Solving ln|x| Improper Integral: -1 to 1

[james5]

hi, why is ln|x|, from -1 to 1, converging?

is "0" the bad point, and must i break up the integral from 1 to $, where $ = 0, and from $ to -1... so i have xlnx-x as my derivative... and i get -2?

thanks

 

[HallsofIvy]

Since this has nothing to do with "differential equations" I am moving it to "calculus".

I can't tell you "why is ln|x|, from -1 to 1, converging" or even the question I think you intended: "why is the integral of ln|x|, from -1 to 1, converging", because it isn't! The definition of an improper integral of f(x), from -1 to 1 where there is a singularity at x= 0, is
$$\lim_{h\rightarrow 0}\int_{-1}^h f(t)dt+ \lim_{k\rightarrow 0}\int_k^1 f(t)dt$$
Notice that the two limits are taken separately. Yes, an anti-derivative of ln x is x ln x- x (for x> 0). An anti-derivative of ln(-x), for x< 0 is x ln(-x)+ x. Therefore
$$\int_{-1}^1 ln|x|dx= \lim_{h\rightarrow 0^+}(-hln(h)-h)+\lim_{k\rightarrow 0^+}(kln(k)-k)$$
and neither of those limits exist.
What you found, by taking both limits with "h" so that they canceled out is the "Cauchy Principal Value" which has some occaisional uses but I don't think gives any good information here.

 

[tacman]

The reason why ln|x|, from -1 to 1, is converging is because it is a proper integral. A proper integral is one where the limits of integration are finite and the function being integrated is bounded on the interval. In this case, the limits of integration are -1 and 1, which are both finite, and the function ln|x| is bounded on the interval since it is a logarithmic function and has a horizontal asymptote at x=0. Therefore, the integral is converging.

As for your question about breaking up the integral, it is not necessary in this case since the function is already bounded on the interval. Breaking it up may be necessary for other functions that are not bounded on the interval or have a singularity at one of the points of integration.

In regards to your derivative, it is not clear what you are trying to calculate or what the function xlnx-x represents. Please provide more context or information for a better understanding of your calculation.