How to show $\int_{0}^{1}\frac{\log(x)}{\sqrt{x}}$ converges? Want to show $\int_{0}^{1}\frac{\log(x)}{\sqrt{x}}dx$ converges.
I understand how to solve the improper integral and got the fact that $\int_{0}^{1}\frac{\log(x)}{\sqrt{x}}dx = -4$.  However, I'm stumped in how to show the integral exists in the first place, since I understand the integral test for convergence requires the function to be non-negative on the interval in question, however, our function here is non-positive.
 A: With the substitution $$x=t^2$$
the integral  $$\int_{0}^{1}\frac{\log(x)}{\sqrt{x}} dx $$ transforms to  $$4\int_0^1  \log t dt = 4(t\log t -t)|_0^1 $$ which is convergent. 
A: Integration by parts $\left(f=\log(x), g'=\frac{1}{\sqrt{x}} \Rightarrow f'=\frac{1}{x}, g=2\sqrt{x}\right)$:
$$\int \frac{\log(x)}{\sqrt{x}} = 2\sqrt{x}\log(x) - \int \frac{2}{\sqrt{x}}$$
$$= 2\sqrt{x}\log(x) - 4\sqrt{x} + C$$
$$= 2\sqrt{x}(\log(x) - 2) + C$$
So the definite integral is:
$$\int_{0}^{1} \frac{\log(x)}{\sqrt{x}} = (2\sqrt{x}(\log(x) - 2) + C)|_{0}^{1} = -4$$
which is convergent
A: Integrating by parts: $\displaystyle \int_0^1 \frac{\ln x}{\sqrt{x}} \, dx =  2\sqrt{x} \ln x \bigg\vert_0^1 - \int_0^1 2 \sqrt{x} \frac{1}{x} \, dx = -2 \int_0^1 \frac{1}{\sqrt{x}} \, dx$ which is convergent by computing. (Hence giving the value of integral equals to $-4$.)  
Limit Comparison: As you can see above, the integral is somewhat equal to integral of the form $\displaystyle \int_0^1 \frac{1}{x^p} \, dx$ with $0 < p <1$. Let choose $p=3/4$ (you will see any $1/2 < p < 1$ also works). Since $\ln x$ is negative for $0<x<1$, we should consider $\displaystyle \int_0^1 \frac{-\ln x}{\sqrt{x}} \, dx$ to perform limit comparison with $\displaystyle \int_0^1 \frac{1}{x^{3/4}} \, dx$. Note that the problematic point is $0$ so 
$\displaystyle \lim_{x \to 0^+} \frac{\frac{- \ln x}{\sqrt{x}}}{\frac{1}{x^{3/4}}} = \lim_{x \to 0^+} - x^{1/4} \ln x = 0$. This says, for some $c>0$, $\dfrac{-\ln x}{\sqrt{x}} < \dfrac{1}{x^{3/4}}$ holds for all $0<x<c$. Thus we can write for some $0<d< \min\{1,c\}$
$\displaystyle \int_0^1 \frac{-\ln x}{\sqrt{x}} \, dx = \int_0^d \frac{-\ln x}{\sqrt{x}} \, dx + \int_d^1 \frac{-\ln x}{\sqrt{x}} \, dx  < \int_0^d \frac{1}{x^{3/4}} \, dx + \int_d^1 \frac{-\ln d}{\sqrt{x}} \, dx$ which implies convergence of $\displaystyle \int_0^1 \frac{-\ln x}{\sqrt{x}} \, dx$ since both two terms are convergent. As a consequence $\displaystyle \int_0^1 \frac{\ln x}{\sqrt{x}} \, dx$ is convergent.
