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Subject: Seeking Help for a Computer Science Contest - Integral Estimation

Hello everyone,

I hope this message finds you well. I am currently preparing for an ongoing computer science contest, and I have encountered a challenging problem related to estimating integrals. The specific problem involves calculating the integral:

$ I_n = \int_0^1 \frac{x^n}{6 + x - x^2} \,dx $

The initial values are given as $ I_0 = \frac{2}{3}\ln\left(\frac{3}{2}\right) $ and $ I_1 = \frac{1}{5}\ln\left(\frac{3}{2}\right) $.

The task is to show that for $ n \geq 2 $, $ I_n $ can be expressed through the following recurrence relation:

$ I_n = \alpha I_{n-1} + \beta I_{n-2} + \gamma_n $

where $ \alpha + \beta $ are constants to be determined, and $ \gamma_n $ is a sequence that needs to be explicitly defined.

I've been struggling to prove this recurrence relation, and any guidance or assistance would be greatly appreciated. Additionally, I am curious to determine the values of $\alpha,\beta$, and the explicit form of $ \gamma_n \ $ for this recurrence relation.

Thank you in advance for your help!

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1 Answer 1

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There is a standard way to approach this sort of problems. Note that $$\begin{align} 6 I_n + I_{n + 1} - I_{n + 2} &= \int_{0}^1 \frac{6 x^n + x^{n + 1} - x^{n + 2}}{6 + x - x^2} dx \\ &= \int_0^1 \frac{x^n(6 + x - x^2)}{6 + x - x^2} dx\\ &= \int_0^1 x^n dx = \frac{1}{n + 1}.\end{align}$$ Thus we have that $I_{n + 2} = 6 I_n + I_{n + 1} - \frac{1}{n + 1},$ i.e. $\alpha = 1, \beta = 6 $ and $\gamma_{n+2} = \frac{1}{n+1}.$ I hope this helps. :)

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