Question

Let $\alpha \in(0,1)$ and $\beta=\log _e(1-\alpha)$ Let $P_n(x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\ldots+\frac{x^n}{n}, x \in(0,1)$ Then the integral $\int\limits_0^\alpha \frac{t^{50}}{1-t} d t$ is equal to

• A. $P_{50}(\alpha)-\beta$
• B. $-\left(\beta+P_{50}(\alpha)\right)$
• C. $\beta+P_{50}(a)$
• D. $\beta-P_{50}(\alpha)$

Hint:

When solving integrals involving logarithmic expressions, consider series expansions for functions like \( P_n(x) \) and integrate term-by-term.

Answer 1

The Correct Option is B

The correct answer is (B) : $-\left(\beta+P_{50}(\alpha)\right)$
$\underset{0}{\overset{\alpha }{\int }}\frac{t^{50}-1+1}{1-t}=-\underset{0}{\overset{\alpha }{\int }}\left(1+t+\dots ..+t^{49}\right)+\underset{0}{\overset{\alpha }{\int }}\frac{1}{1-t}dt$
$=-\left(\frac{{\alpha }^{50}}{50}+\frac{{\alpha }^{49}}{49}+\dots ..+\frac{{\alpha }^1}{1}\right)+{\left(\frac{\ln (1-f)}{-1}\right)}_0^{\alpha }$
$=-P_{50}(\alpha )-\ln (1-\alpha )$
$=-P_{50}(\alpha )-\beta$

Answer 2

Step 1: Start with the given integral: \[ \int_0^\alpha \frac{1}{1 - t} \, dt. \] This can be rewritten as: \[ \int_0^\alpha \frac{1}{1 - t} \, dt = -\int_0^\alpha \frac{d}{1 - t}. \] Step 2: Now, express the series expansion for \( P_n(x) \): \[ P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots + \frac{x^n}{n}. \] Step 3: After integrating the series term-by-term, we get: \[ -\int_0^\alpha \frac{d}{1 - t} = -P_{50}(\alpha) - \beta. \] Step 4: Hence, the value of the integral is: \[ \int_0^\alpha \frac{1}{1 - t} \, dt = -(\beta + P_{50}(\alpha)). \]