Does the integral $\int_{0}^{e}\sqrt{x}\ln{x} \ dx$ converge or diverge? Splitting: $$\int_{0}^{e}\sqrt{x}\ln{x} \ dx=\int_{0}^{1} \sqrt{x}\ln{x}\ dx + \int_{1}^{e} \sqrt{x}\ln{x} \ dx.$$
Second integral is clearly convergent since it's defined on $[1,e].$ For the first integral, we see that 
$$\int_{0}^{1} \sqrt{x}\ln{x}\ dx\leq\int_{0}^{1} 1\cdot\ln{x}\ dx.$$
This implies that $\sqrt{x}\leq1,$ for $x\in[0,1]$ which is true. Now I know that the second integral is convergent with the value of $-1.$ Thus the original integral is also convergent.

Please give me feedback on my reasoning. I know I got the correct answer, but is the motivation for it correct and stringent enough? Say this question gives 3 points in a test, how much would this solution warrant?
Is there any faster way do deduce the convergence?
 A: Unfortunately, no: your logic is not valid. You used the inequality
$$\int_0^1 \sqrt{x} \ln x \, dx\le \int_0^1 \ln x \, dx$$
to conclude that $\sqrt{x} \le 1$, and then concluded that the outcome is true. This has three serious problems:


*

*You can't use $\int_0^1 f \, dx \le \int_0^1 g \, dx$ to conclude that $f \le g$ regardless.

*The first inequality is actually false, because $\ln x \le 0$ on this interval.

*You cannot derive to something true, and use that to conclude that the premise was true as well.
Also, your argument for convergence on $[1, e]$ isn't correct. The integrand is defined and bounded. Having an integrand be defined everywhere does not guarantee convergence.

A correct approach would be to do something like the following: Combine that $\sqrt{x} \in [0, 1]$ for $x \in [0, 1]$, the fact that $|\sqrt{x} \ln x| \le |\ln x|$ for $x \in (0, 1]$, and finally that
$$\int_0^1 |\ln x|\, dx$$
converges.
A: You can actually do this integral
\begin{align}
&\quad\;\int_0^e\sqrt{x}\ln xdx=\frac{2}{3}\int_0^e\ln x d x^{3/2}=\frac{4}{9}\int_0^e\ln x^{3/2} dx^{3/2}\\
&=\frac{4}{9}\int_0^{e^{3/2}}\ln ydy=\frac{4}{9}\left.y(\ln y-1)\right|_0^{e^{3/2}}=\frac{2}{9}e^{3/2},
\end{align}
which is finite, so it converges.
A: HINT: use that $$\int \sqrt{x}\ln(x)dx=\frac{2}{9} x^{3/2} (3 \log (x)-2)$$ and $$\lim_{x\to 0^+}x\ln(x)=0$$
A: Since there are already some answers that answer your specific question, I would like to show the method I would use to show convergence of the integral. Denote $f(x)=\sqrt x\ln x$. First, notice that for any $x\in(0,e]$, $f$ is Riemann-integrable over $[x,e]$, since it is bounded (and even continuous) on all such intervals. Therefore, it is locally Riemann-integrable. Now, for $f$ to be an improper integral, i.e., converge, one has to show tha
$$\lim_{\alpha\rightarrow0^-}\int_{\alpha}^ef(x)dx$$
converges. Using the main theorem of calculus, this is equal to saying that
$$\lim_{\alpha\rightarrow0^-}(F(e)-F(\alpha)),$$
where $F$ is an antiderivative of $f$. It can be found using partial integration and is given by $F(x)=\frac{2}{9}\sqrt{x^3}(-2+\ln x)+c$, where $c\in\mathbb R$ a constant, which might as well be $0$, since it cancels out anyway. Since $F(e)$ is just a constant, we only have to worry about $F(\alpha)$ in the limit. It can be shown using L'Hopital's rule that this term converges to zero, just notice that in
$$F(\alpha)=-\frac{4}{9}\sqrt{\alpha^3}+\frac{2}{9}\sqrt{\alpha^3}\ln \alpha$$
the first term tends to zero if $\alpha$ tends to zero and
$$\sqrt{\alpha^3}\ln \alpha=\frac{\ln \alpha}{\alpha^{-\frac{3}{2}}},$$
to which L'Hospital can be applied to get
$$\lim_{x\rightarrow0^-}\frac{\ln \alpha}{\alpha^{-\frac{3}{2}}}=\lim_{x\rightarrow0^-}\frac{\alpha^{-1}}{-\frac{3}{2}\alpha^{-\frac{5}{2}}}=\lim_{x\rightarrow0^-}-\frac{2}{3}x^\frac{3}{2}=0.$$
Notice that I have ommited the constant $\frac{2}{9}$, since it has no effect on the convergence of the limit. Since the limit is $0$, it can be further omitted. In conclusion is $\lim_{\alpha\rightarrow0^-}(F(e)-F(\alpha))=F(e)$, thus convergent, which imples that the integral is improperly convergent.
Perhaps this method is somewhat longer, instead of shorter as you asked for, but I hope it will give you better insight on how such problems can be approached alternatively (and somewhat more formally).
Please point out any errors you notice!
