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. $$\lim_{n\rightarrow\infty}\int_0^1\frac{(\lambda_1+\lambda_2x)^n-x^n}{1-x}\,dx.$$

  • 2
    $\begingroup$ Write the integral as the sum of two integrals $A = \int_{0}^{1}\frac{1-x^{n}}{1-x}dx$ and $B=\int_{0}^{1}\frac{(1-\lambda_{2} (1-x))^{n}-1}{1-x}dx = \int_{0}^{1}\frac{(1-\lambda_{2} t)^{n}-1}{t}dt$. Now calculate explicitly $A = 1+\frac{1}{2}+...+\frac{1}{n}$. And the second integral $B = \sum_{k=0}^{n-1}\frac{(1-\lambda_{2})^{k+1}-1}{k+1}$. Then $A+B=\sum_{k=0}^{n-1}\frac{(1-\lambda_{2})^{k+1}}{k+1} = \ln(\lambda_{2})$ $\endgroup$ – Paata Ivanishvili Jul 9 at 3:24
  • $\begingroup$ Can that be right? Since $\lambda_{1} + \lambda_{2} x > x$, it seems to me the integrand is positive. But $\ln \lambda_{2} $ is negative. $\endgroup$ – Michael Engelhardt Jul 9 at 4:25
  • 1
    $\begingroup$ I am sorry, I forgot to put negative sign in front. It should be $-\ln(\lambda_{2})$. $\endgroup$ – Paata Ivanishvili Jul 9 at 4:31

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*}


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