# Improper Integral Question $\int_0^1 \ln\sqrt{x}dx$

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.

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please write the integral as lnx with the 1/2 upfront. The anti derivative of lnx is standard: xlnx - x Can you take it from here? Don't forget to multiply your final answer by half – imranfat May 22 '13 at 16:37
Towards the end of the post, you have $\epsilon\ln(\sqrt{x})$. You would want to find the limit of $\epsilon \ln(\sqrt{\epsilon})$. – André Nicolas May 22 '13 at 16:43

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 $\lim_{\epsilon\to 0^+}\epsilon\ln \epsilon=0$.

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Darn, beat me to it @AndréNicolas ;) – Alex Wertheim May 22 '13 at 16:35

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:

$$\lim_{x \rightarrow 0^{+}} \frac{1}{2}x\ln x = \lim_{x \rightarrow 0^{+}} \frac{1}{2} * \frac{\ln x}{\frac{1}{x}} = \lim_{x \rightarrow 0^{+}}\frac{1}{2} \frac{\frac{1}{x}}{-\frac{1}{x^{2}}} = \lim_{x \rightarrow 0^{+}} -\frac{1}{2}x = 0$$

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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...

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