/A problem from the 2012 MIT Integration Bee is $$ \int_0^1\frac{x^7-1}{\log(x)}\mathrm dx $$ The answer is $\log(8)$. Wolfram Alpha gives an indefinite form in terms of the logarithmic integral function, but times out doing the computation. Is there a way to do it by hand?

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    $\begingroup$ Generally speaking, $$\int_0^1\frac{x^n-1}{\ln x}dx=\ln(n+1)$$ $\endgroup$ – Lucian Nov 14 '13 at 7:18
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    $\begingroup$ @Lucian That's an interesting identity, why is that? $\endgroup$ – YoniY Nov 14 '13 at 7:19
  • $\begingroup$ Make the change of variables $\ln(x)=-u$. $\endgroup$ – Mhenni Benghorbal Nov 14 '13 at 7:21
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    $\begingroup$ A magician, uhm, I mean, mathematician NEVER betrays his tricks! Especially when he doesn't know why either. :-) $\endgroup$ – Lucian Nov 14 '13 at 7:22
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    $\begingroup$ Might be relevant: fy.chalmers.se/~tfkhj/FeynmanIntegration.pdf $\endgroup$ – Miguelgondu Dec 20 '14 at 5:14

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{\pp\pars{\mu} \equiv \int_{0}^{1}{x^{\mu} - 1 \over \ln\pars{x}}\,\dd x}$

$$ \pp'\pars{\mu} \equiv \int_{0}^{1}{x^{\mu}\ln\pars{x} \over \ln\pars{x}}\,\dd x = \int_{0}^{1}x^{\mu}\,\dd x = {1 \over \mu + 1} \quad\imp\quad \pp\pars{\mu} - \overbrace{\pp\pars{0}}^{=\ 0} = \ln\pars{\mu + 1} $$

$$ \pp\pars{7} = \color{#0000ff}{\large\int_{0}^{1}{x^{7} - 1 \over \ln\pars{x}} \,\dd x} = \ln\pars{7 + 1} = \ln\pars{8} = \color{#0000ff}{\large 3\ln\pars{2}} $$


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} \mathcal I'(\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 $\alpha = 7$ will of course solve our original problem: \begin{align} \int_0^1\frac{x^\alpha-1}{\ln x}\ dx&=\ln(\alpha+1)\\ \int_0^1\frac{x^7-1}{\ln x}\ dx&=\large\color{blue}{\ln8}. \end{align}

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    $\begingroup$ Lol, aren't you a year late? And isn't this basically a copy of the accepted answer? $\endgroup$ – Simply Beautiful Art Jul 23 '17 at 18:22

Change of variables $\log(x) = -t$ makes this into $$ \int_0^\infty \dfrac{1 - e^{-7t}}{t} e^{-t}\ dt $$ More generally, for $\alpha \ge 0$ let $$f(\alpha) = \int_0^\infty \dfrac{1-\exp(-\alpha t)}{t} e^{-t}\ dt$$ Then $f(0) = 0$ while $$f'(\alpha) = \int_0^\infty \exp(-(\alpha+1) t)\ dt = \dfrac{1}{1+\alpha}$$ from which $$f(\alpha) = \ln(1+\alpha)$$


Yet another way direct forward is to use Frullani's Integral. To that end, let $I(a)$ be the integral given by

$$I(a)=\int_0^1 \frac{x^a-1}{\log x}\,dx$$

Enforcing the substitution $\log x \to -x$ yields

$$\begin{align} I(a)&=\int_{0}^{\infty} \frac{e^{-ax}-1}{x}\,e^{-x}\,dx\\\\ &=-\int_{0}^{\infty} \frac{e^{-(a+1)x}-e^{-x}}{x}\,dx \end{align}$$

whereupon using Frullani's Integral we obtain

$$\bbox[5px,border:2px solid #C0A000]{I=\log(a+1)}$$

For $a=7$, we have $$\bbox[5px,border:2px solid #C0A000]{I(7)=\log (8)}$$as expected!

  • $\begingroup$ +1 this was my initial approach before I learned the Feynman way. $\endgroup$ – Simply Beautiful Art Jul 23 '17 at 18:23
  • $\begingroup$ @SimplyBeautifulArt Thank you for the up vote. I posted another solution on this page that uses an approach that relies on Fubini-Tonneli. $\endgroup$ – Mark Viola Jul 23 '17 at 18:25
  • $\begingroup$ Yup, I already saw :-) $\endgroup$ – Simply Beautiful Art Jul 23 '17 at 18:27

I thought it might be instructive to present yet another approach.

Note that using $\int_0^1 x^t \,dt=\frac{x-1}{\log(x)}$ we can write

$$\begin{align} \int_0^1 \frac{x^7-1}{\log(x)}\,dx&=\int_0^1 (x^6+x^5+x^4+x^3+x^2+x+1)\left(\int_0^1 x^t\,dt\right)\,dx\tag1\\\\ &=\int_0^1\int_0^1 (x^{t+6}+x^{t+5}+x^{t+4}+x^{t+3}+x^{t+2}+x^{t+1}+x^t)\,dx\,dt\tag2\\\\ &=\int_0^1 \left(\frac{1}{t+7}+\frac{1}{t+6}+\frac{1}{t+5}+\frac{1}{t+4}+\frac{1}{t+3}+\frac{1}{t+2}+\frac{1}{t+1}\right)\,dt\\\\ &=\log(8) \end{align}$$

as expected, where the Fubini-Tonelli Theorem guarantees the legitimacy of interchanging the order of integration in going from $(1)$ to $(2)$.

Note that if we first enforce the substitution $x^7\to x$, we obtain

$$\begin{align} \int_0^1 \frac{x^7-1}{\log(x)}\,dx&=\int_0^1 x^{1/7-1}\left(\frac{(x-1)}{\log(x)}\right)\\\\ &=\int_0^1 x^{1/7-1}\left(\int_0^1 x^t\,dt\right)\,dx\\\\ &=\int_0^1\int_0^1 x^{t+1/7-1} \,dx\,dt\\\\ &=\int_0^1 \frac{1}{t+1/7}\,dt\\\\ &=\log(8) \end{align}$$

as expected!


$$\int_{0}^{1} \frac{u^7-1}{\ln u} du$$

Let $y=\ln u$ we get,

$$=\int_{-\infty}^{0} \frac{e^{8y}-e^{y}}{y} dy$$

Notice we are interesting in integrating over $u \in (0,1)$, so we after the substitution we are integrating over $y \in (-\infty,0)$.

$$=- \int_{-\infty}^{0} \int_{y}^{8y} \frac{e^x}{y} dx dy$$

The reason I wrote it as above is to to change the order of integration.

$$=-\int_{-\infty}^{0} \int_{x}^{\frac{1}{8}x} \frac{e^x}{y} dy dx$$

Because $y \in (-\infty,0)$ and $x \in [y,8y]$ for our purposes, it is safe to say $x \in (-\infty,0)$ so that as we vary $x$, $x \neq 0$ always holds. Furthermore $\ln |\frac{1}{8}x|-\ln |x|=\ln \frac{1}{8}$ for $x \neq 0$.

$$=-\int_{-\infty}^{0} e^x \ln \frac{1}{8} dx$$

$$=\ln 8$$


Substitution $x=e^{-u}$, then \begin{align} \int_0^1\frac{x^7-1}{\log(x)}\mathrm dx &= \int_\infty^0\frac{e^{-7u}-1}{-u}(-e^{-u})du \\ &= -\int_0^\infty\frac{e^{-8u}-e^{-u}}{u}du \\ &= -\int_0^\infty\frac{1}{s+8}-\frac{1}{s+1}ds \\ &= \ln\dfrac{s+1}{s+8}\Big|_0^\infty \\ &= \color{blue}{\ln8} \end{align} which I used this property of Laplace transform: $$\int_0^\infty\frac{f(t)}{t}dt=\int_0^\infty{\cal L}(f)(s)ds$$


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