How to evaluate this integral : $I(a,b)=\int^{a}_0 \ln^2(bx + \sqrt{1+b^2x^2} ) \ln \left({\frac{a+\sqrt{a^2-x^2}}{x}}\right)dx$

Question

I need to evaluate this integral : $$I(a,b)=\int^{a}_0 \ln^2(bx + \sqrt{1+b^2x^2} ) \ln \left({\frac{a+\sqrt{a^2-x^2}}{x}}\right)dx$$

where : $a\ge 0$

I tried to solve the integral by setting $ a=b=1$

\begin{align*} I &= \int^{1}_0 \ln^2(x + \sqrt{1+x^2} ) \ln \left({\frac{1+\sqrt{1-x^2}}{x}}\right)dx \\ &= \int^{1}_0 \ln^2(x + \sqrt{1+x^2} ) \ln \left({\frac{x}{1-\sqrt{1-x^2}}}\right)dx \end{align*} Therfore: \begin{align*} 2I &= \int^{1}_0 \ln^2(x + \sqrt{1+x^2} ) \ln \left({\frac{1+\sqrt{1-x^2}}{1-\sqrt{1-x^2}}}\right)dx \\ &= 2\int^1_0 \tanh^{-1}(\sqrt{1-x^2})\left({\sinh^{-1}(x)}\right)^2 dx \\ &\overset{x=\sin(t)}{=} 2\int_0^{\frac{\pi}{2}}\cos(t)\tanh^{-1}(\cos(t))\left({\sinh^{-1}(\sin(t))}\right)^2 dt \end{align*}

How can I complete the solution. I need help with this.

This problem is from a book special function by Yury

In the book this answer was given.

$$I(a,b)=\frac{\pi a }{4}\left[{4-2\ln(1+a^2b^2)-\text{Li}_2(-a^2b^2)}\right]-\frac{\pi}{b}\arctan(ab)$$

and therfore : $$I =I(1,1)= \pi -\frac{\pi^2}{4}+\frac{\pi^3}{48}-\frac{\pi}{2}\ln(2)$$

Answer 1

Here is an elementary approach. Rescale the integral with $x\to ax$ to have $$\int^{a}_0 \ln^2(bx + \sqrt{1+b^2x^2} ) \ln \left({\frac{a+\sqrt{a^2-x^2}}{x}}\right)dx=a K(ab )$$ where $K(p)= \int_0^1 \text{sech}^{-1}x \ (\sinh^{-1}px)^2dx$ \begin{align} K’(p) = &\int^{1}_0 \text{sech}^{-1}x \ \frac{2x \sinh^{-1}px}{\sqrt{1+p^2x^2}}dx =\int^{1}_0 \frac{x \ \text{sech}^{-1}x }p d\left[(\sinh^{-1}px)^2 \right]\\ \overset{ibp}=&-\frac1p K(p)+\frac1p \int_0^1\frac{(\sinh^{-1}px)^2}{\sqrt{1-x^2}}\overset{x=\cos t}{dx} \end{align} which leads to \begin{align} & [pK(p)]’=\int_0^{\pi/2}{[\sinh^{-1}(p\cos t)]^2}dt\\ &[ pK(p)]’’ = 2\int_0^{\pi/2}\frac{\sinh^{-1}(p\cos t) \cos t }{\sqrt{1+p^2\cos^2t}}\ dt\\ =&\ \frac2p \int_0^p \bigg(\frac d{ds}\int_0^{\pi/2}\frac{s \sinh^{-1}(s\cos t) \cos t}{\sqrt{1+s^2\cos^2t}}\ dt\bigg)ds\\ =& \ \frac2p \int_0^p \int_0^{\pi/2} \bigg[\frac{\sinh^{-1}(s\cos t)}{1+s^2}\overset{ibp}d\bigg(\frac{\sin t}{\sqrt{1+s^2\cos^2t}}\bigg)+\frac{s\cos t}{1+s^2\cos^2t} dt\bigg] ds\\ =&\ \frac2p \int_0^p\int_0^{\pi/2}\frac {s}{1+s^2}dt\ ds= \frac\pi{2p}{\ln(1+p^2)} \end{align} and \begin{align} K(p) =&\ \frac1p \int_0^p \int_0^v [uK(u)]’’ du \ dv = \frac1p \int_0^p \int_u^p [uK(u)]’’ dv \ du\\ = &\ \frac{\pi}{2} \int_0^p \frac {\ln(1+u^2)}{u}du-\frac\pi{2p}\int_0^p {\ln(1+u^2)}du\\ =&-\frac\pi4\text{Li}_2(-p^2) -\frac\pi2\ln(1+p^2)-\frac\pi p\tan^{-1} p+\pi \end{align} where $\int_0^p \frac{\ln(1+u^2)}u du=-\frac12 \text{Li}_2(-p^2) $.

Answer 2

$$\begin{align*} I(a,b) &= \int_0^a \operatorname{arsech}\frac xa \operatorname{arsinh}^2(bx) \, dx \\ &= \int_0^1 a \operatorname{arsech} x \operatorname{arsinh}^2(abx) \, dx & x\to ax \\ &= \int_0^1 \int_x^1 \cdots \, dy \, dx = \int_0^1 \int_0^y \frac{a \operatorname{arsinh}^2(abx)}{y\sqrt{1-y^2}} \, dx \, dy \\ &= a \int_0^1 \frac{2y - \frac2{ab} \sqrt{1+a^2b^2y^2} \operatorname{arsinh}(aby) + y \operatorname{arsinh}^2(aby)}{y\sqrt{1-y^2}} \, dy & \text{by parts} \\ &= a\pi - \frac2b \underbrace{\int_0^1 \frac{\sqrt{1+a^2b^2y^2}}{y\sqrt{1-y^2}} \operatorname{arsinh}(aby) \, dy}_{J(a,b)} + a \underbrace{\int_0^1 \frac{\operatorname{arsinh}^2(aby)}{\sqrt{1-y^2}} \, dy}_{K(a,b)} \end{align*}$$

The integrals $J$ and $K$ can be evaluated by invoking the Taylor expansion of $\operatorname{arsinh}^2z=-\arcsin^2(iz)$.

$$\operatorname{arsinh}^2z = \sum_{n\ge1} \frac{(-1)^{n+1} 2^{2n-1}}{n^2 \binom{2n}n} z^{2n} \\ \implies \frac{\operatorname{arsinh}z}{\sqrt{1+z^2}} = \frac12 \frac d{dz} \operatorname{arsinh}^2z = \sum_{n\ge1} \frac{(-1)^{n+1} 2^{2n-1}}{n \binom{2n}n} z^{2n-1}$$

For example:

$$\begin{align*} K(a,b) &= \sum_{n\ge1} \frac{(-1)^{n+1} 2^{2n-1} (ab)^{2n}}{n^2 \binom{2n}n} \int_0^1 \frac{y^{2n}}{\sqrt{1-y^2}} \, dy \\ &= \sum_{n\ge1} \frac{(-1)^{n+1} 2^{2n-2} (ab)^{2n}}{n^2 \binom{2n}n} \underbrace{\int_0^1 y^{n-1/2} (1-y)^{-1/2} \, dy}_{\text{beta function}} & y\to\sqrt y \\ &= \frac{\sqrt\pi}4 \sum_{n\ge1} \frac{(-1)^{n+1} (2ab)^{2n}}{n^2} \frac{n!\left(n-\frac12\right)!}{(2n)!} \\ &= -\frac\pi4 \sum_{n\ge1} \frac{\left(-a^2b^2\right)^n}{n^2} & (*) \\ &= -\frac\pi4 \operatorname{Li}_2\left(-a^2b^2\right) \end{align*}$$

provided $\lvert ab\rvert<1$, where $(*)$ follows from the dimidiation formula for the Pochhammer symbol,

$$\left(n-\frac12\right)! = \left(-\frac12\right)! \cdot \frac{(2n)!}{2^{2n} n!}$$

By the same process, we find

$$\begin{align*} J(a,b) &= \frac\pi2 \sum_{n\ge1} \frac{(-1)^{n+1} (ab)^{2n-1}}{2n-1} \left(1+a^2b^2-\frac{a^2b^2}{2n}\right) \\ &= \frac\pi2 \left(\sum_{n\ge1} \frac{(-1)^{n+1} (ab)^{2n-1}}{2n-1} + \sum_{n\ge1} \frac{(-1)^{n+1} (ab)^{2n+1}}{2n}\right) \\ &= \frac\pi2\arctan(ab) + \frac{\pi ab}4 \log\left(1+a^2b^2\right) \end{align*}$$