Evaluate $\displaystyle \int_0^1\ln x\,dx$
$\displaystyle \int_0^1ln(x)~dx = \lim_{a \to 0}\int_a^1ln(x)~dx$
$\displaystyle = \lim_{a \to 0}(x~ln(x) - x)|_a^1 = -1 - \lim_{a \to 0}(a~ln(a) - a)$
So what is $\displaystyle \lim_{a \to 0}(a~ln(a) - a)$? Obviously this translates to the problem of finding
$\displaystyle \lim_{a \to 0}a~ln(a)$
This is a $\displaystyle 0 \cdot \infty$ form, so try the following:
$\displaystyle \lim_{a \to 0}a~ln(a) = \lim_{a \to 0}\frac{ln(a)}{\frac{1}{a}}$
By L'Hopital's rule
$\displaystyle = \lim_{a \to 0} \frac{\frac{1}{a}}{- \frac{1}{a^2}} = \lim_{a \to 0}(-a) = 0$
So finally:
$\displaystyle \int_0^1ln(x)~dx = \lim_{a \to 0}\int_a^1ln(x)~dx = -1 - \lim_{a \to 0}(a~ln(a) - a) = -1 - 0 = -1$
-Dan
$\displaystyle \int_0^1 \ln x dx = \int_0^1 \ln [(x-1)+1] dx, \mbox{ let }t=x-1$
$\displaystyle \int_{-1}^0 \ln (t+1) dt = \int_{-1}^0 t - \frac{t^2}{2}+\frac{t^3}{3} - ... dt$
$\displaystyle \frac{t^2}{1\cdot 2} - \frac{t^3}{2\cdot 3} + \frac{t^4}{3\cdot 4}-... \big|_{-1}^0 = -\left( \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+... \right)$
$\displaystyle = - \sum_{n=1}^{\infty} \frac{1}{n(n+1)} = -1$