Let (0,1) and =_e(1-).

Question

Let $\alpha\in(0,1)$ and $\beta=\log_e(1-\alpha)$. Let $P_n(x) = x+\dfrac{x^2}2+\dfrac{x^3}3+\cdots+\dfrac{x^n}n$, $x\in(0,1)$. Then the integral $\displaystyle\int_0^\alpha\frac{t^{50}}{1-t}\,dt$ is equal to:

Accepted Answer

$\int_{0}^{\alpha} \frac{t^{50}}{1-t} d t$

$$
\begin{aligned}
& = \int_{0}^{\alpha} \frac{t^{50}-1+1}{1-t}\\
& =-\int_{0}^{\alpha}\left(\frac{1-t^{50}}{1-t}-\frac{1}{1-t}\right) d t\\
& =-\int_{0}^{\alpha}\left(1+t+\ldots . .+t^{49}\right)+\int_{0}^{\alpha} \frac{1}{1-t} d t
\end{aligned}
$$

$=-\left(\frac{\alpha^{50}}{50}+\frac{\alpha^{49}}{49}+\ldots . .+\frac{\alpha^{1}}{1}\right)+\left(\frac{\ln (1-\mathrm{f})}{-1}\right)_{0}^{\alpha}$

$=-\mathrm{P}_{50}(\alpha)-\ln (1-\alpha)$

$=-\mathrm{P}_{50}(\alpha)-\beta$

Solutions · 1