Assume $\lambda\_1+\lambda\_2=1$ and both $\lambda\_1$ and $\lambda\_2$ are positive reals.



> 
> **QUESTION.** What is the value of this limit? It seems to exist.
>  $$\lim\_{n\rightarrow\infty}\int\_0^1\frac{(\lambda\_1+\lambda\_2x)^n-x^n}{1-x}\,dx.$$
> 
> 
> 



Limits of a family of integrals
The limit is $-\ln(\lambda\_{2})$.  

\begin{align\*}
\int\_{0}^{1}\frac{(\lambda\_{1}+\lambda\_{2}x)^{n}-x^{n}}{1-x}dx 
&= \int\_{0}^{1}\frac{(\lambda\_{1}+\lambda\_{2}x)^{n}-1}{1-x}dx + \int\_{0}^{1}\frac{1-x^{n}}{1-x}dx \\
&= \int\_{0}^{1}\frac{(1-\lambda\_{2}t)^{n}-1}{t}dt + \int\_{0}^{1}\sum\_{k=0}^{n-1} x^{k-1}dx \\
&=\int\_{0}^{1}-\lambda\_{2}\sum\_{k=0}^{n-1}(1-\lambda\_{2}t)^{k}dt + \sum\_{k=0}^{n-1}\frac{1}{k+1} \\
&=\sum\_{k=0}^{n-1}\frac{(1-\lambda\_{2})^{k+1}-1}{k+1} + \sum\_{k=0}^{n-1}\frac{1}{k+1} = \sum\_{k=0}^{n-1}\frac{(1-\lambda\_{2})^{k+1}}{k+1} \\
&\to -\ln(\lambda\_{2}).
\end{align\*}


