Define $x_n=\int_{0}^{1}t^n(t+5)^{-1}dt$. Show that $x_0=\ln 1,2$ and that $x_n=n^{-1}-5x_{n-1}$ for $n=1,2,...$. Compute $x_0, x_1, ..., x_{10}$

Question

Define $x_n=\int_{0}^{1}t^n(t+5)^{-1}dt$. Show that $x_0=\ln 1.2$ and that $x_n=n^{-1}-5x_{n-1}$ for $n=1,2,...$. Compute $x_0, x_1, ..., x_{10}$ using this recurrence formula and estimate the accuaracy of $x_{10}$.

$x_0=\int_{0}^{1}(t+5)^{-1}dt=\ln (t+5)]_0^1=\ln(6)-\ln(5)=\ln 1. 2$

How can I prove that $x_n=n^{-1}-5x_{n-1}$ for $n=1,2,...$? Could I do it by induction?

How can I calculate $x_0$ with this recurrence formula if $0^{-1}$ and $x_{-1}$ are not defined and $x_0=0^{-1}-5x_{-1}$?

Answer 1

$\displaystyle \int_0^1 \frac{t^n}{t+5} dt =\int_0^1 \frac{t^n+5t^{n-1}}{t+5} dt - 5\int_0^1 \frac{t^{n-1}}{t+5} dt$. So, this recurrence formula holds.

$x_0$ can't be calculated like this, since the first term above does not make sense.(which is $\displaystyle \int_0^1 \frac 1t dt$)

Answer 2

\begin{align} x_n &=\int_{0}^{1}\frac{t^n}{(t+5)}dt\\ &= \int_0^1\frac{ t^{n-1} \cdot t}{t+5} dt\\ &= \int_0^1\frac{ t^{n-1} \cdot (t\color{blue}{+5-5)}}{t+5} dt\\ &= \int_0^1\frac{ t^{n-1} \cdot( t+\color{blue}{5})}{t+5}-\frac{\color{blue}{5} t^{n-1}}{t+5} dt\\ &= \int_0^1 t^{n-1} dt- 5 \int_0^1 t^{n-1} (t+5)^{-1} dt\\ &= \frac{1}{n} - 5 \underbrace{\int_0^1 t^{n-1} (t+5)^{-1} dt}_{=x_{n-1}}\\ &=n^{-1} - 5 x_{n-1}\\ \end{align}