/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?

  • 11
    $\begingroup$ Generally speaking, $$\int_0^1\frac{x^n-1}{\ln x}dx=\ln(n+1)$$ $\endgroup$
    – Lucian
    Nov 14 '13 at 7:18
  • 1
    $\begingroup$ @Lucian That's an interesting identity, why is that? $\endgroup$
    – YoniY
    Nov 14 '13 at 7:19
  • $\begingroup$ Make the change of variables $\ln(x)=-u$. $\endgroup$ Nov 14 '13 at 7:21
  • 12
    $\begingroup$ 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
  • 1
    $\begingroup$ Might be relevant: fy.chalmers.se/~tfkhj/FeynmanIntegration.pdf $\endgroup$ Dec 20 '14 at 5:14

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

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

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


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 direct way 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. $\endgroup$ 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 :-) $\endgroup$ 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

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

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: $$\int_0^\infty\frac{f(t)}{t}dt=\int_0^\infty{\cal L}(f)(s)ds$$


enter image description here

Application of Leibniz's rule for integration

  • $\begingroup$ This is exactly the same as the accepted answer, posted 7 years ago. $\endgroup$
    – Mark Viola
    Jan 20 '21 at 17:58
  • $\begingroup$ Nice integral though $\endgroup$ Jan 21 '21 at 19:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.