I'm trying to compute this integral and check if it's an improper integral.
What I did so far is to write the limit.
$$\begin{align*}
\int_0^1\ln\sqrt x\, dx &= x\ln\sqrt x - \int \frac12\, dx = x\ln\sqrt x - \frac12x + C \\
&=\lim_{\epsilon\to0^+}\Bigl(x\ln\sqrt x - \frac12x \Bigr)\Bigr\rvert_\epsilon^1
\end{align*}$$
after I set the epsilon and the value 1, I got the expression : $\epsilon \ln \sqrt{x}$ and I want to know if this expression say that this integral is not improper.
Thanks.
It would have been slightly more pleasant to rewrite the integrand as $\frac{1}{2}x\ln x$.
After the integration, the issue is whether $\lim_{\epsilon\to 0^+}\epsilon\ln \epsilon$ exists. It does, and is equal to $0$.
One way of showing it is by writing $\epsilon=e^{-t}$. Then we are interested in the limit as $t\to\infty$ of $e^{-t}(-t)$.
The fact that $$\lim_{t\to\infty} -\frac{t}{e^t}=0$$ can be proved in various ways. It follows that $\lim_{\epsilon\to 0^+}\epsilon\ln \epsilon=0$.
Well, for starters, I think you can make your life easier in evaluating the limit (and in integrating) by rewriting $\ln(\sqrt{x})$ as $\frac{1}{2}\ln x$. $\frac{1}{2}x$ is obviously easy to deal with, so you are interested in the term $$\lim_{x \rightarrow 0^{+}} \frac{1}{2}x\ln x$$ This is a standard indeterminate form $0 * -\infty$, so we apply L'Hospital's rule to conclude:
$$ \lim_{x \rightarrow 0^{+}} \frac{1}{2}x\ln x = \lim_{x \rightarrow 0^{+}} \frac{1}{2} * \frac{\ln x}{\frac{1}{x}} = \lim_{x \rightarrow 0^{+}}\frac{1}{2} \frac{\frac{1}{x}}{-\frac{1}{x^{2}}} = \lim_{x \rightarrow 0^{+}} -\frac{1}{2}x = 0$$
Check the second example in the "examples section in this wikipedia article.
I think that this is an improper integral by definition, because $\ln(\sqrt x) $ is not defined when $x=0$, so you have to use the limit to approach this point during the integration. Hope, this helps...
