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Apostol's Calculus Vol.1 6.26 #1

  1. Jul 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Let f(x)=∫1x (logt)/(t+1)dt if x>0. Compute f(x) + f(1/x). As a check, you should obtain f(2) + f(1/2) = 1/2(log2)2.

    3. The attempt at a solution
    I have tried various calculation that I know, such as integration by parts, but after ten hours with pieces of papers, I still have no idea on how to lead to the answer: 1/2(logx)2. It'd be appreciated if you gave me some hints!
     
  2. jcsd
  3. Jul 11, 2012 #2

    SammyS

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    Why do you think the answer is (1/2)(log(x))2 ?
     
  4. Jul 11, 2012 #3
    Let me assume your log is ln. Well, directly integrating the original integral goes nowhere because it can't be represented by elementary functions.

    This problem can be done by first finding f'(x)+f'(1/x) using Fundamental Theorem of Calculus
    [tex]
    f(x)=\frac{d}{dx}\int _{a}^{x}f(\tau)d\tau
    [/tex]
    f'(1/x) needs to be handled with care. use u substitution, then use chain rule for differentiation. After all that, you will be integrating a much nicer function that leads to 0.5ln2(x)

     
    Last edited: Jul 11, 2012
  5. Jul 12, 2012 #4
    hmm, I am afraid that Apostol wants us to solve it by applying integral and differential properties of logarithm function instead of FTC as I suggested in post #3.
     
  6. Jul 12, 2012 #5
    If you brute force the integral (as I've tried in Mathematica), you get the PolyLog special function coming into play, as well as pi. In fact before I got Mathematica to simplify it, the answer it gave me for f(2) + f(1/2) looks like this:
    [tex]\frac{\pi^{2}}{6} - Log[\frac{3}{2}] Log[2] + Log[2] Log[3] + PolyLog[2, -2] +
    PolyLog[2, -1/2][/tex]
    which of course simplifies to [itex]\frac{1}{2}Log[2]^{2}[/itex] but it is not intuitive at all.

    So I'm sure Apostol wants you to employ those tricks.
     
  7. Jul 12, 2012 #6
    klondike, I got it, thanks! I've realized that forgetting to multiply by -1/x2 led me to the wrong way, and made me believe that FTC won't make it. Now I can have sweet dreams.:smile:

    SammyS, Apostol's Calculus has answers to numerical exercises in the back.
     
  8. Jul 13, 2012 #7

    SammyS

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    Integrating to find f(x) directly gives a result which includes the Polylogarithm function.

    Using substitution to find an alternate integral representation for f(1/x) can lead to an integral representation of f(x)+f(1/x) which can be integrated using elementary functions... And, yes, it does give the result that
    [itex]\displaystyle f(x) + f\left(\frac{1}{x}\right)=\frac{\left(\ln(x)\right)^2}{2}[/itex][itex]\frac{}{}[/itex]​
     
  9. Jul 13, 2012 #8
    Would you suggest how did you obtain an alternate representation for f(1/x) that can be added to f(x)? Just substituting u=xt into f(1/x) yields
    [itex]\displaystyle f\left(\frac{1}{x}\right) = \int_1^{1/x} \frac{\ln{t}}{t+1} dt = \ln{x} \int_1^x \frac{dt}{t+x} - \int_1^x \frac{\ln{t}}{t+x}dt[/itex]​
    and adding this to f(x) I get
    [itex]\displaystyle f(x) + f\left(\frac{1}{x}\right) = \ln(x)\int_1^x \frac{dt}{t+x} + (x-1)\int_1^x \frac{\ln{t}}{(t+1)(t+x)}dt[/itex]​
    which I am not sure can be converted into an elementary function...
     
  10. Jul 13, 2012 #9

    Curious3141

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    Consider substituting [itex]y = \frac{1}{t}[/itex] into the expression for f(x). Make sure you figure out the correct bounds. Now do a partial fraction decomposition. One term be f(1/x), the other will be easily integrable by parts. You can then solve a simple equation to derive the result required.
     
  11. Jul 13, 2012 #10
    Aha! It makes sense now!
    [itex]\displaystyle f(x) = \int_1^x \frac{\ln{t}}{t+1} dt = \int_1^{1/x} \frac{-\ln{y}}{\frac{1}{y}+1} \cdot \frac{-1}{y^2} dy = \int_1^{1/x} \frac{\ln{y}}{y(y+1)} dy = - f\left(\frac{1}{x}\right) + \int_1^{1/x} \frac{\ln{y}}{y} dy[/itex]​
    which gives the answer. Wow, I learned a lot from this problem. Thanks, everyone!
     
  12. Jul 13, 2012 #11

    SammyS

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    I don't follow that substitution at all.

    Anyway, here's what I did with this.

    [itex]\displaystyle f\left(\frac{1}{x}\right) = \int_1^{1/x} \frac{\ln(t)}{t+1} dt [/itex]

    Let u = 1/t , then t = 1/u, and [itex]\displaystyle dt=-\frac{1}{u^2}du\,.[/itex]

    Furthermore, when t = 1/x, u = x .

    Therefore, [itex]\displaystyle f\left(\frac{1}{x}\right) = -\int_1^{x} \frac{\ln(1/u)}{u^2(1/u+1)} du [/itex]

    ln(1/u) = -ln(u) .

    u is a "dummy" variable, so change it to t, allowing you to combine the integrals.
     
  13. Jul 13, 2012 #12

    Curious3141

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    Masaki's already successfully solved the problem using this method.
     
  14. Jul 13, 2012 #13

    SammyS

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    Now, I see that!

    Well, I started my previous post more than 3 hours ago, then came back to finish it. I didn't want to kill it after composing part of it, so I didn't look for any more recent posts.

    At any rate, the approach is a little different. I suggested working with f(1/x), you suggested working with f(x). My approach didn't use partial fractions.
     
  15. Jul 13, 2012 #14

    Curious3141

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    Fair enough. It's a teensy weensy "potayto, potahto" sort of difference. :wink:
     
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