Another integral similar to my previous question:
$$\int_0^\infty\frac{\ln\left(\sqrt{x+1\vphantom{x^0}}-1\right)\,\ln\left(\sqrt{x^{-1}+1}+1\right)}{(x+1)^{3/2}}dx$$
Can someone suggest how to evaluate it? Is there a closed form?
[Math] Integral $\int_0^\infty\frac{\ln\left(\sqrt{x+1\vphantom{x^0}}-1\right)\,\ln\left(\sqrt{x^{-1}+1}+1\right)}{(x+1)^{3/2}}dx$
calculusclosed-formimproper-integralsintegrationlogarithms
Related Solutions
Define the function $I(s)$ for $s > 0$ by
$$ I(s) = \int_{0}^{\infty} \frac{\operatorname{arccot}( \sqrt{x} - e^{s}\sqrt{x+1} )}{x+1} \, dx. $$
By observing that $I(\infty) = 0$, we have
$$ I(s) = -\int_{s}^{\infty} I'(t) \, dt. $$
Applying Leibniz's integral rule,
$$ I'(s) = \int_{0}^{\infty} \frac{e^{s}\sqrt{x+1}}{1 + (\sqrt{x} - e^{s}\sqrt{x+1} )^{2}} \, \frac{dx}{x+1}. $$
Now with the substitution $x = \tan^{2}\theta$,
\begin{align*} I'(s) &= \int_{0}^{\frac{\pi}{2}} \frac{\sin\theta}{\cosh s - \sin\theta} = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos\theta}{\cosh s - \cos\theta}. \end{align*}
Then with the substitution $z = e^{i\theta}$, it follows that
\begin{align*} I'(s) &= \frac{i}{2} \int_{\Gamma} \frac{z^{2} + 1}{z^{2} - 2z \cosh s + 1} \frac{dz}{z}, \end{align*}
where the contour $\Gamma$ denotes the counter-clockwise semicircular arc joining from $-i$ to $i$. Note that we have $z^{2} - 2z \cosh s + 1 = 0$ if and only if $\cosh s = \cos \theta$ if and only if $z = e^{i\theta} = e^{\pm s}$. Thus deforming the contour to the straight line joining from $-i$ to $i$ with infinitesimal indent at the origin, we obtain
\begin{align*} I'(s) &= \frac{i}{2} \left\{ \operatorname{PV}\! \int_{-i}^{i} f(z) \, dz + 2\pi i \operatorname{Res} (f, e^{-s}) + \pi i \operatorname{Res} (f, 0) \right\}, \end{align*}
where $f$ denotes the integrand
$$ f(z) = \frac{z^{2} + 1}{z (z^{2} - 2z \cosh s + 1)}. $$
Proceeding the calculation,
\begin{align*} I'(s) &= \frac{i}{2} \left\{ i \operatorname{PV} \! \int_{-1}^{1} f(ix) \, dx + \pi i - 2 \pi i \coth s \right\} \\ &= - \frac{1}{2} \int_{0}^{1} \{ f(ix) + f(-ix) \} \, dx - \frac{\pi}{2} + \pi \coth s \\ &= \cosh s \int_{0}^{1} \frac{\frac{1}{2} (1 - x^{-2}) }{\{ \frac{1}{2}(x + x^{-1}) \}^{2} + \sinh^{2} s} \, dx - \frac{\pi}{2} + \pi \coth s. \end{align*}
Now with the substitution $u = \frac{1}{2} \{ x + x^{-1} \}$, it follows that
\begin{align*} I'(s) &= - \cosh s \int_{1}^{\infty} \frac{du}{u^{2} + \sinh^{2} s} - \frac{\pi}{2} + \pi \coth s \\ &= - \arctan(\sinh s) \coth s - \frac{\pi}{2} + \pi \coth s. \end{align*}
Finally, with the substitution $x = \sinh t$,
\begin{align*} I(s) = - \int_{s}^{\infty} I'(t) \, dt &= \int_{s}^{\infty} \left\{ \arctan(\sinh t) \coth t + \frac{\pi}{2} - \pi \coth t \right\} \, dt \\ &= \int_{\sinh s}^{\infty} \left( \frac{\arctan x}{x} + \frac{\pi}{2\sqrt{x^{2} + 1}} - \frac{\pi}{x} \right) \, dx. \end{align*}
Our problem corresponds to $s = \log 2$, or equivalently, $a := \sinh s = \frac{3}{4}$.
\begin{align*} &\int_{a}^{\infty} \left( \frac{\arctan x}{x} + \frac{\pi}{2\sqrt{x^{2} + 1}} - \frac{\pi}{x} \right) \, dx \\ &= - \int_{a}^{\infty} \frac{\arctan (1/x)}{x} \, dx + \frac{\pi}{2} \int_{a}^{\infty} \left( \frac{1}{\sqrt{x^{2} + 1}} - \frac{1}{x} \right) \, dx \\ &= - \int_{0}^{1/a} \frac{\arctan x}{x} \, dx + \frac{\pi}{2} \lim_{R\to\infty} \left[ \log\left( \frac{x + \sqrt{x^{2} + 1}}{x} \right) \right]_{a}^{R} \\ &= - \operatorname{Ti}\left(\frac{1}{a}\right) - \frac{\pi}{2} \log\left( \frac{a + \sqrt{a^{2} + 1}}{2a} \right), \end{align*}
where the function
$$ \operatorname{Ti}(x) = \frac{\operatorname{Li}_{2}(ix) - \operatorname{Li}_{2}(-ix)}{2i} $$
is the inverse tangent integral. Using the following simple identity
$$ \operatorname{Ti}(x) = \operatorname{Ti}\left(\frac{1}{x}\right) + \frac{\pi}{2}\log x, $$
it follows that
\begin{align*} \int_{a}^{\infty} \left( \frac{\arctan x}{x} + \frac{\pi}{2\sqrt{x^{2} + 1}} - \frac{\pi}{x} \right) \, dx = - \operatorname{Ti}(a) - \frac{\pi}{2} \log\left( \frac{a + \sqrt{a^{2} + 1}}{2a^{2}} \right). \end{align*}
Therefore, plugging $a = \frac{3}{4}$ gives
$$ \int_{0}^{\infty} \frac{\operatorname{arccot}( \sqrt{x} - 2\sqrt{x+1} )}{x+1} \, dx = \pi \log(3/4) - \operatorname{Ti}(3/4) $$
as Cleo pointed out without explanation. More generally, for $k > 1$
$$ \int_{0}^{\infty} \frac{\operatorname{arccot}( \sqrt{x} - k \sqrt{x+1} )}{x+1} \, dx = \frac{\pi}{2} \log \left( \frac{(k^{2} - 1)^{2}}{2k^{3}} \right) - \operatorname{Ti}\left( \frac{k^{2} - 1}{2k} \right). $$
Here is a partial progress report. I am basically repeating Jim Belk's analysis from the previous answer.
Set $F(a) = \int_{x=0}^1 \frac{\log(1+x^a)}{1+x} dx$. Then $$F(a) = \int_{x=0}^1 \int_{y=0}^{x^a} \frac{dx dy}{(1+x)(1+y)} = \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)}$$ so $$F(a) + F(a^{-1}) = \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)} + \int_{0 \leq y \leq x^{1/a} \leq 1} \frac{dx dy}{(1+x)(1+y)}$$ $$= \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)} + \int_{0 \leq y^a \leq x \leq 1} \frac{dx dy}{(1+x)(1+y)} = \int_{0 \leq x,y \leq 1} \frac{dx dy}{(1+x)(1+y)} = (\log 2)^2.$$ (In order to combine the integrals, first switch the names of $x$ and $y$ in the second one.)
So $$F'(a) - a^{-2} F'(a^{-1})=0.$$ This gives a linear relation between $F'(2 + \sqrt{3})$ and $F'(2-\sqrt{3})$. If we find a second one, we can solve the linear equations and be done.
Notice that $$F'(a) = \int_{x=0}^1 \frac{x^a \log x dx}{(1+x)(1+x^a)} = \sum_{m,n=0}^{\infty} \int_{x=0}^1 (-1)^{m+n} x^{m+(n+1) a} \log x dx.$$ Integrating by parts, $\int_{x=0}^1 x^b \log x dx = \frac{-1}{(b+1)^2}$. So, ignoring issues of convergence, we should have $$F'(a) = \sum_{m,n=0}^{\infty} \frac{(-1)^{m+n+1}}{(m+(n+1) a + 1)^2} = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n+1}}{(m+n a)^2}$$ In the last step, we turned $m+1$ and $n+1$ into $m$ and $n$ to make things pretty. My guess is that the convergence issues can be dealt with for any $a>0$, but I haven't thought much about it.
So $$F'(a) + F'(a^{-1}) = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \left( \frac{(-1)^{m+n+1}}{(m+n a)^2} +\frac{(-1)^{m+n+1}}{(m+n a^{-1})^2} \right).$$ Putting $a=2 + \sqrt{3}$, this is $$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} (-1)^{m+n+1} \frac{2 (m^2+4mn+7n^2)}{(m^2+4mn+n^2)^2} $$ $$= 2 \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n+1}}{m^2+4mn+n^2} +12 \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n+1} n^2}{(m^2+4mn+n^2)^2}.$$
Here is where I run out of ideas. The first sum is basically the one at the end of Jim Belk's post, but I have no ideas for the second one.
Best Answer
Yes, there is a closed form: $$\frac{\pi^2}3-\ln^22-4\,G,$$ where $G$ is the Catalan constant: $$G=-\int_0^1\frac{\ln x}{x^2+1}dx.$$