Hints on calculating the integral $\int_0^1\frac{x^{19}-1}{\ln x}\,dx$

I would be happy to get some hints on the following integral: $$ \int_0^1\frac{x^{19}-1}{\ln x}\,dx $$

Solution 1:

Let $x = e^{-y}$, we have

$$\int_0^1 \frac{x^{19} - 1}{\log x} dx = \int_0^\infty \frac{e^{-\color{blue}{1}y} - e^{-\color{orange}{20}y}}{y} dy$$ This is in the form of a Frullani's integral and one can read off the value of the integral as

$$( \color{red}{1} - \color{green}{0} )\log\left(\frac{\color{orange}{20}}{\color{blue}{1}}\right) = \log 20 \quad\text{ since }\quad e^{-y} = \begin{cases}\color{red}{1}, &y = 0\\ \color{green}{0}, & y \to \infty\end{cases}$$

If you really need to perform the integral yourself without using Frullani's integral directly, I'll recommend you look at answers of this question and learn the various proof there. A good exercise is translate the proof there to your particular case. This will get you familiar with the steps that need to evaluate this sort of integral.

Solution 2:

Differentiation of the integrand $$f(x,a) = \frac{x^a-1}{\log x}$$ with respect to $a$ gives $$\frac{\partial f}{\partial a} = x^a.$$ Therefore, $$I(a) = \int_{x=0}^1 f(x,a) \, dx$$ implies $$\frac{d I}{d a} = \int_{x=0}^1 x^a \, dx = \frac{1}{a+1}, \quad a > -1.$$ Integrating with respect to $a$ then yields $$I(a) = \log(a+1), \quad a > -1.$$ There are some omitted details, but this is an outline of the general solution.

Solution 3:

Here is the details from @heropup's answer.

Let us generalize the problem. We will evaluate $$ \mathcal{I}(\alpha)=\int_0^1\frac{x^\alpha-1}{\ln x}\ dx\qquad;\qquad \alpha>-1.\tag1 $$ Now we apply Feynman's method (differentiate under the integral sign). Diferentiating both sides of $(1)$ yields \begin{align} \frac{\partial\mathcal{I}}{\partial\alpha}&=\int_0^1\frac{\partial}{\partial\alpha}\left[\frac{x^\alpha-1}{\ln x}\right]\ dx\\ \mathcal{I}'(\alpha)&=\int_0^1 x^\alpha\ dx\\ &=\left.\frac{x^{\alpha+1}}{\alpha+1}\right|_{x=0}^1\\ &=\frac{1}{\alpha+1}.\tag2 \end{align} Integrating $(2)$ yields \begin{align} \mathcal{I}(\alpha)&=\int\frac{1}{\alpha+1}\ d\alpha\\ &=\ln(\alpha+1)+C.\tag3 \end{align} In order to find out our constant of integration, we let $\alpha = 0$ so that our integrand is $0$, implying that $C = 0$. Letting $\alpha = 19$ will of course solve our original problem: \begin{align} \color{purple}{\int_0^1\frac{x^\alpha-1}{\ln x}\ dx}&\color{purple}{=\ln(\alpha+1)}\\ \int_0^1\frac{x^{19}-1}{\ln x}\ dx&=\large\color{blue}{\ln20}. \end{align}

Solution 4:

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \color{#66f}{\large\int_{0}^{1}{x^{19} - 1 \over \ln\pars{x}}\,\dd x}&= \int_{0}^{1}\pars{x^{19} - 1}\ \overbrace{\pars{-\int_{0}^{\infty}x^{t}\,\dd t}} ^{\ds{=\ {1 \over \ln\pars{x}}}}\ \,\dd x =\int_{0}^{\infty}\int_{0}^{1}\pars{x^{t} - x^{t + 19}}\,\dd x\,\dd t \\[5mm]&=\int_{0}^{\infty}\pars{{1 \over t + 1} - {1 \over t + 20}}\,\dd t =\left.\ln\pars{t + 1 \over t + 20}\right\vert_{\,t\ =\ 0}^{\,t\ \to\ \infty} =0 - \ln\pars{1 \over 20} \\[8mm]& =\color{#66f}{\Large\ln\pars{20}} \approx {\tt 2.9957} \end{align}

In addition, you can see this method.