Does $\int_0^1 (1-x)\ln(x) \ dx$ converge?

Question

I was trying to compute a limit of a sum of a question and I got stuck at this integral. $$\int_0^1 (1-x)\ln(x) \ dx$$

I cannot proceed as I keep getting $\ln(0)$ and cannot solve the limit of integral i.e

$$ \lim_{h \to 0}[ \int_h^1 (1-x)\ln(x) \ dx]$$

So I am asking for solution to this limit.

Question:

Does the area of curve $\displaystyle (1-x)\ln(x)$ converge in 0 to 1. If yes then how and what is its value?

Answer 1

Yes, it converges to $-3/4$.

Use Integration by Parts and techniques of Improper Integration. Following LIATE, let $u=\ln x$ and $dv = (1-x)$, giving $du = \frac1x dx$ and $v = x-\frac12x^2$. This leads to

$$\int_0^1 (1-x)\ln x\ dx = \lim_{a\to 0}\int_a^1 (1-x)\ln x\ dx = \lim_{a\to 0}\left((x-\frac12x^2)\ln x\Big|_a^1-\int_a^1\frac1x(x-\frac12x^2)\ dx\right).$$

The integral above converges to $-3/4$ without difficulty. But the first expression in the limit needs L'Hopital's Rule:

$$\lim_{a\to 0}(x-\frac12x^2)\ln x) = \lim_{a\to 0}x\ln x-\lim_{a\to 0}\frac12x^2\ln x).$$ Both limits converge to 0; we'll show the first.

$$\lim_{a\to 0}x\ln x = \lim_{a\to 0}\frac{\ln x}{1/x} = \text{(by L'Hopital's)}= \lim_{a\to 0} \frac{1/x}{-1/x^2} = \lim_{a\to 0} -x = 0.$$

In summary, $$\int_0^1 (1-x)\ln x\ dx = \lim_{a\to 0}\left((x-\frac12x^2)\ln x\Big|_a^1-\int_a^1\frac1x(x-\frac12x^2)\ dx\right) = 0-3/4 =-3/4.$$

Answer 2

Intergation by parts: $\int_0^1 (1-x)\ln(x) \ dx=(1-x)(x\ln(x) -x)|_0^1 +\int_0^1 x\ln(x)-x \;dx=[(1-x)(x\ln(x)-x)-\dfrac{x^2}2]|_0^1+\int_0^1 x\ln(x) \;dx=[(1-x)(x\ln(x)-x)-\dfrac{x^2}2+\dfrac{x^2}2\ln(x)|_0^1-\int_0^1 \dfrac{x^2}2\dfrac1x=[(1-x)(x\ln(x)-x)+\dfrac{x^2}2\ln(x)-\dfrac{3x^2}4|_0^1$

L'Hopital's Rule: $\lim_{x\to 0}x\ln(x)=\lim_{x\to 0}\dfrac{\ln(x)}{1/x}=\lim_{x\to 0}\dfrac{1/x}{-1/x^2}=\lim_{x\to 0}(-x)=0$. Hence $\lim_{x\to 0}x^2\ln(x)=0$ as well.

So the integral converges to $\dfrac{-3}{4}$.

Answer 3

Hint

If $0<s<t$, a simple integration by part yields $$\int_s^t(x-1)\ln(x)\,\mathrm d s=\left[\frac{(x-1)^2}{2}\ln(x)\right]_{s}^t-\int_s^t\frac{(x-1)^2}{2x}\,\mathrm d x.$$

Answer 4

For the antiderivative first

$$I=\int (1-x)\log (x) \, dx$$ use integration by parts $$u=\log (x)\implies u'=\frac {dx} x$$ $$v'=(1-x)\,dx\implies v=x-\frac 12 x^2$$ $$I=\left(x-\frac 12 x^2 \right)\log(x)-\int \frac 1 x\left(x-\frac 12 x^2 \right)\,dx=\left(x-\frac 12 x^2 \right)\log(x)-\left(x-\frac 14 x^2 \right)$$ Uing the bounds, the first terms is $0$ at both ends and it just remains $-\frac 34$.

Answer 5

Define for $x\in [0;1]$, $f(x)=x\ln x$. $f$ is a continuous function on $[0;1]$ because $\lim_{x\rightarrow 0} x\ln x=0$.

Moreover, for $\displaystyle \epsilon>0 \int_{\epsilon}^{1}\ln xdx=\Big[x\left(\ln x-1\right)\Big]_{\epsilon}^{1}=-1-\epsilon\ln \epsilon+\epsilon$.

Therefore, $\displaystyle\int_{0}^{1}\ln xdx= -1$.

Therefore, $\displaystyle \int_0^1 (1-x)\ln(x) \ dx$ converges since for $x\in]0;1],(1-x)\ln x=\ln x-f(x)$.

$\displaystyle \int_0^1 f(x) \ dx=\int_0^1 x\ln(x) \ dx=\Big [\frac{x^2}{2}\ln x\Big]_0^1-\int_0^1 \frac{x^2}{2x}dx=\left[-\frac{x^2}{4}\right]_0^1=-\frac{1}{4}$.

Therefore,

$\boxed{\displaystyle \int_0^1 (1-x)\ln(x) \ dx=-\frac{3}{4}}$.