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

• Dell
In summary, the conversation discusses different methods for solving the integral \intln2|x|dx (from -1 to 1). One method involves integration by parts, while another uses symmetry to simplify the integral. The conversation also mentions using the formula \int_{0}^{1}x^{p}dx=\frac{1}{1+p} to solve the integral, and discusses the use of power series and the deformation of Mcloren to find the value of the integral. The final result is that the integral is equal to 2.

#### Dell

as part of a question i need to integrate

$$\int$$ln2|x|dx (from -1 to 1)

what i did was integration in parts

$$\int$$ln2|x|dx =x*ln2|x| - $$\int$$2(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

Dell said:
as part of a question i need to integrate
$$\int$$ln2|x|dx (from -1 to 1)

(snip)

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

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

where d the value of the integral, or $$\pm \infty$$ if it diverges.

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

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.

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)

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

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