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Indefinite Integration

  1. Aug 31, 2012 #1
    1. The problem statement, all variables and given/known data
    The value of [tex]\int_0^{1} (\prod_{r=1}^{n} (x+r))(\sum_{k=1}^{n} \frac{1}{x+k}) dx[/tex] equals:
    a)n
    b)n!
    c)(n+1)!
    d)n.n!

    (Can someone tell me how to make bigger parentheses using latex?)

    2. Relevant equations



    3. The attempt at a solution
    I know that the question becomes a lot easier if i put n=1 or 2 and then integrate. But i was wondering if there is any proper way to solve it. I can't go further after expanding the given expression.
    [tex]\int_{0}^{1} (x+1)(x+2)......(x+n)(\frac{1}{x+1}+\frac{1}{x+2}+.....\frac{1}{x+n})dx[/tex]
    which is equal to
    [tex]\int_{0}^{1} \sum_{r=1}^n \frac{(x+n)!}{x+r}[/tex]
    I am stuck now, i can't find any way further.

    Any help is appreciated. :smile:
     
  2. jcsd
  3. Aug 31, 2012 #2
    I still don't have any clue.
     
  4. Aug 31, 2012 #3

    ehild

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    What is the derivative of a product? What is the derivative of (x+1)(x+2)? and (x+1)[(x+2)(x+3)] ?

    ehild
     
  5. Aug 31, 2012 #4
    Derivative of (x+1)(x+2)=2x+3
    Derivative of (x+1)(x+2)(x+3)=3x2+12x+11

    I still don't get any idea how this would help?
     
  6. Aug 31, 2012 #5

    ehild

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    Do not simplify.

    d/dx[(x+1)(x+2)]=(x+2)+(x+1)=(x+1)(x+2)/(x+1)+(x+1)(x+2)/(x+2)=[(x+1)(x+2)][1/(x+1)+1/(x+2)]

    d/dx[(x+1)(x+2)(x+3)]=(x+2)(x+3)+(x+1)(x+3)+(x+1)(x+2)=[(x+1)(x+2)(x+3)][1/(x+1)+1/(x+2)+1(x+3)]

    Do you see?

    ehild
     
  7. Aug 31, 2012 #6
    Wait, i guess i have got it, i will be back in a few hours with a solution.
    Thanks for the help! :smile:

    EDIT: Oops, seems like you posted just a few seconds before me.
     
  8. Aug 31, 2012 #7

    ehild

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    Well, I looking forward to the solution:smile:

    ehild
     
  9. Aug 31, 2012 #8
    So here's the solution:
    Let
    [tex]t=(x+1)(x+2)(x+3).......(x+n)[/tex]
    [tex]\ln t=\ln(x+1)+\ln(x+2)+\ln(x+3)......+\ln(x+n)[/tex]
    [tex]\frac{1}{t}\frac{dt}{dx}=\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}......\frac{1}{x+n}[/tex]
    [tex]dt=(x+1)(x+2)(x+3)....(x+n)(\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}........+\frac{1}{x+n})dx[/tex]
    Therefore the original question becomes ∫dt, upper limit is (n+1).n! and lower limit is n!.
    Solving, i get the answer as d) option.

    Thanks for the help! :smile:
     
  10. Aug 31, 2012 #9

    Mentallic

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    By the way, since it seems like your little sub-question about the larger brackets was missed, I'll chime in on that.

    I myself use /left( and /right) which looks like: [tex]\left(\frac{1}{1}\right)[/tex] but there are other more specific "fonts" you can use, which ranges from smallest to largest:

    /big( [tex]\big(\frac{1}{1}\big)[/tex]
    /Big( [tex]\Big(\frac{1}{1}\Big)[/tex]
    /bigg( [tex]\bigg(\frac{1}{1}\bigg)[/tex]
    /Bigg( [tex]\Bigg(\frac{1}{1}\Bigg)[/tex]

    And there could be others, but that should about cover it.
     
  11. Aug 31, 2012 #10

    ehild

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    That is ingenious! You are really cool:cool:

    Very good. You are better and better every day!

    ehild
     
  12. Aug 31, 2012 #11
    Thanks Mentallic, really helpful. :smile:
    I did nothing, you almost solved it by giving me a simple hint. Thanks! :smile:
     
  13. Aug 31, 2012 #12

    ehild

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    That method of using logarithm was entirely your idea :cool:. I did not think of it.

    ehild
     
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