The integral \begin{align} J_{m} = \int_{0}^{1} \frac{t^{m}}{1+t} \, \ln(1+t) \, dt \end{align} has the general form \begin{align} J_{m} = (-1)^{m} \left[ A_{m} - B_{m} \, \ln(2) + C_{m} \, \ln^{2}(2) \right] \end{align}. Is there a general form for the coefficients $A_{m}$, $B_{m}$, and $C_{m}$ ?
2026-03-28 18:12:53.1774721573
Log integrals III
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The generating function is, according to Maple, $$ \sum_{m=0}^\infty J_m x^m = \int_0^1 \dfrac{\ln(1+t)}{(1+t)(1-xt)}\; dt = (1+x)^{-1} \left(\dfrac{1}{2} \ln(2)^2 + \ln \left(\dfrac{x}{1+x}\right) \ln(1-x) - \text{dilog} \left(\dfrac{x}{1+x}\right) + \text{dilog} \left(\dfrac{2x}{1+x}\right) \right)$$ but I don't know if you can get closed forms from that.
It appears (using the gfun package) that $(-1)^n A_n$ satisfies the recurrence $$ \left( -{n}^{2}-5\,n-6 \right) a \left( n+1 \right) + \left( -{n}^{2} -4\,n-3 \right) a \left( n+2 \right) + \left( {n}^{2}+9\,n+19 \right) a \left( n+3 \right) + \left( {n}^{2}+8\,n+16 \right) a \left( n+4 \right) =0$$ while $(-1)^{n+1} B_n$ satisfies $$ \left( -n-2 \right) b \left( n+1 \right) + \left( -n-2 \right) b \left( n+2 \right) + \left( n+4 \right) b \left( n+3 \right) + \left( n+4 \right) b \left( n+4 \right) =0$$ and $C_n = 1/2$.