Solve this improper integral: $\int_0^1 \ln{(x - 1)} \ dx$

Question

$$\int_0^1 \ln{(x - 1)} \ dx$$

I don't know how to solve this integral.

My teacher says the solution is $-1$ but I don't know how to reach this result.

Answer 1

The function $\log(x-1)$ is not defined on $(0,1)$, just take $x=\tfrac{1}{2}$ and you get a logarithm of a negative number.

Let us suppose then that you wanted to evaluate, $$ \int_0^1 \log(1-x) ~ dx $$ You can do this by using infinite series, $$ \log(1-x) = -\sum_{n=1}^{\infty} \frac{x^n}{n} \implies \int_0^1 \log(1-x)~ dx = -\sum_{n=1}^{\infty} \int_0^1 \frac{x^n}{n} ~ dx = - \sum_{n=1}^{\infty} \frac{1}{n(n+1)} $$ Now you can evaluate the infinite series by telescoping it, $$ \sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \sum_{n=1}^{\infty} \left\{ \frac{1}{n} - \frac{1}{n+1} \right\} = 1$$ So the integral evaluates to $-1$ as you wanted to show.

Answer 2

Hints:

Assuming you meant a real function then it could be $\;\log(1-x)\;$:

$$u=1-x\implies -du=dx\implies$$

$$\int\limits_0^1\log(1-x)\,dx=-\int\limits_{1}^0\log u\,du=\int\limits_0^1\log u\,du=$$

$$=\left.\left(u\log u-u\right)\right|_0^1=-1-\underbrace{\lim_{\epsilon\to 0}\epsilon\log\epsilon}_{=0}=-1$$