Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 6 down vote accepted

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 $$ \begin{eqnarray} \int_0^1 \frac{1-x^k}{\ln(x)} \mathrm{d}x &\stackrel{x = \exp(-t)}{=}& -\int_0^\infty \mathrm{e}^{-t} \frac{1-\exp(-k t)}{t} \mathrm{d}t \\ &=& \int_0^{k} \frac{\mathrm{d}}{\mathrm{d}m } \left(-\int_0^\infty \mathrm{e}^{-t} \frac{1-\exp(-m t)}{t} \mathrm{d}t\right)\mathrm{d}m \\ &=& -\int_0^k \int_0^\infty \exp(-t(m+1)) \mathrm{d}m \\ &=& -\int_0^k \frac{\mathrm{d}m}{m+1} = -\log(k+1) \end{eqnarray} $$

share|cite|improve this answer
2  
nice answer! Since $\int_0^1 \frac{1-x^k}{\ln(x)} \mathrm{d}x$ is a Frullani integral, the answer is straightforward. – I'm an artist Dec 26 '12 at 23:46
1  
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 $|ax|<1$ ? – I'm an artist Dec 27 '12 at 9:15
    
Yes, but $|a x| < |a| < 1$ – Sasha Dec 27 '12 at 13:51
    
how about $a<-1$ ? – I'm an artist Dec 27 '12 at 13:54

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

share|cite|improve this answer
    
Nice solution (+1) – I'm an artist Dec 27 '12 at 8:40
    
@Chris'ssister: Thanks, another interesting question! (+1) – user26872 Dec 27 '12 at 8:53

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.