galactus wrote:

\(\displaystyle \int_{0}^{1/2}\frac{log^{2}(x+\sqrt{x^{2}+1})}{x}dx\)

I worked out a generalized (parametric) closed form for integrals of this type a few years back. Let me see now...

EDIT: I'd forgotten just how long and laborious the proof is, so I'll sketch out how it's done, but not add the entire proof (it'll be enough to demonstrate everything that's required for the full result).

Part 1:

Define

\(\displaystyle \mathcal{S}_m(z)=\int_0^z\frac{\log^m(x+\sqrt{1+x^2})}{x}\,dx=\int_0^z\frac{(\sinh^{-1}x)^m}{x}\,dx\)

for \(\displaystyle 0 < z\le 1\)

Make the sub \(\displaystyle y=x+\sqrt{1+x^2}\,;\, x=\frac{y^2-1}{2y}\,;\, dx=\left(\frac{y^2+1}{2y^2}\right)\,dy\, \Rightarrow\)

\(\displaystyle \mathcal{S}_m(z)=\int_1^w\frac{(y^2+1)}{y\,(y^2-1)}(\log y)^m\,dy\)

where \(\displaystyle w=z+\sqrt{1+z^2}\)

Next, make the reciprocal sub \(\displaystyle y \to 1/x\) to get

\(\displaystyle (-1)^{m+1}\int_1^{1/w}\frac{(1+x^2)}{x\, (1-x^2)}\,(\log x)^m\,dx=\)

\(\displaystyle (-1)^{m+1}\int_1^{1/w}\frac{[(1-x^2)+x^2]}{x\, (1-x^2)}\,(\log x)^m\,dx+(-1)^{m+1}\int_1^{1/w}\frac{[1-(1-x^2)]}{x\, (1-x^2)}\,(\log x)^m\,dx\)

Or

\(\displaystyle (-1)^{m+1}\mathcal{S}_m(z)=\)

\(\displaystyle \int_1^{1/w}\frac{(\log x)^m}{x}\,dx+\int_1^{1/w}\frac{x\,(\log x)^m}{(1-x^2)}\,dx+\)

\(\displaystyle \int_1^{1/w}\frac{(\log x)^m}{x\,(1-x^2)}\,dx-\int_1^{1/w}\frac{(\log x)^m}{x}\,dx=\)

\(\displaystyle \int_1^{1/w}\frac{x\,(\log x)^m}{(1-x^2)}\,dx+\int_1^{1/w}\frac{(\log x)^m}{x\,(1-x^2)}\,dx=\)

\(\displaystyle \frac{1}{2}\int_1^{1/w}\frac{x\,(\log x)^m}{(1-x)}\,dx+\frac{1}{2}\int_1^{1/w}\frac{x\,(\log x)^m}{(1+x)}\,dx+\)

\(\displaystyle \frac{1}{2}\int_1^{1/w}\frac{(\log x)^m}{x\,(1-x)}\,dx+\frac{1}{2}\int_1^{1/w}\frac{(\log x)^m}{x\,(1+x)}\,dx=\)

\(\displaystyle \frac{1}{2}\left(I_1+I_2+I_3+I_4\right)\,\) or

\(\displaystyle 2(-1)^{m+1}\mathcal{S}=\left(I_1+I_2+I_3+I_4\right)\)

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My comp's being a tad slow today, so I'll post the proof in stages...