1. Not finding help here? Sign up for a free 30min 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!

Logarithm defined as integral

  1. Apr 25, 2006 #1
    Confused, but tried it this way:

    Use u-substitution to show that (for y a positive number and x>0)

    [tex]\int_{x}^{xy} \frac{1}{t} dt = \int_{1}^{y} \frac{1}{t} dt [/tex]

    so, u=t and du=dt
    if x=1
    t=xy u=y(1)=y
    t=x u=1

    or
    u=1/t and du/ln [t] = dt
    if x=1
    t=xy u=1/y
    t=x y=1

    Thanks for your help
    Todd
     
  2. jcsd
  3. Apr 25, 2006 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    That's not a very useful substitution, is it?
    but x is not 1.

    No, if u= 1/t, then du= -dt/t2
    Again, you cannot just say "if x= 1"- it's not, it's a variable. Also you haven't used those substitutions- you haven't put them into either integral.

    Look at the upper limits on each integral. On one it is xy, on the other it is just y. To show that the two integrals are equal, you need to change one into the other by some substitution. Okay,xy/x= y so we need to divide by x. Try u= t/x in the left integral only.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Logarithm defined as integral
  1. Logarithm Integral (Replies: 2)

  2. Logarithmic Integration (Replies: 17)

  3. Logarithmic Integral (Replies: 5)

Loading...