Jtdylktuy

Evaluate $$I=int_0^1frac{ln(1+x-x^2)}xdx$$ without using polylog functions.

This integral can be easily solved by factorizing $1+x-x^2$ and using the values of dilogarithm at some special points.

The motivation of writing this post is someone said that this integral cannot be solved without using special functions.

Another alternative solutions will be appreciated.

Put
begin{equation*}
I=int_{0}^1dfrac{ln(1+x-x^2)}{x},mathrm{d}x = int_{0}^1dfrac{ln(1+x(1-x))}{x}mathrm{d}x.
end{equation*}
If we change $x$ to $ 1-x $ we get
begin{equation*}
I=int_{0}^1dfrac{ln(1+x-x^2)}{1-x},mathrm{d}x.
end{equation*}
Consequently
begin{equation*}
2I = int_{0}^1ln(1+x-x^2)left(dfrac{1}{x}+dfrac{1}{1-x}right),mathrm{d}x.
end{equation*}
The next step will be integration by parts.
begin{equation*}
2I= underbrace{left[ln(1+x-x^2)lndfrac{x}{1-x}right]_{0}^{1}}_{=0} -int_{0}^1dfrac{1-2x}{1+x-x^2}lndfrac{x}{1-x}, mathrm{d}x
end{equation*}
Thenbegin{equation*}
I=dfrac{1}{2}int_{0}^1dfrac{2x-1}{1+x-x^2}lndfrac{x}{1-x}, mathrm{d}x.
end{equation*}
If we substitute $ z=dfrac{x}{1-x} $ we get
begin{equation*}
I = int_{0}^{infty}dfrac{(z-1)ln z}{2(z+1)(z^2+3z+1)},mathrm{d}z.
end{equation*}
In order to evaluate this integral we integrate $displaystyle dfrac{(z-1)log^2 z}{2(z+1)(z^2+3z+1)}$ along a keyhole contour and use residue calculus. We get that
begin{equation*}
I = 2ln^2varphi
end{equation*}
where $ varphi = dfrac{1+sqrt{5}}{2}. $

$$begin{aligned}
I&=int_0^1frac{ln(1+x-x^2)}xmathrm{d}x\
&overset{(1)}{=}int_0^1sum_{n=1}^inftyfrac{(-1)^{n-1}(x-x^2)^n}{nx}mathrm{d}x\
&overset{(2)}{=}sum_{n=1}^inftyfrac{(-1)^{n-1}}nint_0^1x^{n-1}(1-x)^nmathrm{d}x\
&overset{(3)}{=}sum_{n=1}^inftyfrac{(-1)^{n-1}}nfrac{(n-1)!n!}{(2n)!}\
&=sum_{n=0}^inftyfrac{(-1)^{n}(n!)^2}{(2n+2)!}\
&=sum_{n=0}^inftyfrac{(-1)^{n}(1times2timescdotstimes n)(1times2timescdotstimes n)}{1times2timescdotstimes (2n+2)}\
&=sum_{n=0}^inftyfrac{(-1)^{n}(1times2timescdotstimes n)}{1times3times5timescdotstimes(2n+1)times (2n+2)2^n}\
&=sum_{n=0}^inftyfrac{(-1)^nn!}{(2n+1)!!(2n+2)2^n}
end{aligned}$$
Explanation

(1) Using the Maclaurin series of $ln(1+w)$, where $w=x-x^2$.

(2) It is legal to change the position of $sum$ and $int$.

(3) Integrate by parts $n-1$ times.
Notice that $$sum_{n=0}^inftyfrac{(2n)!!}{(2n+1)!!}x^{2n+1}=frac{arcsin x}{sqrt{1-x^2}},$$
integrate both sides from $0$, we have $$sum_{n=0}^inftyfrac{(2n)!!}{(2n+1)!!(2n+2)}x^{2n+2}=frac12arcsin^2x.$$
Letting $x=frac i2$ leads to
$$sum_{n=0}^inftyfrac{(-1)^{n+1}(2n)!!}{(2n+1)!!(2n+2)}2^{-2n-2}=frac12arcsin^2frac i2.$$
Combining with $(2n)!!=2^{n}n!$, we have $$-frac14I=-frac12operatorname{arccsch}^22,$$
or $I=2ln^2varphi,$ where $varphi$ denotes the golden ratio.