# Evaluating $\int_{0}^{1}\frac{1-x}{1+x}\frac{\mathrm dx}{\ln x}$

Some time ago I came across to the following integral:

$$I=\int_{0}^{1}\frac{1-x}{1+x}\frac{\mathrm dx}{\ln x}$$ What are the hints on how to compute this integral?

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What happens if you substitute $u=\ln x$? (On second thought, it doesn't look too useful.) – Harald Hanche-Olsen Jan 16 '13 at 15:30
@HaraldHanche-Olsen: Isn't it an improper integral? – Babak S. Jan 16 '13 at 15:39
Fwiw Wolfram alpha couldn't find an indefinite integral, although it can find the definite integral, presumably by approximation. – JSchlather Jan 16 '13 at 15:40
@JacobSchlather: But it converges. – Babak S. Jan 16 '13 at 15:40
Maple says the value is $-\ln(\pi/2)$. – Harald Hanche-Olsen Jan 16 '13 at 15:42

Make the substitution $x=e^{-y}$ and do a little algebra to get the value of the integral to be

$$\int_0^{\infty} \frac{dy}{y} \frac{e^{-y} - e^{-2 y}}{1+e^{-y}}$$

Now Taylor expand the denominator and get

$$\int_0^{\infty} \frac{dy}{y} (e^{-y} - e^{-2 y}) \sum_{k=0}^{\infty} (-1)^k e^{-k y}$$

If we can reverse the order of sum and integral, we get

$$\sum_{k=0}^{\infty} (-1)^k \int_0^{\infty} \frac{dy}{y} (e^{-(k+1) y} - e^{-(k+2) y})$$

The integral inside may be evaluated exactly, and the result is the sum

$$\sum_{k=0}^{\infty} (-1)^k \log{\frac{k+1}{k+2}}$$

$$= \lim_{n \rightarrow \infty} \; \log{\frac{\frac{1}{2} \frac{3}{4} \ldots \frac{2 n-1}{2 n}}{\frac{2}{3} \frac{5}{6} \ldots \frac{2 n+1}{2 n+2}}}$$

$$= \log{\left ( \frac{2}{\pi} \right )}$$

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Seems reasonable. Thank you! – Martin Gales Jan 16 '13 at 16:27
Thank you for asking it - this was fun! – Ron Gordon Jan 16 '13 at 16:40
Yeah, math is about having fun! :-) (+1) – user 1618033 Jan 16 '13 at 16:54
To reverse the order of sum and integra, I understand that you appeal to the absolute convergence of sum and integral, that is, to the convergence of the positive series $\sum\limits_{k\geqslant0}\int\limits_0^{+\infty}(e^{-(k+1)y}-e^{-(k+2)y})dy/y$. But this series diverges. Maybe I did not understand the result you are applying, could you explain? – Did Jan 21 '13 at 9:17
Maybe we are speaking across each other. Look, do you agree that the sum in Line 2 absolutely converges (in this case, to $(1-e^{-y})^{-1})$? And that the integral absolutely converges (in this case, to $\log{\frac{k+1}{k+2}}$). Because each integral is also continuous, then we get a convergent result in exchanging the order of the sum and integral in Line 2. As I said, this does not apply to Line 3 because, as you point out, that the sum does not converge absolutely. But I am not seeking to reverse the order of integration here, just in Line 2. – Ron Gordon Jan 21 '13 at 14:38

I tried with "Differentiation under integration sign":

$$J(\alpha)=\int_0^1\frac{1-x^\alpha}{1+x}\frac{dx}{\ln x}\quad \alpha>0$$

Then, as usual, $$\frac{dJ}{d\alpha}=-\int_0^1\frac{x^\alpha}{1+x}dx=-f(\alpha),\text{(say)}$$

Then, integrating $$J(\alpha)=-\int f(\alpha)d\alpha+c$$ I used Mathematica to evaluate $f(\alpha)$ (it gives difference of two Harmonic numbers) and its integral. The integral is just $\ln\frac{\Gamma(\frac{2m+1}{2})}{\Gamma(\frac{m+1}{2})}$. Note that $J(0)=0$ and putting $\alpha=1$, the result follows.

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Very interesting approach! Thanks! – Martin Gales Jan 17 '13 at 16:05
Thanks, Martin! – Tapu Jan 17 '13 at 17:13
Where does $m$ come from in $\ln\dfrac{\Gamma(\frac{2m+1}{2})}{\Gamma(\frac{m+1}{2})}$? – Alex Schiff Aug 1 '14 at 14:33

Consider $$\mathcal{I}(\alpha)=\int_0^1\frac{1-x^\alpha}{(1+x)\ln x}\ dx.\tag1$$ Differentiating $(1)$ with respect $\alpha$ yields \begin{align} \frac{d\mathcal{I}}{d\alpha}&=\int_0^1\frac{\partial}{\partial\alpha}\left[\frac{1-x^\alpha}{(1+x)\ln x}\right]\ dx\\ &=-\int_0^1\frac{x^\alpha}{1+x}\ dx\\ &=-\int_0^1\sum_{k=0}^\infty(-1)^kx^{\alpha+k}\ dx\\ &=-\sum_{k=0}^\infty(-1)^k\int_0^1x^{\alpha+k}\ dx\\ &=-\sum_{k=0}^\infty\frac{(-1)^k}{\alpha+k+1}.\tag2 \end{align} Now consider polygamma function $$\psi_n(z)=\frac{d^{n+1}}{dz^{n+1}}\ln\Gamma(z)\tag3$$ and $$\psi_n(z)=(-1)^{n+1}n!\sum_{k=0}^\infty\frac{1}{(z+k)^{n+1}}.\tag4$$ Hence by using $(4)$ we obtain $$\sum_{k=0}^\infty\frac{(-1)^{k}}{(z+k)^{n+1}}=\frac1{(-2)^{n+1}n!}\left[\psi_n\left(\frac{z}{2}\right)-\psi_n\left(\frac{z+1}{2}\right)\right].\tag5$$ Using $(3)$ and $(5)$ then $(2)$ becomes \begin{align} \frac{d\mathcal{I}}{d\alpha}&=\frac12\left[\psi_0\left(\frac{\alpha+1}{2}\right)-\psi_0\left(\frac{\alpha+2}{2}\right)\right]\\ \mathcal{I}(\alpha)&=\frac12\int\left[\psi_0\left(\frac{\alpha+1}{2}\right)-\psi_0\left(\frac{\alpha+2}{2}\right)\right]\ d\alpha\\ &=\ln\Gamma\left(\frac{\alpha+1}{2}\right)-\ln\Gamma\left(\frac{\alpha+2}{2}\right)+C\\ &=\ln\Gamma\left(\frac{\alpha+1}{2}\right)-\ln\Gamma\left(\frac{\alpha+2}{2}\right)-\frac12\ln\pi,\tag6 \end{align} where $\mathcal{I}(0)=0$ and $C=-\ln\Gamma\left(\frac{1}{2}\right)$.

Thus \begin{align} \int_0^1\frac{1-x}{(1+x)\ln x}\ dx&=\mathcal{I}(1)\\ &=\ln\Gamma\left(1\right)-\ln\Gamma\left(\frac{3}{2}\right)-\frac12\ln\pi\\ &=\color{blue}{\ln\left(\frac{2}{\pi}\right)}. \end{align}

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{I\equiv\int_{0}^{1}{1 - x \over 1 + x}\,{\dd x \over \ln\pars{x}}:\ {\large}}$

Lets consider $\ds{\fermi\pars{\mu}\equiv \int_{0}^{1}{1 - x^{\mu} \over 1 + x}\,{\dd x \over \ln\pars{x}}}$ such that $\ds{I = \fermi\pars{1}:\ {\large ?}}$. Note that $\ds{\fermi\pars{0} = 0}$.

Then, \begin{align} \fermi'\pars{\mu}&=-\int_{0}^{1}{x^{\mu} \over 1 + x}\,\dd x= \int_{0}^{1}{1 - x^{\mu} \over 1 + x}\,\dd x - \int_{0}^{1}{\dd x \over 1 + x} \\[3mm]&=\int_{0}^{1} \pars{1 - x^{\mu}}\pars{{1 \over 1 + x} + {1 \over 1 - x}}\,\dd x -\int_{0}^{1}{1 - x^{\mu} \over 1 - x}\,\dd x - \ln\pars{2} \\[3mm]&=2\int_{0}^{1}{1 - x^{\mu} \over 1 - x^{2}}\,\dd x -\int_{0}^{1}{1 - x^{\mu} \over 1 - x}\,\dd x - \ln\pars{2} \\[3mm]&=\int_{0}^{1}{x^{-1/2} - x^{\pars{\mu - 1}/2} \over 1 - x}\,\dd x -\int_{0}^{1}{1 - x^{\mu} \over 1 - x}\,\dd x - \ln\pars{2} \\[3mm]&=\int_{0}^{1}{1 - x^{\pars{\mu -1}/2} \over 1 - x}\,\dd x -\int_{0}^{1}{1 - x^{-1/2} \over 1 - x}\,\dd x -\int_{0}^{1}{1 - x^{\mu} \over 1 - x}\,\dd x - \ln\pars{2} \end{align}

$$\fermi'\pars{\mu}=\Psi\pars{\mu + 1 \over 2} -\Psi\pars{\mu + 1} + \ln\pars{2}$$ where we used the Digamma $\ds{\Psi\pars{z}}$ identity $\ds{\bf\mbox{6.3.22}}$ and the identity $\ds{\bf\mbox{6.3.3}}$.

Since $\ds{\Psi\pars{z} = \totald{\ln\pars{\Gamma\pars{z}}}{z}}$ and $\ds{\fermi\pars{0} = 0}$: $$\fermi\pars{\mu} =\ln\pars{{2^{\mu}\,\Gamma^{2}\pars{\bracks{\mu + 1}/2} \over \Gamma\pars{\mu + 1}}\, {\Gamma\pars{1} \over 2^{0}\,\Gamma^{2}\pars{1/2}}} =\ln\pars{{2^{\mu} \over \pi}\, {\Gamma^{2}\pars{\bracks{\mu + 1}/2} \over \Gamma\pars{\mu + 1}}}$$ with the identities $\quad\ds{\Gamma\pars{1} = 1\quad\mbox{and}\quad\Gamma\pars{\half} = \root{\pi}}$.

With $\ds{\Gamma\pars{2} = 1}$: $$I=\color{#66f}{\large\int_{0}^{1}{1 - x \over 1 + x}\,{\dd x \over \ln\pars{x}}}= \fermi\pars{1} = \color{#66f}{\large\ln\pars{2 \over \pi}} \approx {\tt -0.4515}$$

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You may like this method. Note $$\int_0^1x^ada=-\frac{1-x}{\ln x}, \text{ for }x>0, H_a=\sum_{k=1}^\infty\frac{a}{k(a+k)},$$ and hence \begin{eqnarray} \int_0^1\frac{1-x}{1+x}\frac{dx}{\ln x}&=&-\int_0^1\int_0^1\frac{x^a}{1+x}dadx\\ &=&-\int_0^1\int_0^1\frac{x^a}{1+x}dxda\\ &=&-\frac{1}{2}\int_0^1(H_{\frac{a}{2}}-H_{\frac{a-1}{2}})da\\ &=&-\frac{1}{2}\int_0^1\sum_{k=1}^\infty\left(\frac{a}{k(a+2k)}-\frac{a-1}{k(a+2k-1)}\right)da\\ &=&-\frac{1}{2}\sum_{k=1}^\infty\left(4\ln k-2\ln(k^2-\frac{1}{4})\right)\\ &=&-\frac{1}{2}\ln\prod_{k=1}^\infty\frac{k^4}{(k^2-\frac{1}{4})^2}\\ &=&-\prod_{k=1}^\infty\frac{k^2}{k^2-\frac{1}{4}}\\ &=&-\ln\left(\frac{\pi}{2}\right). \end{eqnarray} In the last step, we used the Wallis Formula from here.

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