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Integral Suppose f(x)=int(1/t.dt)

  1. Jul 28, 2003 #1
    Hello everyone, this one might seem easy to you but it's driving me insane.

    Q) Suppose f(x)=int(1/t.dt)
    where the upper limit=x, lower limit=1 ; for x>0.

    Without evaluating the integral show that
    for any x,y>0, f(x)+f(y)=f(xy).
    where you may consider a substitution s=xt in the left-hand side.

    Thanks for your help in advance.
     
    Last edited by a moderator: Feb 5, 2013
  2. jcsd
  3. Jul 28, 2003 #2
    I don't know what you mean by "consider the substitution," but the best I can do anyway is show that the most they differ by is a constant. I at first assumed you didn't want me to differentiate either so I tried separating the integrals (for instance, instead of from 1 to y, I tried 1 to x plus x to y) but I'm not sure that helped.
    If you differentiate f(x)+f(y)=f(xy)
    you get
    1/x*x' + 1/y*y'=1/(xy)(x'y+y'x)=1/x*x'+1/y*y'
    where x' and y' are the derivatives of x and y with respect to t.
    So since the derivatives are equal, they can only differ by a constant.
    To show that C=0, you could pick some values of x and y. I wanted to stay somewhat general, so I picked x=y. It boils down to showing that
    2*S(1/t)dt=S(1/t)dt
    if the limits on the left are 1 and x
    and on the right 1 and x2
     
    Last edited: Jul 28, 2003
  4. Jul 28, 2003 #3

    mathman

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    int(1,x)dt/t=int(y,xy)ds/s, where s=yt.
    However, s is dummy. Therefore
    int(1,x)dt/t + int(1,y)dt/t = int(1,xy)dt/t
     
  5. Jul 28, 2003 #4
    mathman, I'd be interested in seeing a little more work (so I can understand what you did), if you don't mind.
     
  6. Jul 28, 2003 #5

    HallsofIvy

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    You will FIRST need to show that f(1/x)= -f(x):
    f(1/x)= int (t=1 to 1/x) 1/t dt.

    Let u= xt so that t= u/x. Then dt= du/x and 1/t= x/u so that
    (1/t) dt= (x/u)du/x= (1/u) du. When t= 1, u= x and when t= 1/x,
    u= x(1/x)= 1. The integral becomes int(from u= x to u= 1)(1/u)du
    which is -int (from u=1 to u= x)(1/u)du= -f(x).

    Now, f(xy)= int (t=1 to xy) 1/t dt.

    Let u= t/x so that t= xu. Then dt= xdu and 1/t= 1/(xu) so that
    (1/t)dt= (1/(xu))(xdu)= (1/u)du. When t= 1, u= 1/x and when t= xy,
    u= y so the integral becomes int (from u= 1/x to y)(1/u)du
    = int (from u= 1/x to 1)(1/u)du+ int (from u= 1 to y) (1/u)du
    = -int (from u= 1 to 1/x)(1/u)du+ int (from u= 1 to y) (1/u)du
    = -(-f(x))+ f(y)= f(x) + f(y).

    f(x)= int(from t=1 to t= x)(1/t)dt is actually a fairly standard definition of ln(x).
     
  7. Jul 29, 2003 #6
    So essentially the idea is to make a substitution for t that will preserve the integrand but change the limits. Very clever.
     
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