# Prove that $\int_{0}^{1} \ln\left(\frac{1-a x}{1-a}\right) \frac{1}{\ln x} \mathrm{dx} = -\sum_{k=1}^{\infty} a^{k} \frac{\ln(1+k)}{k}, \space a<1$

Prove that $$\int_{0}^{1} \ln\left(\frac{1-a x}{1-a}\right) \frac{1}{\ln x} \mathrm{dx} = -\sum_{k=1}^{\infty} a^{k} \frac{\ln(1+k)}{k}, \space a<1$$ I find this question rather troublesome since the both sides seem hard to compute.
I appreciate any hint/suggestion. Thanks!

-

Develop series expansion of the logarithm of the quotient: $$\ln \left( \frac{1-a x}{1-a} \right) = \ln\left(1-a x\right) - \ln\left(1-a\right) = \sum_{k=1}^\infty \frac{1}{k} a^k \left( 1-x^k \right)$$ It is now down to evaluation of $$\begin{eqnarray} \int_0^1 \frac{1-x^k}{\ln(x)} \mathrm{d}x &\stackrel{x = \exp(-t)}{=}& -\int_0^\infty \mathrm{e}^{-t} \frac{1-\exp(-k t)}{t} \mathrm{d}t \\ &=& \int_0^{k} \frac{\mathrm{d}}{\mathrm{d}m } \left(-\int_0^\infty \mathrm{e}^{-t} \frac{1-\exp(-m t)}{t} \mathrm{d}t\right)\mathrm{d}m \\ &=& -\int_0^k \int_0^\infty \exp(-t(m+1)) \mathrm{d}m \\ &=& -\int_0^k \frac{\mathrm{d}m}{m+1} = -\log(k+1) \end{eqnarray}$$

-
nice answer! Since $\int_0^1 \frac{1-x^k}{\ln(x)} \mathrm{d}x$ is a Frullani integral, the answer is straightforward. –  Chris's sis the artist Dec 26 '12 at 23:46
Thanks. For those uninitiated, here is a link to Frullani's integral page on MathWorld. –  Sasha Dec 26 '12 at 23:48
isn't the Taylor form you used above available only for the case when $|ax|<1$ ? –  Chris's sis the artist Dec 27 '12 at 9:15
Yes, but $|a x| < |a| < 1$ –  Sasha Dec 27 '12 at 13:51
how about $a<-1$ ? –  Chris's sis the artist Dec 27 '12 at 13:54

Here's another way to handle the integral appearing in @Sasha's answer. $$\begin{eqnarray*} \int_0^1 dx \, \frac{1-x^k}{\log x} &=& \int_\infty^{0} d\alpha\, \frac{\partial}{\partial \alpha} \int_0^1 dx \, \frac{1-x^k}{\log x} x^\alpha \\ &=& \int_\infty^{0} d\alpha\, \int_0^1 dx \, (1-x^k)x^\alpha \\ &=& \int_\infty^{0} d\alpha\, \left(\frac{1}{\alpha+1} - \frac{1}{k+\alpha+1}\right) \\ &=& \left.\log\frac{\alpha+1}{k+\alpha+1}\right|_\infty^{0} \\ &=& -\log(k+1) \end{eqnarray*}$$

-
Nice solution (+1) –  Chris's sis the artist Dec 27 '12 at 8:40
@Chris'ssister: Thanks, another interesting question! (+1) –  user26872 Dec 27 '12 at 8:53