Evaluate $\int_0^1\frac{x^3 - x^2}{\ln x}\,\mathrm dx$?

Question

How do I evaluate the following integral?

$$\int_0^1\frac{x^3 - x^2}{\ln x }\,\mathrm dx$$

Answer 1

Sub $x=e^{-u}$, $dx = -e^{-u} du$. Then the integral is

$$\int_0^1 dx \frac{x^3-x^2}{\log{x}} = \int_0^{\infty} du \, \frac{e^{-3 u} - e^{-4 u}}{u} = \int_0^{\infty} du \, \int_3^4 dt \, e^{-u t} \\ = \int_3^4 dt \,\int_0^{\infty} du \, e^{-u t} = \int_3^4 \frac{dt}{t} = \log{\frac{4}{3}}$$

The change in the order of integration is justified by Fubini's Theorem.

Answer 2

$$ \begin{align} \int_0^1\frac{x^3-x^2}{\log(x)}\mathrm{d}x &=\int_0^1\frac{x^4-x^3}{\log(x)}\frac{\mathrm{d}x}{x}\\ &=\lim_{a\to0}\int_a^1\frac{x^4-x^3}{\log(x)}\frac{\mathrm{d}x}{x}\\ &=\lim_{a\to0}\left(\int_a^1\frac{x^4-1}{\log(x)}\frac{\mathrm{d}x}{x} -\int_a^1\frac{x^3-1}{\log(x)}\frac{\mathrm{d}x}{x}\right)\\ &=\lim_{a\to0}\left(\int_{a^4}^1\frac{x-1}{\log(x)}\frac{\mathrm{d}x}{x} -\int_{a^3}^1\frac{x-1}{\log(x)}\frac{\mathrm{d}x}{x}\right)\\ &=\lim_{a\to0}\int_{a^4}^{a^3}\frac{x-1}{\log(x)}\frac{\mathrm{d}x}{x}\\ &=\lim_{a\to0}\int_{a^4}^{a^3}\frac1{\log(x)}\mathrm{d}x -\lim_{a\to0}\int_{a^4}^{a^3}\frac1{\log(x)}\frac{\mathrm{d}x}{x}\\ &=0-\Big[\log(\log(x))\Big]_{a^4}^{a^3}\\[6pt] &=\log(4)-\log(3)\\[12pt] &=\log(4/3) \end{align} $$

Answer 3

$$\begin{align} \int_0^1\frac{x^3-x^2}{\ln x }\mathrm{d}x &=\int_0^1\frac{x^3-1-x^2+1}{\ln x }\mathrm{d}x\tag{1}\\ &=\int_0^1\frac{x^\color{blue}{3}-1}{\ln x }\mathrm{d}x-\int_0^1\frac{x^\color{red}{2}-1}{\ln x}\mathrm{d}x\tag{2}\\ &=\ln(\color{blue}{3}+1)-\ln(\color{red}{2}+1)\tag{3}\\ &=\ln\frac43\tag{4}\\ \end{align}$$

$$\large\int_0^1\frac{x^3-x^2}{\log(x)}\mathrm{d}x=\ln\frac43$$

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$\text{Explanation : } (3 )$

Consider parametric integral $\displaystyle\quad\quad\quad I(\alpha)=\int_0^1{\frac{x^{\alpha}-1}{\ln x}\mathrm dx}$

We have $I(0)=0$, By differentiating w.r.t $\alpha$ we get

$$ I'(\alpha) =\int_0^1{\frac{x^{\alpha}\ln x}{\ln x}\,\mathrm dx} =\int_0^1{x^{\alpha}\,\mathrm dx} =\frac{x^{\alpha+1}}{\alpha+1} =\frac{1}{\alpha+1}$$ Integrating w.r.t. $\alpha$ and using $I(0)=0$ we get $$I(\alpha)=\int_0^1{\frac{x^{\alpha}-1}{\ln x}\,\mathrm dx}=\ln(\alpha+1)$$

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \int_{0}^{1}{x^{3} - x^{2} \over \ln\pars{x}}\,\dd x} =\int_{0}^{1}\pars{x^{2} - x^{3}}\ \overbrace{\int_{0}^{\infty}x^{t}\,\dd t}^{\dsc{-\,{1 \over \ln\pars{x}}}}\ \,\dd x\ =\ \int_{0}^{\infty}\int_{0}^{1}\pars{x^{2 + t} - x^{3 + t}}\,\dd x\,\dd t \\[5mm]&=\int_{0}^{\infty}\pars{{1 \over t + 3} - {1 \over t + 4}}\,\dd t =\left.\ln\pars{t + 3 \over t + 4}\right\vert_{\, t\ =\ 0}^{\, t\ \to\ \infty} =-\ln\pars{3 \over 4}=\color{#66f}{\large\ln\pars{4 \over 3}} \end{align}

Answer 5

You may like this method. Note $$ \lim_{n\to\infty} n(x^{\frac{1}{n}}-1)=\ln x, \text{ for }x>0 $$ and hence \begin{eqnarray} \int_0^1\frac{x^3-x^2}{\ln x}dx&=&\int_0^1\lim_{n\to\infty}\frac{x^3-x^2}{n(x^{\frac{1}{n}}-1)}dx\\ &=&\lim_{n\to\infty}\int_0^1\frac{x^2(x-1)}{n(x^{\frac{1}{n}}-1)}dx\\ &=&\lim_{n\to\infty}\int_0^1x^2\frac{1}{n}\sum_{k=0}^{n-1}x^{\frac{k}{n}}dx\\ &=&\lim_{n\to\infty}\int_0^1\frac{1}{n}\sum_{k=0}^{n-1}x^{\frac{k}{n}+2}dx\\ &=&\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\frac{k}{n}+3}\\ &=&\int_0^1\frac{1}{x+3}dx\\ &=&\ln\frac{4}{3}. \end{eqnarray}

Answer 6

We have:

$$I = \int_{0}^{1} \frac{x^2(x - 1)}{\log(x)}$$

Consider:

$$I(b) = \int_{0}^{1} \frac{x^b(x - 1)}{\log(x)}$$

Recognize that $I(0) = \int_{0}^{1} (x-1)/\log(x) dx = \log(2)$, this is a particular solution for later.

Differentiate partially with respect to $b$

$$I'(b) = \int_{0}^{1} \frac{\log(x)\cdot x^b(x - 1)}{\log(x)} dx$$

$$I'(b) = \int_{0}^{1} \frac{x^b(x - 1)}{1} dx = \int_{0}^{1} x^b(x - 1) dx $$

$$I'(b) = \int_{0}^{1} x^{b+1} - x^b dx$$

$$I'(b) = \frac{1}{b+2} - \frac{1}{b+1}$$

Integrate this with respect to $db$

$$I(b) = \int \frac{1}{b+2} - \frac{1}{b+1} db$$

$$I(b) = \log(b+2) - \log(b+1) + C$$

$C= 0$

$$I(2) = \log(4) - \log(3) = \log(4/3)$$