0

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

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.

Salech Alhasov
  • 6,780
  • 2
  • 29
  • 47
Zaida
  • 1
  • You forgot $dx$ – user116017 Dec 18 '13 at 17:44
  • 2
    Hi, and welcome! I've edited your post to include mathjax formatting; please make sure it says what you intended. Can you please share your thoughts on the problem, and explain what you've tried? For example, do you know how to find an antiderivative of $\ln$? –  Dec 18 '13 at 17:44
  • 2
    It's not even defined on $]0,1[$. – Git Gud Dec 18 '13 at 17:44
  • @GitGud: What does $]a,b[$ mean? I am not familiar with that notation. – Sujaan Kunalan Dec 18 '13 at 17:46
  • 1
    @SujaanKunalan You may know it as $(a,b)$, the open interval. The notation $]a,b[$ is used because $(a,b)$ may look like an ordered pair. – angryavian Dec 18 '13 at 17:47
  • First, you need to consider the function $\ln(1-x)$, as the function you have isn't defined. Better yet, consider $$\int_0^1\ln(x),dx,.$$ Do you not have any idea how to approach this? – Ted Shifrin Dec 18 '13 at 17:48
  • 2
    @SujaanKunalan: It's because you are not born in France. It mean the open interval $(a,b)$. – Salech Alhasov Dec 18 '13 at 17:48
  • Ah, yes I am more accustomed to $(a,b)$. Thanks for the clarification. – Sujaan Kunalan Dec 18 '13 at 17:49
  • @T.Bongers, I do this lim t->1 [-1/2 ln[(1-x)]^2] which is not -1 – Zaida Dec 18 '13 at 17:52
  • We have: $\int_0^1 \ln{(x - 1)} \ dx=i\pi-1$ – AHH Dec 18 '13 at 17:54
  • It must be the $\int_0^1\ln(1-x)$ that is $1$, because the value of $\int_0^1\ln(x-1)$ as a definite integration of complex function $\ln(x-1)$ is equal to $-1+\pi i$. And in that case you can use integration by part. – Farshad Nahangi Dec 18 '13 at 17:58
  • If the integrand is $\ln(1-x)$ you will get the value $-1$ for the integral. – Mhenni Benghorbal Dec 18 '13 at 17:59
  • Either this is meant to be taken as a complex integral or else it makes no sense as the real integrand $;\log(x-1);$ is not even defined on $;(0,1);$ ... – DonAntonio Dec 18 '13 at 18:01
  • @DonAntonio : You must consider the integrand as a complex function so it make sense. Consider $\ln z=\ln|z|+i\theta$, then for example $\ln(-1)=\ln(1)+\pi= i\pi i$ – Farshad Nahangi Dec 18 '13 at 18:05
  • I know that, @FarshadNahangi: thanks. I don't think the OP meant that, though, as the question seems to be pretty elementary...and assuming we already chose a definite branch cut for the complex logarithm (say, the non-positive imaginary axis), then $;\text{Log},(-1)=\log|-1|+i\arg(-1)=\pi i;$ – DonAntonio Dec 18 '13 at 18:06

2 Answers2

2

Hints:

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

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

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

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

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
0

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.

Nicolas Bourbaki
  • 5,064