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.
 A: Hints:
$$\int\limits_0^1\log x\,dx=\lim_{b\to 0^+}\int\limits_b^1\log x\,dx=\left.\lim_{b\to 0^+}\left(x\log x-x\right)\right|_b^1=\ldots\ldots$$
A: Alternatively, your area is the same as $$-\int_{-\infty}^0 e^x\,dx$$
Can you see why?
A: I expect you found, perhaps by integration by parts, that  $x\ln x-x$ is an antiderivative of $\ln x$..
Imagine calculating $\int_t^1\ln x\,dx$, where $t$ is a small positive number. We get $-1-(t\ln t-t)=-1+t-t\ln t$.  
Now let $t$ approach $0$ through positive values. We want to find out what happens to $-t\ln t$. Of course, $t$ becomes very small positive, and $\ln t$ becomes very large negative, so it is not clear what happens to $-t\ln t$.
Let $t=\frac{1}{e^w}$, where $w$ is large. Then $-\ln t=w$, so we are interested in the behaviour of $\frac{w}{e^w}$ as $w$ gets large. It is a probably familiar fact that
$$\lim_{w\to\infty} \frac{w}{e^w}=0. $$
Alternately, rewrite $-t\ln t$ as
$$\frac{-\ln t}{\frac{1}{t}}$$
and use L'Hospital's Rule to calculate the limit as $t$ approaches $0$ through positive values. 
Remark: Because $\ln x$ blows up (or is it down?) as $x$ approaches $0$ from the right, our integral is an improper integral, so in principle it cannot be evaluated by "plugging in." The limit process that we used is a built-in part of the defintion of convergence of an improper integral. 
A: Yet another approach:
\begin{align}
\int_0^{1}\ln x\,dx=\left[\frac{\partial}{\partial s}\left(\int_0^1x^s dx\right)\right]_{s=0^+}=\left(\frac{\partial}{\partial s}\left[\frac{x^{s+1}}{s+1}\right]_0^1\right)_{s=0}=\left[-\frac{1}{(s+1)^2}\right]_{s=0}=-1.
\end{align}
A: $\int_0^1 \log x dx=\lim_{a\to 0^+}\int_a^1\log x dx=\lim_{a\to 0^+}(x\log x-x|_a^1)=\lim_{a\to 0^+}(a-1-a\log a)=\lim_{a\to 0^+}(a-1) -\lim_{a\to 0^+}a\log a=-1-\lim_{a\to 0^+}a\log a$
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, $$\int_0^1 \log x dx= -1-\lim_{a\to 0^+}a\log a =-1$$
A: The only place there is a problem
is as $x \to 0$,
so let's look at
$\int_{c}^1 \log x \,\mathrm dx$
and see what happens as 
$c \to 0$.
Since $(x \ln x)' = 1+\ln x$,
$(x \ln x - x)' = \ln x$,
so
$\int_{c}^1 x \ln x dx = 
x \ln x - x |_c^1
= c-1-c \ln c
$.
As you have shown,
$\lim_{x \to 0} x \ln x
= \lim_{x \to 0} -x
= 0
$,
so
$\lim_{c \to 0}\int_{c}^1 x \ln x dx = 
x \ln x - x |_c^1
= \lim_{c \to 0}c-1-c \ln c
= -1
$.
A: for every 's' we know that $ \int_{0}^{1}dxx^{s}=\frac{1}{1+s}$
take the derivative with respec to 's' at $ s=0 $ so $ \int_{0}^{1}dxlogx=-\frac{1}{(1+0)^{2}}=-1$
A: $$\int_{0}^{1}\ln x dx=\int_{0}^{1}(x+\ln x+\frac{x+\ln x}{x}-\frac{x+\ln x}{x}-x)dx $$
$$\int_{0}^{1}(x+\ln x+\frac{x+\ln x}{x}-\frac{x+\ln x}{x}-x)dx =\int_{0}^{1}((x+\ln x)(1+\frac{1}{x})-1-x-\frac{\ln x}{x}) dx$$
$$\int((x+\ln x)(1+\frac{1}{x})-1-x-\frac{\ln x}{x}) dx=\frac{(x+\ln x)^2}{2}-x-\frac{x^2}{2}-\frac{(\ln x)^2}{2}$$
$$\frac{(x+\ln x)^2}{2}-x-\frac{x^2}{2}-\frac{(\ln x)^2}{2}=x \ln x -x$$
$$\int_{0}^{1}\ln x dx=1 \ln 1 -1 -(\lim_{x\rightarrow 0} x \ln x -x)$$
$$\int_{0}^{1}\ln x dx=0-1-(0-0)$$
$$\int_{0}^{1}\ln x dx=-1$$
A: I may have missed the direct application of integration by parts,
$$
\begin{align}
\int_0^1\log(x)\,\mathrm{d}x
&=\lim_{a\to0}\int_a^1\log(x)\,\mathrm{d}x\\
&=\lim_{a\to0}a\log(a)-\lim_{a\to0}\int_a^1x\,\mathrm{d}\log(x)\\
&=0-\int_0^11\,\mathrm{d}x\\[6pt]
&=-1
\end{align}
$$
