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I think this involves Leibniz theorem

  1. Apr 20, 2015 #1
    1. The problem statement, all variables and given/known data
    For x> 0 , let f(x) = $$\int _1^x \frac{log t dt }{1+t}$$
    Then ## f(x) + f(1/x)## is equal to :

    A. ##¼ (log x)^2 ##

    B. ## ½ (log x)^2 ##

    C. ##log x ##

    D. ## ¼ log x^2 ##
    2. Relevant equations
    Suppose f(x) = $$\int _1^x g(t)$$
    Then by Leibniz rule ,
    f' (x) = g(x)

    3. The attempt at a solution
    I found f' (x) = logx/(1+x)
    f' ( 1/x) = logx/x(x+1)
    f' (x) + f'(1/x) = logx/x
    Now what to do?
    We can integrate both sides but a constant will appear which we don't know?
     
  2. jcsd
  3. Apr 20, 2015 #2
    Hi!!!
    I think this question doesn't require the leibnitz rule. I think it CAN be done by :
    f(x)=∫1x(logt/1+t)dt
    Substituting t=1/k
    dt=-dk/k2
    1 will be 1
    And x will become 1/x
    Then f(x)=∫11/x[log k/k(k+1)]dk
    Then you can find f(1/x) from this equation and add that to the original f(x).
    But I am getting an answer that is not any of the options . I think I am getting a wrong f(1/x). If this method works , can you give the steps? I would be very thankful.
     
  4. Apr 20, 2015 #3
    I am getting solving your attempt,
    f(x) + f(1/x) = $$ \int_1^{1/x} \frac{logk dk}{k(k+1)} + \int_1^{x} \frac{logk dk}{k(k+1)} $$
    How we'll solve further?
     
  5. Apr 20, 2015 #4
    Instead of using the new f(x) , use the original one( given in the question).
     
  6. Apr 20, 2015 #5
    I am then getting,
    same result.
    How you have evaluated the integral?
     
  7. Apr 20, 2015 #6
    I am saying that:
    f(x)+f(1/x)=
    1x(log t/1+t)dt[i.e the original integrand]+∫1xlog[log t /t(t+1)]dt [i have replaced k with t]
     
  8. Apr 20, 2015 #7
    Mistake in bold part.
    That bold part should not be there.
    Then we'll simply get,
    1x (logt/t)
    = (logt)2/2
    Putting limits,
    1/2 (logx)2
    Got it
    Thanks buddy.
    Can you tell what was wrong in my attempt?
     
  9. Apr 20, 2015 #8
    Are you talking about post #1? No ., sorry I dont know how to do this question by that method. You are right about the unavaibility of constant of integration that would be needed....
     
  10. Apr 20, 2015 #9
    But we are getting that answer only if we don't take into account constant of integration, in post 1 attempt.
     
  11. Apr 20, 2015 #10

    Dick

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    You can determine the constant of integration. Put ##x=1##.
     
  12. Apr 20, 2015 #11
    We'll get 2f(1) = 0 + c
    Now f(1) = 0
    So c= 0.
    How did you think that we should put x = 1 ?
     
  13. Apr 20, 2015 #12

    Dick

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    So ##f(1)=0## tells you the constant of integration must be zero. What's the question?
     
  14. Apr 20, 2015 #13
    But instead of writing question again, I think you could see it yourself in post 1.
    I think you mean something else.
    Integral from 1 to 1 is zero.
     
  15. Apr 20, 2015 #14

    Dick

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    Integrating both sides as in post 1 gives you ##\frac{1}{2} (log(x))^2+C=f(x)+f(1/x)## where ##C## is the constant of integration. Putting ##x=1## shows that ##C=0##. Isn't that what you wanted to know?
     
  16. Apr 20, 2015 #15
    Yeah. Thanks.
     
  17. Apr 20, 2015 #16
    Yes, but How you got the idea that we'll find C if we'll put x = 1?
     
  18. Apr 20, 2015 #17

    Dick

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    It's the same way you eliminate a constant of integration when you are solving other differential equations. You substitute a point where the functions of ##x## are known and find ##C##.
     
    Last edited: Apr 20, 2015
  19. Apr 22, 2015 #18
    Shouldn't ##f'(1/x) =-xlog x/1+x ##
     
  20. Apr 22, 2015 #19
    Why it should be?
    Apply Leibniz rule. Can you show what you have done?
    For x> 0 , let f(x) = $$\int _1^x \frac{log t dt }{1+t}$$
    Then ## f(x) + f(1/x)## is equal to :


    Suppose f(x) = $$\int _1^x g(t)$$
    Then by Leibniz rule ,
    f' (x) = g(x)

    3. The attempt at a solution
    I found f' (x) = logx/(1+x)
    f' ( 1/x) = logx/x(x+1)
    f' (x) + f'(1/x) = logx/x
     
  21. Apr 22, 2015 #20

    Ray Vickson

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    The notation ##f'(1/x)## should be avoided in this problem; it means ##\left. f'(t) \right|_{t = 1/x}## in standard mathematical notation. That is very different from ##(d/dx) f(1/x)##, which is equal to ##-f'(1/x)/x^2##.
     
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