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Evaluate

$$\int_{0}^{1}\left [ \frac{1+\sqrt{1-x}}{x} +\frac{2}{\ln\left ( 1-x \right )}\right ]\, \mathrm{d}x$$

I tried to let $1-x\rightarrow x$ ,and got $$\int_{0}^{1}\left [ \frac{1+\sqrt{1-x}}{x} +\frac{2}{\ln\left ( 1-x \right )}\right ]\, \mathrm{d}x=\int_{0}^{1}\left ( \frac{1+\sqrt{x}}{1-x}+\frac{2}{\ln x}\right )\, \mathrm{d}x$$ but I have been stuck here for a long time, any idea on how to go futher?

Edit:with the help of Mathematica,I got an answer below

enter image description here

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  • $\begingroup$ if have closed form first step you can $\sqrt{x}=t$ $\endgroup$ – math110 Jan 10 '17 at 8:47
  • $\begingroup$ Integral $\int_0^1\frac{2}{\ln x}dx$ is unbounded $\endgroup$ – Nosrati Jan 10 '17 at 8:49
  • $\begingroup$ For first fraction let $x=\sin^4 t$ and obtain $4\ln|\cos t|+\cos2t$ $\endgroup$ – Nosrati Jan 10 '17 at 8:51
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$$\gamma = \sum_{n\geq 1}\left(\frac{1}{n}-\log\frac{n+1}{n}\right) \stackrel{F}{=} \int_{0}^{+\infty}\sum_{n\geq 1}\left(e^{-nx}-\frac{e^{-nx}-e^{-(n+1)x}}{x}\right)\,dx \tag{1}$$ where $F$ stands for Frullani's theorem. That leads to: $$ \gamma = \int_{0}^{+\infty}\left(\frac{1}{e^x-1}-\frac{1}{x e^x}\right)\,dx = \int_{0}^{1}\left(\frac{1}{\log u}+\frac{1}{1-u}\right)\,du \tag{2}$$ so the given integral equals: $$ \int_{0}^{1}\left(\frac{1+\sqrt{u}}{1-u}+\frac{2}{\log u}\right)\,du = 2\gamma+\int_{0}^{1}\frac{-1+\sqrt{u}}{1-u}\,du\tag{3}$$ or: $$ 2\gamma - \int_{0}^{1}\frac{du}{1+\sqrt{u}} = 2\gamma -\int_{0}^{1}\frac{2v\,dv}{1+v} = \color{red}{2(\gamma+\log 2-1)}\tag{4}$$ as wanted.

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  • 2
    $\begingroup$ This is nice, but it begs the question: how did you know $\;\gamma\;$ was involved in this mess before solving the integral? I first made two substitutions and simplifcations until I did $\;e^{-u}:=1-x\;$ and got something close to an integral expression of $\;\gamma\;$, and even heavily suspected that something like $\;2\gamma\;$ is there, and then your answer came in so I stopped. $\endgroup$ – DonAntonio Jan 10 '17 at 9:57
  • $\begingroup$ @DonAntonio: the presence of $\gamma$ is suggested by the presence of the term $\frac{1}{\log(1-x)}$ in the integrand function, whose singularity in the origin is compensated by the presence of the other term. The RHS of $(2)$ is pretty well-known. $\endgroup$ – Jack D'Aurizio Jan 10 '17 at 11:31
  • $\begingroup$ @JackD'Aurizio Thanks. I know about (1) and (2), yet it would be smoother, or perhaps more "educative", to begin with the integral expression and then compare with the known integral expression for $\;\gamma\;$, otherwise it looks, imo, as hokus pokus. Very nice answer, though. +1 $\endgroup$ – DonAntonio Jan 10 '17 at 11:38
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Let us consider $$I\left(\epsilon\right)=\int_{\epsilon}^{1}\frac{1+\sqrt{1-x}}{x}dx+2\int_{\epsilon}^{1-\epsilon}\frac{1}{\log\left(1-x\right)}dx $$ we get $$\int_{\epsilon}^{1}\frac{1+\sqrt{1-x}}{x}dx=-\log\left(\epsilon\right)+2\int_{0}^{\sqrt{1-\epsilon}}\frac{u^{2}}{1-u^{2}}du $$ $$=-\log\left(\epsilon\right)-2\sqrt{1-\epsilon}-\log\left(\frac{1-\sqrt{1-\epsilon}}{1+\sqrt{1-\epsilon}}\right) $$ and $$2\int_{\epsilon}^{1-\epsilon}\frac{1}{\log\left(1-x\right)}dx=2\int_{\epsilon}^{1-\epsilon}\left(\frac{1}{\log\left(x\right)}+\frac{1}{1-x}\right)dx+2\log\left(\epsilon\right)-2\log\left(1-\epsilon\right) $$ then $$I=\int_{0}^{1}\left(\frac{1+\sqrt{1-x}}{x}+\frac{2}{\log\left(1-x\right)}\right)dx=\lim_{\epsilon\rightarrow0^{+}}\left(-2\sqrt{1-\epsilon}-\log\left(\frac{\left(1-\epsilon\right)\left(1-\sqrt{1-\epsilon}\right)}{\left(1+\sqrt{1-\epsilon}\right)\epsilon}\right)+2\int_{\epsilon}^{1-\epsilon}\left(\frac{1}{\log\left(x\right)}+\frac{1}{1-x}\right)dx\right)\tag{1} $$ $$=\color{blue}{-2+\log\left(4\right)+2\gamma}.$$ The calculation of the integral in the RHS of $(1)$ is classical and can be found here.

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  • $\begingroup$ Nice approach! Got it,thx! $\endgroup$ – Renascence_5. Jan 10 '17 at 11:48
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I have been stuck here for a long time, any idea on how to go further?

Yes. For starters, rewrite $~\dfrac{1+\sqrt x}{1-x}~$ as $~\dfrac1{1-\sqrt x}~,~$ then let $x=t^2,~$ and rewrite $~\dfrac t{1-t}~$ as

$1-\dfrac1{1-t}.~$ You'll be eventually left with $\displaystyle\int_0^1\left(\frac1{1-x}~+~\frac x{\ln x}\right)~dx,~$ which is much better

looking. Taking each part individually, and inserting certain helpful parameters, we have

$\displaystyle\int_0^1\frac{x^\epsilon}{(1-x)^{1-\epsilon}}~dx~=~B(\epsilon,~1+\epsilon),~$ and $~\displaystyle\int_0^1\frac{x}{\ln^{1-\epsilon}x}~dx~=~-\left(-\frac12\right)^\epsilon~\Gamma(\epsilon),~$ ultimately

leaving us with evaluating $~\displaystyle\lim_{\epsilon\to0^+}~B(\epsilon,~1+\epsilon)~-~\frac{\Gamma(\epsilon)}{2^\epsilon}~=~\gamma~+~\ln2.$

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  • $\begingroup$ It's very nice! thx! $\endgroup$ – Renascence_5. Jan 10 '17 at 11:47
  • $\begingroup$ You said you are stuck ? where ? $\endgroup$ – Zaid Alyafeai Jan 10 '17 at 14:47
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int_{0}^{1}\bracks{{1 + \root{1 - x} \over x} + {2 \over \ln\pars{1 - x}}} \,\dd x \,\,\,\stackrel{\root{1 - x}\ \mapsto\ x}{=}\,\,\, 2\int_{0}^{1}{x \over 1 - x}\ \overbrace{\bracks{1 + {1 - x \over \ln\pars{x}}}} ^{\ds{\int_{0}^{1}\pars{1 - x^{t}}\,\dd t}}\ \,\dd x \\[5mm] = &\ 2\int_{0}^{1}\int_{0}^{1}{x - x^{t + 1} \over 1 - x}\,\dd x\,\dd t = 2\int_{0}^{1}\pars{\int_{0}^{1}{1 - x^{t + 1} \over 1 - x}\,\dd x - \int_{0}^{1}\dd x}\dd t \\[5mm] = &\ 2\int_{0}^{1}\braces{\vphantom{\Large A}\bracks{\vphantom{\large A} \Psi\pars{t + 2} + \gamma} - 1}\,\dd t\qquad \pars{~\Psi:\ Digamma\ Function~}\label{1}\tag{1} \\[5mm] = &\ 2\bracks{\ln\pars{\Gamma\pars{3} \over \Gamma\pars{2}} + \gamma - 1}\qquad \pars{~\Gamma:\ Gamma\ Function~} \\[5mm] = &\ \bbx{\ds{2\bracks{\vphantom{\large A}\gamma + \ln\pars{2} - 1}}} \end{align}

In \eqref{1}, I used the well known identity $\ds{\mathbf{6.3.22}}$ from A & S Table.

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