Show that $I_{12}\ne-0.0189$

Question

For $n=1,2,3,\ldots$, let $$I_n=\int_0^1 \frac{x^{n-1}}{2-x} \,dx.$$ The value taken for $I_1=\ln2$ is $0.6931.$ If the recurrence relation is now used to calculate successive values of $I_n$ we find $I_{12}=-0.0189$ (you are not required to confirm the calculation). Explain carefully both why this cannot be correct and the source of the error, given that all intermediate arithmetical operations have been performed exactly.

I am not sure how to answer the question?

Answer 1

HINT:

For $x\ne2,$$$\frac{x^{n-1}}{2-x}=\frac{x^{n-1}-2x^{n-2}+2x^{n-2}}{2-x}=-2x^{n-2}+\frac{2x^{n-2}}{2-x}$$

Answer 2

Hint:

$$I_n= \int_{0}^1\frac{x^{n-1}}{2-x}dx= \frac{1}{2}\int_0^1 x^{n-1}\left(\frac{1}{1-\frac{x}{2}}\right)dx =\frac{1}{2}\int_0^1 x^{n-1}\left(1+\frac{x}{2}+\frac{x^2}{4}+..+\frac{x^n}{2^n}+...\right)dx $$

$$I_n = \frac{1}{2}\left[\frac{x^n}{n}+\frac{x^{n+1}}{2(n+1)}+...+ \frac{x^{2n}}{2n\cdot2^n}+...\right]^1_0$$

$$I_n= \frac{1}{2}\left(\frac{1}{n}+\frac{1}{2(n+1)}+...+\frac{1}{2n\cdot2^n}+...\right)$$

Answer 3

Another hint: Substituting first $t=1-x$ followed by $u=1+t$, and then expanding the numerator via the binomial theorem, the integral becomes:

$$\begin{align} I_{n} &=\int_{0}^{1}\frac{x^{n-1}}{2-x}\,\mathrm{d}x\\ &=\int_{0}^{1}\frac{(1-t)^{n-1}}{1+t}\,\mathrm{d}t\\ &=\int_{1}^{2}\frac{(2-u)^{n-1}}{u}\,\mathrm{d}u\\ &=\int_{1}^{2}\frac{1}{u}\sum_{k=0}^{n-1}\binom{n-1}{k}2^{n-1-k}(-1)^ku^{k}\,\mathrm{d}u\\ &=\sum_{k=0}^{n-1}\binom{n-1}{k}2^{n-1-k}(-1)^k\int_{1}^{2}u^{k-1}\,\mathrm{d}u \end{align}$$