# Thread: improper integrals

1. ## improper integrals

Hey everyone, i had a question but whenever i do the integral i keep on getting 0 but the answer is supposed to be -2

the question is the integral ln(abs(x)) from -1 to 1

i divided the integral into ln(-x) form -1 to 0 and ln(x) from 0 to 1
i evaluated it but i keep on getting 1 + -1 = 0

can anyone help me with this

2. ## Re: improper integrals

Originally Posted by ubhutto
Hey everyone, i had a question but whenever i do the integral i keep on getting 0 but the answer is supposed to be -2
the question is the integral ln(abs(x)) from -1 to 1
i divided the integral into ln(-x) form -1 to 0 and ln(x) from 0 to 1
i evaluated it but i keep on getting 1 + -1 = 0
Note that $\displaystyle \log(|x|)$ is an even function so $\displaystyle \int_{ - 1}^1 {\log (|x|)dx} = 2\int_0^1 {\log (x)dx}$.

Look at this webpage calculation.

3. ## Re: improper integrals

ʃ-11ln|x|dx=2 ʃ01lnxdx

ʃ01lnxdx =limx→0 ʃx1lnxdx

ʃx1lnxdx=xlnx-x|x1=-1-( xlnx-x)

limx→0 ʃx1lnxdx = -1

(limx→0xlnx=0)

(I didn't think the web reference was that helpful)

4. ## Re: improper integrals

Originally Posted by Hartlw
ʃ-11ln|x|dx=2 ʃ01lnxdx

ʃ01lnxdx =limx→0 ʃx1lnxdx

ʃx1lnxdx=xlnx-x|x1=-1-( xlnx-x)

limx→0 ʃx1lnxdx = -1

(limx→0xlnx=0)

(I didn't think the web reference was that helpful)
You are correct, but I would have chosen a different variable for the limit. x is confusing being both inside the integral and also as part of the limit.

-Dan