Finding the definite integral $\int_0^1 \log x\,\mathrm dx$ $$\int_{0}^1 \log x \,\mathrm dx$$
How to solve this? I am having problems with the limit $0$ to $1$. Because $\log 0$ is undefined.

$\int_0^1 \log x dx=\lim_{a\to 0^+}\int_a^1\log x dx=\lim_{a\to 0^+}(x\log x-x|_a^1)=\lim_{a\to 0^+}(a-1-a\log a)=\lim_{a\to 0^+}(a-1) -\lim_{a\to 0^+}a\log a=-1-\lim_{a\to 0^+}a\log a$
Now $$\lim_{a\to 0^+}a\log a=\lim_{a\to 0^+}\frac{\log a}{1/a}$$
Using L' hopital's rule( which is applicable here), $$\lim_{a\to 0^+}\frac{\log a}{1/a}=\lim_{a\to 0^+}\frac{1/a}{1/a^2}=\lim_{a\to 0^+}(-a)=0$$
Therefore, $$\int_0^1 \log x dx= -1-\lim_{a\to 0^+}a\log a =-1$$
