Showing the existence of $\int_0^1\sqrt{x}\ln (x)\, dx$ and calculating it 
Show that the integral $$\int_0^1\sqrt{x}\log (x)\, dx$$ exists and calculate that.

To calculate that integral I have done the following :
We substitute $z=\ln (x) \Rightarrow x=e^z$.
When $x=0$ then $z=-\infty$
When $x=1$ then $z=0$
We have that $\frac{dz}{dx}=\frac{1}{x} \Rightarrow dx=xdz=e^zdz$
Then we get $$\int_0^1\sqrt{x}\ln (x)\, dx=\int_{-\infty}^0\sqrt{e^z}z\, e^zdz=\int_{-\infty}^0e^{z/3}z\, dz$$
Is that the correct way to calculate the integral?
At the beginning where we have to show the existence, do we show that just by calculating the integral or is there also an other way to show the existence?
 A: By using this substitution, you are using the fact that the map $x \mapsto e^x$ is differentiable with a continuous derivative.
But no, this is not the most efficient way. You may use integration by parts immediately here. The reason is that product of the integral of $\sqrt x$ and the derivative of $\log x$ is readily integrable: $x^{3/2} \cdot \frac 1 x = x^{1/2}$.
You may read about the two topics in depth here: Integration by parts
, and substitution.
A: You could make it simpler using $x=t^2$
$$I=\int_0^1\sqrt{x}\log (x)\, dx=4\int_0^1 t^2\log(t) \,dt$$ and we know that $\lim_{t\to 0} \, t\log(t)=0$.
Now, one integration by parts to obtain the result.
A: 
Then we get $$\int_0^1\sqrt{x}\log (x)\, dx=\int_{-\infty}^0\sqrt{e^z}z\, e^zdz=\int_{-\infty}^0e^{z/3}z\, dz$$

The last step should be
$$I=\int_{-\infty}^0e^{\frac{3}2z}z\, dz$$
you can define: $z=-\frac{2}3u$
$$I=-\frac{4}9\int_0^\infty ue^{-u}du=-\frac{4}9\Gamma(2)=-\frac{4}9$$
A: One way to show just existence is as follows:
Note that $\ln(0) = -\infty$ so we need to show that
$$\lim_{c \to 0^+} \int_c^1 \sqrt{x}\ln(x)dx\;\;\text{exists}$$
We can use the fact that $\sqrt{x},\ln(x)$ are continuous on $(0,1]$, so their product is continuous on $(0,1]$
Let $0<a<1$, then by the mean value theorem for definite integrals:
$$\exists k \in (a,1]: \sqrt{k}\ln(k) = \frac{1}{1-a}\int_a^1 \sqrt{x}\ln(x)dx$$
So in the case where $a>0$ we know the integral exists.
Now we need to show that $\sqrt{x}\ln(x)$ is right-continuous at $x=0$.
$$\lim_{x \to 0^+} \sqrt{x}\ln(x) = 0\cdot -\infty \implies \lim_{x \to 0^+} \frac{\ln(x)}{x^{-\frac12}} =  \frac{-\infty}{-\infty}$$
$$ \stackrel{\text{L'Hopital}}{\implies} \lim_{x \to 0^+}\frac{\frac{1}{x}}{-\frac12 x^{-\frac32}}=\lim_{x \to 0^+}\frac{-\frac12 x^{\frac32}}{x} = -\lim_{x \to 0^+}\frac12 x^{\frac12} = 0$$
So it $\sqrt{x}\ln(x)$ is (right)-continuous on $[0,1]$ so by above the integral exists.
