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How can we evaluate $$\int_0^1 x\ln \left ( \sqrt{1+x}+\sqrt{1-x}\right)\ln \left ( \sqrt {1+x} -\sqrt{1-x} \right)\mathrm{d}x?$$

Usually when having as the integrand a logarithmic function, the first thing would be to try to integrate by parts, however in this case there's a product of logarithms and trying to integrate by parts would just produce two harder integrals and it might not be a good approach.

There is also An integral by O. Furdui $\int_0^1 \log^2(\sqrt{1+x}-\sqrt{1-x}) \ dx$ which is quite similar to this one, however the substitution $\sqrt{1+x}-\sqrt{1-x}=2\sin t$ doesn't seem to be useful due to the $\sqrt{1+x}+\sqrt{1-x}$ term.

Is there a better approach to this integral that reduces it to something simpler?

Zacky
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tjujystsll
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1 Answers1

10

One may prove that

$$ \int_0^1 x\ln \left( \sqrt{1+x}+\sqrt{1-x}\right)\ln \left( \sqrt {1+x} -\sqrt{1-x}\right)\:\mathrm{d}x=\frac{\ln^2 2}4-\frac{3\ln2}4+\frac1{16}. \tag1 $$

Hint. By using $$ \begin{align} 4ab&=(a+b)^2-(a-b)^2 \end{align} $$ giving $$ \begin{align} 4\ln b \ln c&=\ln^2 (bc)-\ln^2 \left(\frac bc\right) \end{align} $$ and by observing that $$ \begin{align} &\left ( \sqrt{1+x}+\sqrt{1-x}\right)\left ( \sqrt {1+x} -\sqrt{1-x} \right)=2x \\ &\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt {1+x} +\sqrt{1-x}}=\frac{x}{1+\sqrt{1-x^2}},\quad 0<x<1, \end{align} $$ one gets $$ \begin{align} 4&\int_0^1 x\ln \left ( \sqrt{1+x}+\sqrt{1-x}\right)\ln \left ( \sqrt {1+x} -\sqrt{1-x} \right)\:\mathrm{d}x\\=&\int_0^1 x\ln^2 \left ( 2x \right)\:\mathrm{d}x -\int_0^1 x \ln^2\left( \frac{x}{1+\sqrt{1-x^2}}\right)\:\mathrm{d}x. \tag2 \end{align} $$ Then integrating by parts twice one gets $$ \int_0^1 x\ln^2 \left ( 2x \right)\:\mathrm{d}x=\frac{\ln^2 2}2-\frac{\ln2}2+\frac14, \tag3 $$ by the change of variable $u=\dfrac{x}{1+\sqrt{1-x^2}}$, one has $$ \begin{align} \int_0^1 x \ln^2\left( \frac{x}{1+\sqrt{1-x^2}}\right)\:dx&=\int_0^1\frac{4u(1-u^2)}{(1+u^2)^3}\cdot \ln^2 u\:du \\&=\frac12\int_0^1\frac{(1-v)}{(1+v)^3}\cdot \ln^2 v\:du \\&= \ln 2,\tag4\end{align} $$ obtained by parts. By inserting $(4)$ and $(3)$ into $(2)$ one gets $(1)$.

Olivier Oloa
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