Not the toughest integral, not the easiest one

Question

Perhaps it's not amongst the toughest integrals, but it's interesting to try to find an elegant
approach for the integral $$I_1=\int_0^1 \frac{\log (x)}{\sqrt{x (x+1)}} \, dx$$ $$=4 \text{Li}_2\left(-\sqrt{2}\right)-4 \text{Li}_2\left(-1-\sqrt{2}\right)+2 \log ^2\left(1+\sqrt{2}\right)-4 \log \left(2+\sqrt{2}\right) \log \left(1+\sqrt{2}\right)-\frac{\pi ^2}{3}$$

Do you see any such a way? Then I wonder if we can think of some elegant ways for the evaluation of the quadratic and cubic versions, that is

$$I_2=\int_0^1 \frac{\log^2 (x)}{\sqrt{x (x+1)}} \, dx$$ $$I_3=\int_0^1 \frac{\log^3 (x)}{\sqrt{x (x+1)}} \, dx.$$

How far can we possibly go with the generalization such that we can get integrals in closed form?

Answer 1

By making the Euler substitution $\sqrt{x^{2}+x} = x+t$, we find

$$ \begin{align} \int_{0}^{1} \frac{\log(x)}{\sqrt{x^{2}+x}} \, dx &= 2 \int_{0}^{\sqrt{2}-1} \frac{\log \left(\frac{t^{2}}{1-2t}\right)}{1-2t} \, dt \\ &= 4 \int_{0}^{\sqrt{2}-1} \frac{\log t}{1-2t} \, dt - 2\int_{0}^{\sqrt{2}-1} \frac{\log(1-2t)}{1-2t} \, dt. \end{align}$$

The first integral can be evaluated by integrating by parts.

$$\begin{align} \int_{0}^{\sqrt{2}-1} \frac{\log t}{1-2t} \, dt &= -\frac{1}{2}\log(t) \log(1-2t) \Bigg|^{\sqrt{2}-1}_{0} + \frac{1}{2}\int_{0}^{\sqrt{2}-1} \frac{\log(1-2t)}{t} \, dt \\ &= -\frac{1}{2}\log(\sqrt{2}-1) \log(3 - 2 \sqrt{2}) - \frac{1}{2}\text{Li}_{2} \big(2(\sqrt{2}-1)\big) \end{align}$$

Therefore,

$$\begin{align}\int_{0}^{1} \frac{\log(x)}{\sqrt{x^{2}+x}} \, dx &= -2 \log(\sqrt{2}-1) \log(3 - 2 \sqrt{2}) - 2 \text{Li}_{2} \big( 2(\sqrt{2}-1) \big) + \frac{1}{2} \log^{2}(3- 2 \sqrt{2}) \\ &\approx -3.8208072259. \end{align}$$

EDIT:

David H used an Euler substitution to evaluate a similar but much more difficult integral here.

Answer 2

Let: $$ f(s)=\int_{0}^{1}\frac{x^s}{\sqrt{1+x}}\,dx. $$ We have, by integration by parts: $$ f(s+1)+f(s) = \int_{0}^{1} x^s\sqrt{1+x}\,dx = \frac{1}{s+1}\left(\sqrt{2}-\frac{1}{2}\,f(s+1)\right)$$ hence: $$ \left(2s+3\right)\, f(s+1)+(2s+2)\,f(s) = 2\sqrt{2},$$ $f(0)=2\sqrt{2}-1$ and $\lim_{s\to +\infty}f(s)=0$. We have:

$$ f(s)=\sum_{n\geq 0}\frac{(-1)^n (n-1/2)!}{\sqrt{\pi}(s+n+1)n!} $$ and to compute: $$ \int_{0}^{1}\frac{\log^k x}{\sqrt{x(x+1)}}\,dx $$ boils down to computing $f^{(k)}\left(-\frac{1}{2}\right)$. That gives: $$ 2^{k+1}\phantom{}_{k+2} F_{k+1}\left(\frac{1}{2},\frac{1}{2},\ldots,\frac{1}{2};\frac{3}{2},\frac{3}{2},\ldots,\frac{3}{2};-1\right),$$ so $k=1$ and $k=2$ are essentially the only cases we are able to manage through hypergeometric identities.

Answer 3

As an alternative, we may apply the substitution $x\mapsto\sinh^2{x}$. \begin{align} \int^1_0\frac{\ln{x}}{\sqrt{x(x+1)}}{\rm d}x &=4\int^{-\ln(\sqrt{2}-1)}_0\ln(\sinh{x})\ {\rm d}x\\ &=4\int^{-\ln(\sqrt{2}-1)}_0\left(\color{green}{\ln\left(1-e^{-2x}\right)}-\ln{2}+x\right)\ {\rm d}x\\ &=-4\sum^\infty_{n=1}\frac{1}{n}\int^{-\ln(\sqrt{2}-1)}_0e^{-2nx}\ {\rm d}x+4\ln{2}\ln\left(\sqrt{2}-1\right)+2\ln^2\left(\sqrt{2}-1\right)\\ &=2\sum^\infty_{n=1}\frac{\left(3-2\sqrt{2}\right)^n-1}{n^2}+4\ln{2}\ln\left(\sqrt{2}-1\right)+2\ln^2\left(\sqrt{2}-1\right)\\ &=\color{red}{2{\rm Li}_2\left(3-2\sqrt{2}\right)-\frac{\pi^2}{3}+4\ln{2}\ln\left(\sqrt{2}-1\right)+2\ln^2\left(\sqrt{2}-1\right)} \end{align}

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Similarly, for the second integral, \begin{align} \int^1_0\frac{\ln^2{x}}{\sqrt{x(x+1)}}\ {\rm d}x &=8\int^{-\ln(\sqrt{2}-1)}_0\ln^2(\sinh{x})\ {\rm d}x=8(A+B+C) \end{align} where \begin{align} A &=\int^{-\ln(\sqrt{2}-1)}_0\left(x^2-2x\ln{2}-2\ln{2}\color{green}{\ln\left(1-e^{-2x}\right)}+\ln^2{2}\right)\ {\rm d}x\\ &=-{\rm Li}_2\left(3-2\sqrt{2}\right)\ln{2}+\frac{\pi^2}{6}\ln{2}-\ln^2{2}\ln\left(\sqrt{2}-1\right)-\ln{2}\ln^2\left(\sqrt{2}-1\right)-\frac{1}{3}\ln^3\left(\sqrt{2}-1\right) \end{align} and \begin{align} B &=\int^{-\ln(\sqrt{2}-1)}_02x\ln\left(1-e^{-2x}\right)\ {\rm d}x\\ &=-2\sum^\infty_{n=1}\frac{1}{n}\int^{-\ln(\sqrt{2}-1)}_0xe^{-2nx}\ {\rm d}x\\ &=2\sum^\infty_{n=1}\frac{1}{n}\left[\frac{xe^{-2nx}}{2n}+\frac{e^{-2nx}}{4n^2}\right]^{-\ln(\sqrt{2}-1)}_0\\ &=-\ln\left(\sqrt{2}-1\right)\sum^\infty_{n=1}\frac{\left(3-2\sqrt{2}\right)^n}{n^2}+\frac{1}{2}\sum^\infty_{n=1}\frac{\left(3-2\sqrt{2}\right)^n}{n^3}-\frac{1}{2}\sum^\infty_{n=1}\frac{1}{n^3}\\ &=\frac{1}{2}{\rm Li}_3\left(3-2\sqrt{2}\right)-\frac{\zeta(3)}{2}-{\rm Li}_2\left(3-2\sqrt{2}\right)\ln\left(\sqrt{2}-1\right) \end{align} and \begin{align} C &=\int^{-\ln(\sqrt{2}-1)}_0\ln^2\left(1-e^{-2x}\right)\ {\rm d}x=2\sum^\infty_{n=1}\frac{H_{n-1}}{n}\int^{-\ln(\sqrt{2}-1)}_0e^{-2nx}\ {\rm d}x\\ &=-\sum^\infty_{n=1}\frac{\left(H_n-\frac{1}{n}\right)\left(\left(3-2\sqrt{2}\right)^n-1\right)}{n^2} \end{align} Using the well-known generating function $$\sum^\infty_{n=1}\frac{H_n}{n^2}z^n={\rm Li}_3(z)-{\rm Li}_3(1-z)+{\rm Li}_2(1-z)\ln(1-z)+\frac{1}{2}\ln{z}\ln^2(1-z)+\zeta(3)$$ we obtain \begin{align} C &={\rm Li}_3\left(2\sqrt{2}-2\right)-{\rm Li}_2\left(2\sqrt{2}-2\right)\ln\left(2\sqrt{2}-2\right)-\ln\left(\sqrt{2}-1\right)\ln^2{2}-2\ln{2}\ln^2\left(\sqrt{2}-1\right)\\ &\ \ \ \ -\ln^3\left(\sqrt{2}-1\right) \end{align} Therefore, \begin{align} \int^1_0\frac{\ln^2{x}}{\sqrt{x(x+1)}}\ {\rm d}x &=\color{blue}{4{\rm Li}_3\left(3-2\sqrt{2}\right)+8{\rm Li}_3\left(2\sqrt{2}-2\right)-4\zeta(3)-8{\rm Li}_2\left(3-2\sqrt{2}\right)\ln\left(2\sqrt{2}-2\right)}\\ &\ \ \ \ \color{blue}{-8{\rm Li}_2\left(2\sqrt{2}-2\right)\ln\left(2\sqrt{2}-2\right)+\frac{4\pi^2}{3}\ln{2}-24\ln{2}\ln^2\left(\sqrt{2}-1\right)}\\ &\ \ \ \ \color{blue}{-16\ln^2{2}\ln\left(\sqrt{2}-1\right)-\frac{32}{3}\ln^3\left(\sqrt{2}-1\right)} \end{align}

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Since the generating function of $\dfrac{H_n}{n^3}$ is known as well, I believe that one can work out the closed form of $\displaystyle\int^1_0\frac{\ln^3{x}}{\sqrt{x(x+1)}}\ {\rm d}x$ without too much difficulty. Nevertheless, the process will be immensely tedious.