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!

Tricky definite integral

  1. Apr 27, 2014 #1
    I am trying to solve this integral:

    [itex]\int[/itex] [itex] \frac{x^a-1}{ln(x)} dx[/itex] (with the interval from 0 to 1).

    I have tried substitution but I could not find a way to get it to work. Any ideas on how to solve this?

    Thanks!
     
  2. jcsd
  3. Apr 28, 2014 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    [tex]\int_0^1 \frac{x^a-1}{\ln(x)} \;dx[/tex]

    In what context did this integral come up?
    What sort of range is "a"?

    Did you try splitting it up? $$\int_0^1 \frac{x^a}{\ln(x)} dx - \int_0^1 \frac{1}{\ln(x)} \;dx$$

    Did you try substituting: ##x = e^u##

    Note: $$\int \frac{dx}{\ln(x)}=\text{li}(x)$$
    http://mathworld.wolfram.com/LogarithmicIntegral.html
     
  4. Apr 28, 2014 #3

    lurflurf

    User Avatar
    Homework Helper

    try first

    $$\int_0^a \! x^t\, \mathrm{d}t$$
     
  5. May 2, 2014 #4

    benorin

    User Avatar
    Homework Helper

  6. May 2, 2014 #5
    One way to approach this is by defining
    $$I(a)=\int_0^1 \frac{x^a-1}{\ln x}$$
    Differentiate both the sides wrt ##a## to obtain
    $$I'(a)=\int_0^1 \frac{x^a\ln x}{\ln x}=\int_0^1 x^a\,dx$$
    $$I'(a)=\frac{1}{a+1}$$
    $$\Rightarrow I(a)=\ln(a+1)+C$$
    Notice that ##I(0)=0##, hence ##C=0##.

    $$\Rightarrow I(a)=\ln(a+1)$$
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Tricky definite integral
  1. Tricky Definite Integral (Replies: 22)

  2. Tricky integral (Replies: 1)

Loading...