = e^{i \frac{7 \pi}{4}} = \frac{1 - i}{\sqrt{2}}
    [/tex]

    Of these, [itex]z_{4}[/itex] is the complex conjugate of [itex]z_{1}[/itex] and [itex]z_{3}[/itex] is the complex conjugate of [itex]z_{2}[/itex]. Now:
    [tex]
    (z - z_{0})(z - \overline{z}_{0}) = z^{2} - (z + \overline{z}_{0}) z + z_{0} \overline{z}_{0} = z^{2} - 2 \mathrm{Re}(z_{0}) z + |z_{0}|^{2}
    [/tex]
    so:
    [tex]
    (z - z_{1})(z - z_{4}) = z^{2} - \sqrt{2} z + 1
    [/tex]
    and
    [tex]
    (z - z_{2})(z - z_{3}) = z^{2} + \sqrt{2} z + 1
    [/tex]
    So, the partial fraction decomposition is:
    [tex]
    \frac{1}{z^{4} + 1} = \frac{1}{(z^{2} - \sqrt{2} z + 1)(z^{2} + \sqrt{2} z + 1)} = \frac{A z + B}{z^{2} - \sqrt{2} z + 1} + \frac{C z + D}{z^{2} + \sqrt{2} z + 1}
    [/tex]
    You need to find the undetermined coefficients [itex]A, B, C, D[/itex] and then use integrals of quadratic binomials. You will use:
    [tex]
    \frac{d}{d z}\left(z^{2} \mp \sqrt{2} z + 1\right) = 2 z \mp \sqrt{2}
    [/tex]
    and
    [tex]
    z^{2} \mp \sqrt{2} z + 1 = (z \mp \frac{1}{\sqrt{2}})^{2} + \left(\frac{1}{\sqrt{2}}\right)^{2}
    [/tex]
     
  7. Sep 18, 2011 #6
    Which is exactly what I id, if you look at the post above...

    Here is my solution though... Never thought I would do all of it.

    Can anyone spot any mistakes here, as I wen through quite a lot to solve this pesky integral.

    ThKrW.gif
     
  8. Sep 18, 2011 #7
  9. Sep 18, 2011 #8
    If you use:
    [tex]
    \mathrm{arctan}(x) \pm \mathrm(y) = \mathrm{arctan} \left( \frac{x \pm y}{1 \mp x y} \right)
    [/tex]
    then
    [tex]
    \begin{array}{l}
    \mathrm{arctan}(\sqrt{2} x + 1) - \mathrm{arctan}(1 - \sqrt{2} x) = \mathrm{arctan}\left( \frac{(\sqrt{2} x + 1) - (1 - \sqrt{2} x)}{1 + (1 + \sqrt{2} x) (1 - \sqrt{2} x)} \right) \\

    = \mathrm{arctan} \left( \frac{2 \sqrt{2} x}{1 + 1 - 2 x^{2}} \right) = \mathrm{arctan} \left( \frac{\sqrt{2} x}{1 - x^{2}} \right) = -\mathrm{arctan} \left( \frac{\sqrt{2} x}{x^{2} - 1} \right)
    \end{array}
    [/tex]
    which is what you have as the argument of the arc tangent function. I think it's correct.
     
  10. Sep 18, 2011 #9
    The sign of your logarithms seems opposite.
     
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