# How to evaluate $\int_0^1 \frac{1-x}{\ln x}(x+x^2+x^{2^2}+x^{2^3}+x^{2^4}+\ldots) \, dx$?

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.

• 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 $x=1. \qquad$ – Michael Hardy Mar 31 '18 at 2:02
• Where is this question from? – bigfocalchord Mar 31 '18 at 7:51
• Thank you everyone for the replies! The source is a tertiary level mathematics competition question from some country. – NetUser5y62 Mar 31 '18 at 9:49

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).

• Long time no see ! – Claude Leibovici Mar 31 '18 at 5:02

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}

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.

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

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*}

• In particular this uses the fact $f(0)=0$. – J.G. Mar 31 '18 at 9:29
• 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. – Mark Viola Mar 31 '18 at 19:57