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Integration by Substitution fails

  1. Dec 4, 2007 #1
    Question:
    [tex]\int^{1}_{-1} \frac{dx}{(1+x^4)} [/tex]

    I attempt:
    u = x^2,
    so x= u^1/2
    dx= 1/2 u^(-1/2)

    Which gives me [tex]\int^{1}_{1} \frac{1}{(1+x^4)} * \frac{1}{(2u^1/2)}du[/tex], which is 0. Thats not the answer as seen by any graphing utility.
    Where is this error? I do not know integration by parts. I just wish to know why my logic fails.

    Any help? :/
     
  2. jcsd
  3. Dec 4, 2007 #2

    Avodyne

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    Your integral goes from 1 to 0 and back to 1. Write the first integral as twice the integral from 0 to 1.
     
  4. Dec 4, 2007 #3
    Ok so it would be
    2 * [tex]\int^{1}_{0} \frac{1}{(1+x^4)} * \frac{1}{(2u^1/2)}du[/tex]

    Now, is my logic correct when I do?
    u = x^2,
    so x= u^1/2
    dx= 1/2 u^(-1/2)
     
  5. Dec 4, 2007 #4

    Avodyne

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    Yes, your logic is correct.
     
  6. Dec 4, 2007 #5

    dynamicsolo

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    I can't say I'm clear as to how a u-substitution will help here (having just tried it out myself). There's nothing here to grab ahold of for integration by parts and the denominator is an irreducible quartic polynomial, so integration by partial fractions doesn't go anywhere. I want to say that something involving a trigonometric substitution might dent it, but there will be work beyond that. I finally looked up the indefinite integral in Petit Bois and it is not pretty, suggesting that there are multiple layers of substitution involved...

    If an infinite series is acceptable for an answer, it is relatively easy to convert the integrand to a power series which is then integrated term-by-term. Since the limits of integration are 0 and 1, the result simplifies considerably to something related to an alternating harmonic series (thus convergent).
     
  7. Dec 5, 2007 #6

    AKG

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    No, x is not equal to u1/2. x originally goes from -1 to 1, so sometimes x is negative and sometimes it's positive. x = u1/2 when x is positive and x = -u1/2 when x is negative.
     
  8. Dec 5, 2007 #7

    HallsofIvy

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    I'm surprised that no one has mentioned an obvious flaw in the "logic". If you are substituting u= some function of x, then the final integral cannot have both x and u in it!
    If you had taken u= x2, then du= 2x dx so du/2x= du/2u1/2= dx. (as AKG pointed out, that is only true for x positive. Fortunately, the integral is clearly symmetric so you can integrate taking x from 0 to 1 and then double.)
    When x= 0, u= 0, when x= 1, u= 1. The integral becomes
    [tex]\frac{1}{2}\int_0^1 \frac{1}{1+u^2} \frac{du}{\sqrt{u}}[/tex]
    Unfortunately, I don't think that's going to be any easier to integrate!
     
  9. Dec 5, 2007 #8

    dynamicsolo

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    Since it was an intermediate step -- and, in any event, doesn't go anywhere in solving the problem -- it didn't seem worth remarking on that. It certainly wouldn't be left that way in continuing the development.

    Unfortunately, it seems that a number of lecturers nowadays actually write down the first step for substitutions into an integrand this way (to judge from the papers I've been grading this past week). It isn't an utterly terrible abuse of notation, since u is plainly intended as a function of x, but it sure sets my teeth on edge (especially when a student is doing a definite integral like that and then can't seem to decide which set of integration limits to use...).
     
  10. Dec 5, 2007 #9

    Gib Z

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    A quartic polynomial with real coefficients can always be factored into two quadratic polynomials with real coefficients, as a consequence of the conjugate root theorem. And thats actually how to solve this integral, first step is to use partial fractions.

    If someone wants to try a substitution, you'll have trouble unless you can do [tex]\int \frac{1}{\sqrt{\tan x}} dx[/tex]

    EDIT: I must warn, though it is possible to do by partial fractions and successive substitutions that aren't too difficult to spot, I would take a good 30 minutes on it, even if your pretty fast.
     
    Last edited: Dec 5, 2007
  11. Dec 5, 2007 #10

    dynamicsolo

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    The fourth roots of -1 would explain the square roots of 2 in the tabulated integration I looked up.

    Yeah, I thought of [tex]x^2 = tan (u)[/tex] this morning, but the outcome just made it clearer why the final integral function is so messy.

    I suspect the cleanest way to answer the original question is to integrate the power series for [tex]\frac{1}{(1+x^4)} [/tex], which at least gives the nice-looking result for the definite integral

    2 · [ 1 - (1/5) + (1/9) - (1/13) + ... ]
     
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