11
$\begingroup$

Evaluate the definite integral: $$\int_0^1 \frac{1-x}{\ln x}(x+x^2+x^{2^2}+x^{2^3}+x^{2^4}+\ldots) \, dx$$

I think the series involving $x$ converges because $x\in[0,1]$, but I cannot form an expression for the series. If I let $$ u_n=x^{2^{n-1}} \\ \frac{\ln u_n}{\ln x}=2^{n-1} $$ but then this series does not converge. Even WolframAlpha cannot evaluate a definite integral together with an infinite series, so I am stuck on this.

$\endgroup$
  • $\begingroup$ A simple comparison test with $1+x+x^2+x^3+x^4+\cdots$ shows the series converges if $0\le x<1$ but not if $x=1. \qquad$ $\endgroup$ – Michael Hardy Mar 31 '18 at 2:02
  • $\begingroup$ Where is this question from? $\endgroup$ – bigfocalchord Mar 31 '18 at 7:51
  • $\begingroup$ Thank you everyone for the replies! The source is a tertiary level mathematics competition question from some country. $\endgroup$ – NetUser5y62 Mar 31 '18 at 9:49
15
$\begingroup$

Hint. One may recall Frullani's integral, for $a,b>0$, $$ \int_{0}^{1}\frac{x^{a-1}-x^{b-1}}{\ln x}\:dx=\ln\frac ba \tag1 $$ giving, by telescoping terms, for $N=0,1,2,\cdots$, $$ \begin{align} &\int_{0}^{1}{\frac{1-x}{\ln x}(x+x^{2}+x^{2^{2}}+\cdots+x^{2^N})}\:dx \\\\&=\sum_{n=0}^N\int_{0}^{1}\frac{x^{2^n}-x^{2^n+1}}{\ln x}\:dx\\\\ &=\sum_{n=0}^N\ln\frac{2^n+1}{2^n+2}\\\\ &=\sum_{n=0}^N\left[\ln(2^n+1)-\ln(2^{n-1}+1)-\ln 2\right]\\\\ &=\ln(2^N+1)-\ln(2^{-1}+1)-(N+1)\ln 2\\\\ &=-\ln 3+\ln\left(1+1/2^N\right), \end{align} $$ then, by letting $N \to \infty$, one gets

$$ \int_{0}^{1}{\dfrac{1-x}{\ln x}(x+x^{2}+x^{2^{2}}+x^{2^{3}}+\cdots)}\:dx=-\ln 3. \tag2 $$

Edit. One may justify the interchange between $\displaystyle\int$ and $\displaystyle\sum$ by noticing that $$ \left|\frac{1-x}{\ln x}\right|\le1,\qquad 0<x<1,\tag3 $$ and that $$ x+x^{2}+x^{2^{2}}+x^{2^{3}}+\cdots \sim -\frac{1}{\ln 2}\:\ln (-\ln x),\quad \text{as} \quad x \to 1^-,\tag4 $$ proved here (example 12, p.31).

$\endgroup$
  • 2
    $\begingroup$ Long time no see ! $\endgroup$ – Claude Leibovici Mar 31 '18 at 5:02
6
$\begingroup$

Claim 1: For $k\geq 1$, we have that \begin{align} \int^1_0 \frac{1-x}{\log x}x^{2^k}\ dx = -\log \frac{2^k+2}{2^k+1}. \end{align}

Claim 2: We have \begin{align} \prod^\infty_{k=0}\left( 1+\frac{1}{2^k+1}\right) = 3 \end{align}

Using the claims, we have the series \begin{align} -\sum^\infty_{k=0} \log \left(\frac{2^k+2}{2^k+1}\right) =-\log\left(\prod^\infty_{k=0} \left(1+\frac{1}{2^k+1}\right) \right) =-\log 3. \end{align}

$\endgroup$
5
$\begingroup$

First note that

$$\int_0^1 x^y\,dy=\frac{x-1}{\log(x)}\tag1$$

Next, using $(1)$ along with Fubini's Theorem reveals

$$\begin{align} \int_0^1 \frac{1-x}{\log(x)}\,x^{2^n}\,dx&=-\int_0^1 \left(\int_0^1 x^y\,dy\right)\,x^{2^n}\,dx\\\\ &=-\int_0^1 \left(\int_0^1 x^{y+2^n}\,dx\right)\,\,dy\\\\ &=-\int_0^1 \frac{1}{y+2^n}\,dy\\\\ &=-\log(2^n+2)+\log(2^n+1)\\\\ &=-\log(2^n+2)+\log(2^{n+1}+2)-\log(2)\tag2 \end{align}$$

which contains a telescoping term. Finally, summing $(2)$ over $n$ yields

$$\sum_{n=0}^\infty \int_0^1 \frac{1-x}{\log(x)}\,x^{2^n}\,dx=-\log(3)$$

Fubini's Theorem under the counting measure may be used again to justify interchanging the series with the integral.

$\endgroup$
  • $\begingroup$ great answer indeed !! $\endgroup$ – освящение Mar 31 '18 at 5:19
  • $\begingroup$ @PrithiviRaj Thank you! Much appreciate it. $\endgroup$ – Mark Viola Mar 31 '18 at 5:26
4
$\begingroup$

Another idea is to differentiate under the integral sign to kill the logarithm. By inserting a parameter , call it $\alpha$ , and then differentiate under the integral sign we have that , if we define

$$f\left ( \alpha \right ) = \int_{0}^{1} \frac{1-x^\alpha}{\log x} \sum_{n=0}^{\infty} x^{2^n} \, {\rm d}x \quad , \quad \alpha \geq 0$$

then

\begin{align*} \frac{\mathrm{d} }{\mathrm{d} \alpha} f(\alpha) &= \int_{0}^{1} \frac{\partial }{\partial \alpha} \frac{1-x^\alpha}{\log x} \sum_{n=0}^{\infty} x^{2^n} \, {\rm d}x \\ &=-\int_{0}^{1} x^\alpha \sum_{n=0}^{\infty} x^{2^n} \, {\rm d}x \\ &= -\sum_{n=0}^{\infty} \int_{0}^{1} x^{2^n} x^\alpha \, {\rm d}x\\ &= -\sum_{n=0}^{\infty} \int_{0}^{1} x^{2^n +\alpha} \, {\rm d}x \\ &=- \sum_{n=0}^{\infty} \frac{1}{\alpha +2^n +1} \end{align*}

While you cannot find a general form for the derivative ( at least not without special functions anyway ) , you can at least evaluate the original integral. How, you may ask? Just integrate from $0$ to $1$, thus:

$$f(1)=\int_0^1 f'(\alpha) \, {\rm d}\alpha$$

Therefore,

\begin{align*} f(1) &= \int_{0}^{1} f'(\alpha) \, {\rm d}\alpha \\ &= -\sum_{n=0}^{\infty} \int_{0}^{1} \frac{{\rm d}\alpha}{\alpha +2^n +1}\\ &= -\sum_{n=0}^{\infty} \log \left ( \frac{2^n +2}{2^n +1} \right )\\ &= - \lim_{N \rightarrow +\infty} \sum_{n=0}^{N} \log \left ( \frac{2^n +2}{2^n +1} \right ) \\ &= - \lim_{N \rightarrow +\infty} \log \left ( \frac{3 \cdot 2^N}{2^N +1} \right ) \\ &= - \log 3 \end{align*}

$\endgroup$
  • $\begingroup$ In particular this uses the fact $f(0)=0$. $\endgroup$ – J.G. Mar 31 '18 at 9:29
  • $\begingroup$ This is noy substantively different from my solution in which I wrote $\frac{1-x}{\log(x)}=-\int_0^1x^y\,dy$ and used Fubini's Theorem to justify interchanging the order of integration. $\endgroup$ – Mark Viola Mar 31 '18 at 19:57

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.