1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integral Suppose f(x)=int(1/t.dt)
  1. F(x) = 1/(x+1) - 2/9 (Replies: 3)

  2. Integrating f'(x)/f(x) (Replies: 8)

Loading...