What is $ \displaystyle\int_0^1 \frac{\ln(1+bx)}{1+x} dx $?
I call it $f(b)$ and differentiate with respect to be $b,$ a bit of partial fractions and the $x$ integral can be done. Then I integrate with respect to $b$ and get a bit lost.
Can some of the terms be expressed in terms of dilogarithms? I get lost in the details! Could we avoid all this just go straight to dilogarithms (with a cunning substitution)?
If $b<1$, one may try:
$$\int_{0}^{1} \frac{\log(1+bx)}{1+x} dx = \int_{0}^{1} \log(1+bx)\cdot\frac{1}{1+x} dx = \int_{0}^{1} \left(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}(bx)^{n}}{n} \right)\left(\sum_{m=0}^{\infty} (-1)^{m}x^{m}\right) dx$$
Then:
$$ \int_{0}^{1} \left(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}(bx)^{n}}{n} \right)\left(\sum_{m=0}^{\infty} (-1)^{m}x^{m}\right) = \sum_{n=1}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^{m+n+1}b^{n}}{n} \int_{0}^{1} x^{m+n} dx = \sum_{n=1}^{\infty} \sum_{m=0}^{\infty} \frac{(-1)^{m+n+1}b^{n}}{n(m+n+1)} $$
Gaffe: Or it would, if the integral didn't diverge. (the numerical packages I'm using are both giving nonsensical results, which either means its an improper integral beyond their scope of dealing with improper integrals or that it doesn't converge) Does it converge for any $b$?
I mucked up the power series for the denominator. Fixed that.
Let $x = \frac{1}{b} \frac{1-u}{u}$. The integration range $0\leqslant x \leqslant 1$ translates to $\frac{1}{1+b} \leqslant u \leqslant 1$, for $b>0$ $$ \frac{\log(1+ b x)}{1+x}\mathrm{d} x \to \frac{-\log(u)}{1 - (1-b) u} \frac{\mathrm{d} u}{u} = \frac{\log(u)}{u} \left(1 + \frac{(1-b) u}{1-(1-b) u} \right) $$ Now $$ \begin{eqnarray} \int \frac{\log(u)}{1 - (1-b) u} \frac{\mathrm{d} u}{u} &\stackrel{\color\red{\text{by parts}}}{=}& \frac{\log^2(u)}{2} - \log(u) \cdot \log(1-(1-b)u) - \int \frac{\log(1-(1-b)u)}{u} \mathrm{d} u \\ &=& \frac{\log^2(u)}{2} - \log(u) \cdot \log(1-(1-b)u) - \operatorname{Li_2}\left((1-b) u\right) + C \end{eqnarray} $$ where a definition of dilogarithm was used $\operatorname{Li_2}^\prime(z) = -\frac{\log(1-z)}{z}$. Now assuming $b$ is such that the anti-derivatives is smooth: $$ \begin{eqnarray} \int_0^1 \frac{\log(1+b x)}{1+x} \mathrm{d} x &=& \int\limits_{\frac{1}{1+b}}^1 \frac{-\log(u)}{1 - (1-b) u} \frac{\mathrm{d} u}{u} \\ &=& \operatorname{Li_2}\left(1-b\right) - \operatorname{Li_2}\left(\frac{1-b}{1+b}\right) + \frac{1}{2} \log^2\left(1+b\right) +\log\left(1+b\right) \log\left(\frac{2b}{1+b}\right) \end{eqnarray} $$
For $(1+bx) > 0, \ln(1+bx) \leq bx$
If not evaluating the exact value, at least we can find that
$$\int_0^1 \frac{\ln(1+bx)}{1+x} dx \leq \int_0^1 \frac{bx}{1+x} dx = b(1-\ln 2)$$
$$\Phi(b)=\int\limits_0^1 \frac{\log(1+bx)}{x+1}dx$$
$$\Phi'(b)=\int\limits_0^1 \frac{x}{(bx+1)(x+1)}dx$$
$$\Phi'(b)=\int\limits_0^1 \frac{1}{bx+1}dx-\int\limits_0^1 \frac{1}{(x+1)(bx+1)}dx$$
$$\Phi'(b)=\frac{\log(b+1)}{b}-\frac{1}{1-b}\int\limits_0^1 \left(\frac{1}{x+1}- \frac{b}{bx+1}\right) dx$$
$$\Phi'(b)=\frac{\log(b+1)}{b}-\frac{\log2}{1-b}+\frac{\log(b+1)}{1-b}$$
With a different approach:
$$\Phi(b)=\int\limits_0^1 \frac{\log(1+bx)}{x+1}dx=\int\limits_0^1 \int\limits_0^{bx}\frac{dx \cdot dy}{(1+y)(1+x)}$$
Putting $y=bxu$ gives
$$\Phi(b)=\int\limits_0^1 \frac{\log(1+bx)}{x+1}dx=\int\limits_0^1 \int\limits_0^{1}\frac{b \cdot x dx \cdot du}{(1+bxu)(1+x)}$$
Some partial fraction decomposition and trickery will give you
$$\Phi(b)=\int\limits_0^b \frac{\log(1+x)}{x}dx + \log2 \log|b-1|+\int\limits_0^b\frac{\log(1+x)}{1-x}dx$$
Note that the first integral is a Dilogarithm and the latter is most probably one too (there is a comment saying Maple gives the answer in terms of te Dilog)
