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Homework Help: 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

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    [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

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    try first

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

    benorin

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  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##.

    $$\Rightarrow I(a)=\ln(a+1)$$
     
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