How to evaluate $\int_0^1\ln(1-x)dx$

Question

I want to compute $\int_0^1\ln(1-x)dx$.

Using integration by parts we have: $\int_0^1\ln(1-x)dx=\int_0^1(x)'\ln(1-x)dx=x\ln(1-x)|_0^1-\int_0^1\frac{x}{1-x}dx$ but limit at $1$ of $x\ln(1-x)$ is $-\infty$ and $\int_0^1\frac{x}{1-x}dx$ does not converge.

I know the answer is $-1$ because desired integral is equal to $\int_0^1\ln(x)dx$ or using integration by parts but with $-(1-x)$ not $x$ yields $-1$ but why first approach does not work and produce false equality?

Answer 1

You have to do the substitution $y=1-x$. The problem is that in your solution you will have two limits that diverge but cancel each other out. You involuntarily split the limits, which makes it a bit confusing since you get something of the form $\infty-\infty$. This doesn't mean that your integral doesn't exist (it's an indeterminate form).

Answer 2

Integrating by parts works a little better if you make a substitution first:

$$\int_0^1 \ln(1-x) \, dx \stackrel{u=\ln(1-x)}= \int_{-\infty}^0 ue^u\,du \stackrel{\rm IBP}= -\int_{-\infty}^0 e^u\,du$$

Answer 3

First, substitute $u = 1 - x$, which makes $du = -1dx$, and thus $dx = -1du$. Now, the integral becomes $-\ln(u)$, which is equal to $-u \ln(u) + u$. Undo the substitution to get $(-1 + x) \ln(1 - x) + 1 - x$. Finally, plug in $1$ and $0$ into this expression to find that the answer is undefined because $\ln(1 - x)$ is undefined at $1$. However, if you think of $(-1 + 1) \ln(1 - 1) + 1 - 1$ as $0$ because $-1 + 1$, which is being multiplied by the $\ln$, is $0$, then the answer is $-1$ because $(-1 + 1) \ln(1 - 1) + 1 - 1 - (-\ln(1) + 1) = -1$.

Answer 4

$((1-x)\ln(1-x))' =-\ln(1-x)-1 $ so $\int \ln(1-x)dx =(1-x)\ln(1-x)-x+c $.

$\begin{array}\\ \int_0^1 \ln(1-x)dx &=((1-x)\ln(1-x)-x+c)|_0^1\\ &=(\lim_{x \to 1}(1-x)\ln(1-x))-1-0\\ &=-1\\ \text{since}\\ \lim_{y \to 0} y \ln(y) &=\lim_{z \to \infty} \frac1{z} \ln(\frac1{z}) \qquad y=\frac1{z}\\ &=\lim_{z \to \infty} - \dfrac{\ln(z)}{z}\\ &=0\\ \text{or}\\ \lim_{y \to 0} y \ln(y) &=\lim_{z \to \infty} e^{-z}\ln(e^{-z}) \qquad y=e^{-z}\\ &=\lim_{z \to \infty} -z e^{-z}\\ &=0\\ \end{array} $

Answer 5

Because this is an improper integral, if you want to do integration by parts as in the OP, you have to do it like this: $$ \int_0^t \ln(1-x)\;dx = x\ln(1-x)\Big|_0^t + \int_0^t\frac{x}{1-x}\;dx \\ \qquad=t\log(1-t)-t-\ln(1-t) = -t - (1-t)\ln(1-t) $$ and $$ \int_0^1 \ln(1-x)\;dx = \lim_{t\to 1^-}\int_0^t \ln(1-x)\;dx = \lim_{t\to 1^-}\big[-t - (1-t)\ln(1-t)\big] = -1 . $$

Answer 6

Use the Taylor series expansion of $\ln(1-x)$. We could use it because $0\le x<1$: $$\int_0^1\ln(1-x)dx=-\int_0^1\sum_{k=1}^\infty\frac{x^k}kdx$$Given that the integral converges, we could interchange the integral and summation symbols: $$-\int_0^1\sum_{k=1}^\infty\frac{x^k}kdx=-\sum_{k=1}^\infty\frac1k\int_0^1x^kdx=-\sum_{k=1}^\infty\frac1{k(k+1)}$$Using partial fraction decomposition: $$-\sum_{k=1}^\infty\frac1{k(k+1)}=-\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)=-\left(\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+...\right)$$This telescopes, so the answer is $-1$.