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Math Help - Integral

  1. #1
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    Unhappy Integral

    ∫1/(1+x^5)dx

    short question , long ans

    Help!!!!!!
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  2. #2
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    Hi:

    The integral is readily evaluated via 'integration by parts'. Let dv = dx, and u = 1 / (x^5 + 1). The new integral is a snap.

    Regards,

    Rich B.
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  3. #3
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    Yeah, this integral is one of the most nastiest things.

    I couldn't post the result, 'cause the String is too long (530, limit 400).
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  4. #4
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    I think that any integral,
    \int \frac{1}{1+x^n} dx \,
    Is always elementary.
    How do you show it?
    Last edited by ThePerfectHacker; June 12th 2007 at 04:54 PM.
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  5. #5
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    \int\frac1{1+x^n}\,dx=x_2F_1\left(\frac1n,1;1+\fra  c1n;-x^n\right) \,

    What does this mean???

    Last edited by ThePerfectHacker; June 12th 2007 at 04:54 PM.
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  6. #6
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    Quote Originally Posted by Krizalid View Post
    \int\frac1{1+x^n}\,dx=x_2F_1\left(\frac1n,1;1+\fra  c1n;-x^n\right)

    What does this mean???

    Hypergeometric Series.

    But I do not think this proves that is an elementary function.
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  7. #7
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    Evaluate ∫(1/(1+x^5))dx.

    Note that –1 = –1(cos0+i sin0)
    ∴the roots of the equation x^5+1 = 0 are
    (–1)^(1/5) (cos((0+360k)/5)+i sin((0+360k)/5)) , where k = 0, 1, 2, 3, 4
    = –1(cos0+i sin0) , –1(cos72+i sin72) , –1(cos144+i sin144) , –1(cos216+i sin216) , –1(cos288+i sin288)
    = –1 , –cos72–i sin72 , cos36–i sin36 , cos36+i sin36 , –cos72+i sin72
    ∴x^5+1
    = (x+1)(x+cos72+i sin72)(x–cos 36+i sin36)(x–cos 36–i sin36)(x+cos 72–i sin72)
    = (x+1)((x–cos36)+sin36)((x+cos72)+sin72)
    = (x+1)(x–2x cos36+cos36+sin36)(x+2x cos72+cos72+sin72)
    = (x+1)(x–2x sin54+1)(x+2x sin18+1)

    For finding the EXACT value of sin18 and sin54,
    Consider 318 = 90–218
    sin(318) = sin(90–218)
    sin(318) = cos(218)
    3sin18–4 sin18 = 1–2 sin18
    4sin18–2sin18–3sin18+1 = 0
    (sin18–1)(4sin18+2sin18–1) = 0
    sin18 = 1(rej.) or 4sin18+2sin18–1 = 0
    sin18 = (–2√(2–4(4)(–1)))/(24)
    = (–2√20)/8
    = (√5–1)/4 or (–√5–1)/4(rej.)

    ∴sin54
    = sin(318)
    = 3sin18–4sin18
    = (3(√5–1))/4–4((√5–1)/4)
    = (3(√5–1))/4–(4(5√5–15+3√5–1))/64
    = (12√5–12–5√5+15–3√5+1)/16
    = (√5+1)/4

    ∴(x+1)(x–2x sin54+1)(x+2x sin18+1)
    = (x+1)(x–((√5+1)x)/2+1)(x+((√5–1)x)/2+1)

    ∴Let 1/(1+x^5) ≡ A/(x+1)+(Bx+C)/(x–((√5+1)x)/2+1)+(Dx+E)(x+((√5–1)x)/2+1)
    1 ≡ A(x–((√5+1)x)/2+1)(x+((√5–1)x)/2+1)+(Bx+C)(x+1)(x+((√5–1)x)/2+1)+(Dx+E)(x+1)(x–((√5+1)x)/2+1)
    1 ≡ A(x^4–x+x–x+1)+(Bx+C)(x+1)(x+((√5–1)x)/2+1)+(Dx+E)(x+1)(x–((√5+1)x)/2+1)

    Put x = –1,
    1 = 5A
    A = 1/5

    ∴1 ≡ (x^4–x+x–x+1)/5+(Bx+C)(x+1)(x+((√5–1)x)/2+1)+(Dx+E)(x+1)(x–((√5+1)x)/2+1
    1 ≡ (x^4–x+x–x+1)/5+(Bx+C)(x+((√5+1)x)/2+((√5+1)x)/2+1)+(Dx+E)(x–((√5–1)x)/2–((√5–1)x)/2+1)
    1 ≡ (1/5+B+D)x^4+((–2/5+(√5+1)B+2C–(√5–1)D+2E)x)/2+(2/5+(√5+1)B+(√5+1)C–(√5–1)D–(√5–1)E)x)/2+((–2/5+2B+(√5+1)C+2D–(√5–1)E)x)/2+1/5+C+E
      ╭
      │1/5+B+D = 0…………………………………………...(1)
      │(–2/5+(√5+1)B+2C–(√5–1)D+2E)/2 = 0……………....(2)
    ∴─┤(2/5+(√5+1)B+(√5+1)C–(√5–1)D–(√5–1)E)/2 = 0……(3)
      │(–2/5+2B+(√5+1)C+2D–(√5–1)E = 0…………………(4)
      │1/5+C+E = 1…………………………………………...(5)
      ╰
    (3)2–(2)2:4/5+(√5–1)C–(√5+1)E = 0……(6)
    (3)2–(4)2:4/5+(√5–1)B–(√5+1)D = 0……(7)
    (7)–(1)4: (√5–5)B–(√5+5)D = 0……(8)
    From (1), B+D = –1/5……(9)
    (9)(√5–5)–(8):2√5D = –(√5–5)/5
    D = (5–√5)/(10√5)
    = (√5–1)/10
    (8)+(9)(√5+5):2√5B = –(√5+5)/5
    B = –(√5+5)/(10√5)
    = –(√5+1)/10
    (6)–(5)4: (√5–5)C–(√5+5)E = –4……(10)
    From (5), C+E = 4/5……(11)
    (11)(√5–5)–(10):2√5E = –4(√5–5)/5+4 = (40–4√5)/5
    E = (40–4√5)/(10√5)
    = (4√5–2)/5
    (10)+(11)(√5+5):2√5C = –4–4(√5+5)/5 = –(4√5+40)/5
    C = –(4√5+40)/(10√5)
    = –(4√5+2)/5

    ∴∫(1/(1+x^5))dx
    = ∫[(1/5)/(x+1)+((–(√5+1)x)/10–(4√5+2)/5)/(x–((√5+1)x)/2+1)+(((√5–1)x)/10+(4√5–2)/5)/(x+((√5–1)x)/2+1)]dx
    = (ln│x+1│)/5–((√5+1)/10)∫[(x+(8√5+4)/(√5+1))/(x–((√5+1)x)/2+1)]dx+((√5–1)/10)∫[(x+(8√5–4)/(√5–1))/(x+((√5–1)x)/2+1)]dx+C_1
    = (ln│x+1│)/5+((√5–1)/20)∫[(2x+2√5+18)/(x+((√5–1)x)/2+1)]dx–((√5+1)/20)∫[(2x–2√5+18)/(x–((√5+1)x)/2+1)]dx+C_1
    = (ln│x+1│)/5+((√5–1)/20)∫[(2x+√5–1)/(x+((√5–1)x)/2+1)]dx+(((√5–1)(√5+19))/20)∫[dx/(x+((√5–1)x)/2+1)]–((√5+1)/20)∫[(2x–(√5+1))/(x–((√5+1)x)/2+1)]dx+(((√5+1)(√5–19))/20)∫[dx/(x–((√5+1)x)/2+1)]+C_1
    = (ln│x+1│)/5+((√5–1)/20)ln│x+((√5–1)x)/2+1│–((√5+1)/20)ln│x–((√5+1)x)/2+1│+((9√5–7)/10)∫[dx/((x+(√5–1)/4)+1–(√5–1)/16)]–((9√5+7)/10)∫[dx/((x–(√5+1)/4)+1–(√5+1)/16))]+C_2
    = (ln│x+1│)/5+((√5–1)/20)ln│2x+(√5–1)x+2│–((√5+1)/20)ln│2x–(√5+1)x+2│+((9√5–7)/10)∫[dx/((x+(√5–1)/4)+(2√5+10)/16)]–((9√5+7)/10)∫[dx/((x–(√5+1)/4)+(10–2√5)/16)]+C_3
    = (ln│x+1│)/5+((√5–1)/20)ln│2x+(√5–1)x+2│–((√5+1)/20)ln│2x–(√5+1)x+2│+[(9√5–7)/(10√(2√5+10)/4)]tan^(–1) [(x+(√5–1)/4)/(√(2√5+10)/4)]–[(9√5+7)/(10√(10–2√5)/4)]tan^(–1) [(x–(√5+1)/4)/(√(10–2√5)/4)]+C
    = (ln│x+1│)/5+((√5–1)/20)ln│2x+(√5–1)x+2│–((√5+1)/20)ln│2x–(√5+1)x+2│+[(18√5–14)/(5√(2√5+10))]tan^(–1) [(4x+√5–1)/√(2√5+10)]–[(18√5+14)/(5√(10–2√5))]tan^(–1) [(4x–√5–1)/√(10–2√5)]+C




    I did it in this way...............
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  8. #8
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    Using Maple to do the calculation

    The following result is given by Maple.
    Attached Thumbnails Attached Thumbnails Integral-01jun2007.jpg  
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  9. #9
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    This exactly the same result which The Integrator gives.
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  10. #10
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    Jeez. Thanks guys! I kept wondering why I couldn't figure out such a simple looking integral.
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  11. #11
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    Quote Originally Posted by CrazyAsian View Post
    Jeez. Thanks guys! I kept wondering why I couldn't figure out such a simple looking integral.
    The integral
    \int e^{x^2} dx \,

    Is also simple looking. But I dare you to find it!
    ---
    Just because integrals look simple does not mean they are really easy.
    Last edited by ThePerfectHacker; June 12th 2007 at 04:53 PM.
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