Integration in Parts: Solving ln Integration with a Twist | Step-by-Step Guide

[Dell]

as part of a question i need to integrate

\intln²|x|dx (from -1 to 1)

what i did was integration in parts

\intln²|x|dx =x*ln²|x| - \int2(lnx/x)xdx

=x*ln²|x| - 2[xln|x| - x]

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

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

 

[meltingwax]

as part of a question i need to integrate
\intln²|x|dx (from -1 to 1)

(snip)

2nd of all, is there no better way to solve this

Symmetry would work pretty nicely:

\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}} = dwhere d the value of the integral, or \pm \infty if it diverges.

 

[Dell]

thanks, but you would still have to do the whole long integration in parts to achieve d, would you not?

 

[Count Iblis]

Alternative method.

\int_{0}^{1}x^{p}dx=\frac{1}{1+p}

Expand both sides in powers of p. We have:

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

\frac{1}{1+p}=1-p+p^2-\cdots

So, we can immediately read off that the integral from zero to 1 is 2.

 

[tnutty]

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)

 

[Dell]

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

 

[Dell]

how did you get to the integral of x^p if we started with integral of ln^2|x|