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4. integration of ln

  • Thread starter Dell
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  • #1
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as part of a question i need to integrate

[tex]\int[/tex]ln2|x|dx (from -1 to 1)

what i did was integration in parts

[tex]\int[/tex]ln2|x|dx =x*ln2|x| - [tex]\int[/tex]2(lnx/x)xdx

=x*ln2|x| - 2[xln|x| - x]

=lim x(ln2|x|-2ln|x|+ 2)|[tex]^{-1}_{0-\epsilon}[/tex] +(ln2|x|-2ln|x|+ 2)|[tex]^{0+\epsilon}_{1}[/tex]
[tex]\epsilon[/tex]->0

1st of all is this correct,?
2nd of all, is there no better way to solve this
 

Answers and Replies

  • #2
as part of a question i need to integrate
[tex]\int[/tex]ln2|x|dx (from -1 to 1)

(snip)

2nd of all, is there no better way to solve this
Symmetry would work pretty nicely:

[tex]\int^{1}_{-1}{ln^2|x| dx} = 2\int^{1}_{0}{ln^2(x) dx} = 2 \lim_{a \rightarrow 0}{\int^{1}_{a}{ln^2(x) dx}} = d[/tex]


where d the value of the integral, or [tex]\pm \infty[/tex] if it diverges.
 
  • #3
590
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thanks, but you would still have to do the whole long integration in parts to achieve d, would you not?
 
  • #4
1,838
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Alternative method.

[tex]\int_{0}^{1}x^{p}dx=\frac{1}{1+p}[/tex]

Expand both sides in powers of p. We have:

[tex] x^p = \exp\left[p\log(x)\right]= 1+p\log(x) + \frac{p^2}{2}\log^{2}(x)+\cdots[/tex]

[tex]\frac{1}{1+p}=1-p+p^2-\cdots[/tex]

So, we can immediately read off that the integral from zero to 1 is 2.
 
  • #5
327
1
nice count Iblis, I should have thought of using that.

But to answer dell question :

use integration by parts formula -- int(u*dv) = uv - int(v*du)
 
  • #6
590
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how did you read that the integral is 2?

i see that you used deformation of Mcloren, (plnx in place of x) but how do you see that that is equal to 2, how do i use this? do i plug 1 into my x,
exp[plog(1)]??? but that will give me exp[0]=1
 
  • #7
590
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how did you get to the integral of x^p if we started with integral of ln^2|x|
 

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