Check if $\int_{-1}^1-x \ln |x|dx$ converges

Question

Check if the integral converges or not.

$\int_{-1}^1-x \ln |x|dx$

Since $ \ln 0 $ is not defined, we split the given integral into two, one from -1 to 0 and second from 0 to 1

$\lim_{b \to 0^-} \int _{-1}^b -x \ln|x| dx + \lim_{b \to 0^+} \int _{b}^1 -x \ln|x| dx $ $ \lim_{b \to 0^-} -(\ln|b|.\frac{b^2}{2}-\frac{b^2}{4}+\frac{1}{4})+\lim_{b \to 0^+} -(-\ln|b|.\frac{b^2}{2}+\frac{b^2}{4}-\frac{1}{4}) $

$\lim _{b \to 0^-} - \ln (-x) \cdot \frac{ b^2}{2}+\lim _{b \to 0^+} \ln (x) \cdot \frac{ b^2}{2}$

$\infty + \infty$

Therefore the integral diverges, but the correct answer is 0

Pls note, i am using android tablet for the first time and having tough time typing latex, pls let me know how to speed up the process

Answer 1

$\lim _{b \to 0^-} - \ln (-x) \cdot \frac{ b^2}{2}+\lim _{b \to 0^+} \ln (x) \cdot \frac{ b^2}{2}$

You're mixing $b$ and $x$ here. What anti-derivative are you using? It's a bit hard to follow. Because: $$\int x \ln x \,\mbox{d}x = \frac{x^2}{4}\left( 2\ln x -1 \right)$$ You get: $$\int_0^1 -x\ln|x|\,\mbox{d}x= \lim_{b \to 0^+}\int_b^1 -x\ln x \,\mbox{d}x = -\lim_{b \to 0^+}\left( \color{blue}{\frac{1^2}{4}\left( 2\ln 1 -1 \right)}-\color{red}{\frac{b^2}{4}\left( 2\ln b -1 \right)}\right)$$ The first term gives $\color{blue}{\tfrac{1}{4}}$ and the second is $\color{red}{0}$, which is probably where you made a mistake: $$\lim_{b \to 0^+}\left( \frac{b^2}{4}\left( 2\ln b -1 \right)\right)=0$$ Although $\lim_{b \to 0^+} \ln b = +\infty$, the following limits are $0$: $$\lim_{b \to 0^+} b \ln b = 0 \quad \mbox{and} \quad \lim_{b \to 0^+} b^2 \ln b = 0$$ Take a look at:

• Limit of $x\ln{x}$
• $\lim_{x\to0^{+}} x \ln x$ without l'Hopital's rule

Answer 2

Remark that you do not need to calculate explicitely the integral because $f(x)=-x\ln(|x|)$ is odd and bounded on $[-1,1]$ so integral over this symetric domain is zero.

• $f$ is odd

$\forall x>0, f(-x)=-(-x)\ln(|-x|)=x\ln(|x|)=-f(x)$

• $f$ has a limit in $0$

for $x>0,\quad$ $\lim\limits_{x\to 0^+}x\ln(x)=0\ $ so by parity $f$ has a common limit in $0^+$ and in $0^-$ which is $0$.

• $f$ is bounded on $[-1,1]$

since we have erased the difficulty in zero, we have that $f$ is continuous over a compact interval, so it is bounded, and $\int_{-1}^{1}f(x)\;dx$ exists.