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!

Please help me find my mistake in this integration

  1. Apr 12, 2009 #1
    [tex]\int[/tex]dx/x*(x2-1)0.5 (from 1 to infinity)

    i said t=(x2-1)0.5, therefore x2=t2+1
    dt=x/(x2-1)0.5

    so now i have

    [tex]\int[/tex]dx/x*(x2-1)0.5=[tex]\int[/tex]xdx/x2*(x2-1)0.5=[tex]\int[/tex]dt/x2=[tex]\int[/tex]dt/t+1 (now integral from 0 to infinity)

    [tex]\int[/tex]dt/t+1 (from 0 to infinity)

    =lim arctan(t)[tex]^{b}_{0}[/tex]=pi/4
    b-inf


    but the correct answer is pi/2, can ANYONE see where i have gone wrong?
     
  2. jcsd
  3. Apr 12, 2009 #2

    Cyosis

    User Avatar
    Homework Helper

    You didn't adjust your limits after the substitution. If [itex]t=\sqrt{x^2-1}[/itex] and x goes from 1 to infinity then t goes from 0 to infinity.

    ps. It would be nice if you use brackets properly so we don't have to guess which function is actually being integrated.
     
  4. Apr 12, 2009 #3
    if you look at the 5th line of text youll see that i did change my limits,
     
  5. Apr 12, 2009 #4

    Cyosis

    User Avatar
    Homework Helper

    I missed that, still are you sure you didn't use the old limits? As they yields pi/4. So what's arctan(0) and what's arctan(x) with x going to infinity? Hint: your mistake has to do with filling in the limits.
     
    Last edited: Apr 12, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Please help me find my mistake in this integration
Loading...