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I managed to find it using the good ol' Feynman's trick but it was an extremely long process, so i'd like to know if theres more efficient/faster methods out there that you guys could provide. thanks.

Ill share my attempt, but it is ridiculously long.

My attempt.

$$I\left(a\right)=\int _0^1\frac{\ln \left(ax^2+b\right)}{x+1}\:dx$$ $$I'\left(a\right)=\int _0^1\frac{x^2}{\left(x+1\right)\left(ax^2+b\right)}\:dx=\frac{1}{a+b}\int _0^1\frac{b\left(x-1\right)}{ax^2+b}+\frac{1}{x+1}\:dx$$ $$I'\left(a\right)=\frac{1}{a+b}\left(\frac{b}{2a}\ln \left(a+b\right)-\frac{b}{2a}\ln \left(b\right)-\frac{\sqrt{b}}{\sqrt{a}}\arctan \left(\frac{\sqrt{a}}{\sqrt{b}}\right)+\ln \left(2\right)\right)$$ $$\int _b^aI'\left(a\right)\:da=\frac{b}{2}\int _b^a\frac{\ln \left(a+b\right)}{a\left(a+b\right)}\:da-\frac{b}{2}\ln \left(b\right)\int _b^a\frac{1}{a\left(a+b\right)}\:da-\sqrt{b}\underbrace{\int _b^a\frac{\arctan \left(\sqrt{\frac{a}{b}}\right)}{\sqrt{a}\left(a+b\right)}\:da}_{u=\sqrt{\frac{a}{b}}}+\ln \left(2\right)\int _b^a\frac{1}{a+b}\:da$$ Now lets calculate $I\left(b\right)$. $$I\left(b\right)=\ln \left(b\right)\int _0^1\frac{1}{x+1}\:dx+\int _0^1\frac{\ln \left(x^2+1\right)}{x+1}\:dx=\ln \left(b\right)\ln \left(2\right)+\frac{3\ln ^2\left(2\right)}{4}-\frac{\pi ^2}{48}$$ Now resuming on the original expression: $$I\left(a\right)-\ln \left(b\right)\ln \left(2\right)-\frac{3\ln ^2\left(2\right)}{4}+\frac{\pi ^2}{48}=\frac{1}{2}\underbrace{\int _b^a\frac{\ln \left(a+b\right)}{a}\:da}_{a=bt}-\frac{1}{2}\int _b^a\frac{\ln \left(a+b\right)}{a+b}\:da-\frac{\ln \left(b\right)}{2}\int _b^a\frac{1}{a}\:da+\frac{\ln \left(b\right)}{2}\int _b^a\frac{1}{a+b}\:da-2\int _1^{\sqrt{\frac{a}{b}}}\frac{\arctan \left(u\right)}{u^2+1}\:du+\ln \left(2\right)\ln \left(a+b\right)-\ln \left(2\right)\ln \left(2b\right)$$ $$I\left(a\right)=\frac{\ln \left(b\right)}{2}\int _1^{\frac{a}{b}}\frac{1}{t}\:dt+\frac{1}{2}\int _1^{\frac{a}{b}}\frac{\ln \left(t+1\right)}{t}\:dt-\frac{\ln ^2\left(a+b\right)}{4}+\frac{\ln ^2\left(2b\right)}{4}-\frac{\ln \left(b\right)\ln \left(a\right)}{2}+\frac{\ln ^2\left(b\right)}{2}+\frac{\ln \left(b\right)\ln \left(a+b\right)}{2}-\frac{\ln \left(b\right)\ln \left(2b\right)}{2}-\arctan ^2\left(\sqrt{\frac{a}{b}}\right)+\frac{\pi ^2}{16}+\ln \left(2\right)\ln \left(a+b\right)-\ln \left(2\right)\ln \left(2b\right)+\ln \left(b\right)\ln \left(2\right)+\frac{3\ln ^2\left(2\right)}{4}-\frac{\pi ^2}{48}$$ $$I\left(a\right)=\frac{1}{2}\ln \left(b\right)\ln \left(\frac{a}{b}\right)+\frac{1}{2}\underbrace{\int _0^{\frac{a}{b}}\frac{\ln \left(t+1\right)}{t}\:dt}_{t=-t}-\frac{1}{2}\int _0^1\frac{\ln \left(t+1\right)}{t}\:dt-\frac{\ln ^2\left(a+b\right)}{4}+\frac{\ln ^2\left(2\right)}{4}+\frac{\ln \left(2\right)\ln \left(b\right)}{2}+\frac{\ln ^2\left(b\right)}{4}-\frac{\ln \left(b\right)\ln \left(a\right)}{2}+\frac{\ln ^2\left(b\right)}{2}+\frac{\ln \left(b\right)\ln \left(a+b\right)}{2}-\frac{\ln \left(2\right)\ln \left(b\right)}{2}-\frac{\ln ^2\left(b\right)}{2}-\arctan ^2\left(\sqrt{\frac{a}{b}}\right)+\frac{\pi ^2}{24}+\ln \left(2\right)\ln \left(a+b\right)-\ln ^2\left(2\right)-\ln \left(2\right)\ln \left(b\right)+\ln \left(b\right)\ln \left(2\right)+\frac{3\ln ^2\left(2\right)}{4}$$ $$I\left(a\right)=\frac{\ln \left(b\right)\ln \left(a\right)}{2}-\frac{\ln ^2\left(b\right)}{2}+\frac{1}{2}\int _0^{-\frac{a}{b}}\frac{\ln \left(1-t\right)}{t}dt-\frac{\pi ^2}{24}-\frac{\ln ^2\left(a+b\right)}{4}+\frac{\ln ^2\left(b\right)}{4}-\frac{\ln \left(b\right)\ln \left(a\right)}{2}+\frac{\ln \left(b\right)\ln \left(a+b\right)}{2}-\arctan ^2\left(\sqrt{\frac{a}{b}}\right)+\frac{\pi ^2}{24}+\ln \left(2\right)\ln \left(a+b\right)$$ To solve the integral remaning we can use the following identity: $$\text{Li}_2\left(z\right)=-\int _0^z\frac{\ln \left(1-t\right)}{t}\:dt$$ After using it and simplifying a bit more we finally arrive at the solution being: $$\boxed{I\left(a\right)=-\frac{\ln ^2\left(b\right)}{4}-\frac{\text{Li}_2\left(-\frac{a}{b}\right)}{2}-\frac{\ln ^2\left(a+b\right)}{4}+\frac{\ln \left(b\right)\ln \left(a+b\right)}{2}-\arctan ^2\left(\sqrt{\frac{a}{b}}\right)+\ln \left(2\right)\ln \left(a+b\right)}$$ One of the integrals i used can be proved with this: $$\int _0^1\frac{\ln \left(x^2+1\right)}{x+1}\:dx=-\frac{\text{Li}_2\left(-1\right)}{2}-\frac{\ln ^2\left(2\right)}{4}-\arctan ^2\left(1\right)+\ln ^2\left(2\right)=\frac{\pi ^2}{24}+\frac{3\ln ^2\left(2\right)}{4}-\frac{\pi ^2}{16}=\frac{3\ln ^2\left(2\right)}{4}-\frac{\pi ^2}{48}$$

Dennis Orton
  • 2,646

1 Answers1

2

The value of both $a$ and $b$ don't matter as long as $a,b >0$. We can split it up to get that

$$\int_0^1\frac{\log(ax^2+b)}{x+1}\:dx = \log a \log 2 + \int_0^1\frac{\log(x^2+c)}{x+1}\:dx$$

where we have a new parameter $c \equiv \frac{b}{a}$. Then taking the derivative we have that

$$I'(c) = \int_0^1\frac{1}{(x^2+c)(x+1)}\:dx = \frac{1}{c+1}\int_0^1 \frac{1}{x+1}-\frac{x-1}{x^2+c}\:dx$$

$$= \frac{\log 2}{c+1} - \frac{\log c}{2(c+1)} + \frac{\arctan\left(\frac{1}{\sqrt{c}}\right)}{\sqrt{c}(c+1)}$$

which can be solved in similar ways as before, but now it's only one variable.

For slight completeness sake, continuing on a bit further we get that

$$I(a,b) = \log 2 \log(a+b) - \arctan^2\left(\sqrt{\frac{a}{b}}\right)+\int_0^1 \frac{2\log t}{t+1}\:dx - \frac{1}{2}\int_0^{\frac{b}{a}} \frac{\log t}{t+1}\:dt$$

where the value of the last two integrals can be given by special functions.

Ninad Munshi
  • 34,407