Key Concepts and Formulas

• Algebraic Manipulation for Integration: Techniques to rewrite integrands, such as adding and subtracting terms, to simplify integration.
• Geometric Series Sum: The formula for the sum of a finite geometric series is $1 + x + x^2 + \dots + x^{n-1} = \frac{1-x^n}{1-x}$ for $x \ne 1$.
• Integral of $\frac{1}{1-x}$: The integral is $-\ln|1-x| + C$.
• Integral of Polynomial Terms: $\int x^k dx = \frac{x^{k+1}}{k+1} + C$.
• Fundamental Theorem of Calculus: Used to evaluate definite integrals: $\int_a^b f(x) dx = F(b) - F(a)$, where $F'(x) = f(x)$.
• Logarithm Properties: $\ln(1) = 0$.

Step-by-Step Solution

We are asked to evaluate the integral $I = \int\limits_0^\alpha {{{{t^{50}}} \over {1 - t}}dt}$.

Step 1: Algebraic Manipulation of the Integrand The integrand is $\frac{t^{50}}{1-t}$. To simplify this rational function, we can add and subtract 1 in the numerator. This allows us to create a term that is divisible by $(1-t)$, leveraging the geometric series formula. $I = \int\limits_0^\alpha {\frac{{{t^{50}} - 1 + 1}}{{1 - t}}dt}$ Explanation: By adding and subtracting 1, we aim to express the numerator in a form that, when combined with the denominator, leads to simpler terms. Specifically, $t^{50}-1$ is related to the sum of a geometric series.

Step 2: Splitting the Integrand We split the fraction into two parts: $I = \int\limits_0^\alpha {\left( {\frac{{{t^{50}} - 1}}{{1 - t}} + \frac{1}{{1 - t}}} \right)dt}$ To match the standard form of the geometric series sum formula, we rewrite $\frac{t^{50}-1}{1-t}$ as $-\frac{1-t^{50}}{1-t}$. $I = \int\limits_0^\alpha {\left( { - \frac{{1 - {t^{50}}}}{{1 - t}} + \frac{1}{{1 - t}}} \right)dt}$ Now, we separate the integral into two distinct integrals: $I = - \int\limits_0^\alpha {\frac{{1 - {t^{50}}}}{{1 - t}}dt} + \int\limits_0^\alpha {\frac{1}{{1 - t}}dt}$ Explanation: This step decomposes the original integral into two parts. The first part involves the sum of a geometric series, and the second part is a standard logarithmic integral. The negative sign is introduced by factoring out $-1$ from $(t^{50}-1)$.

Step 3: Applying the Geometric Series Formula We use the formula for the sum of a finite geometric progression: $1 + x + x^2 + \dots + x^{n-1} = \frac{1-x^n}{1-x}$. Here, $x=t$ and $n=50$, so $\frac{1-t^{50}}{1-t} = 1 + t + t^2 + \dots + t^{49}$. Substituting this into the first integral: $I = - \int\limits_0^\alpha {\left( {1 + t + {t^2} + \dots + {t^{49}}} \right)dt} + \int\limits_0^\alpha {\frac{1}{{1 - t}}dt}$ Explanation: We have replaced the rational function term with its equivalent polynomial expansion, which is straightforward to integrate.

Step 4: Evaluating the Integrals We now evaluate each integral.

For the first integral: $\int\limits_0^\alpha {\left( {1 + t + {t^2} + \dots + {t^{49}}} \right)dt} = \left[ {t + \frac{{{t^2}}}{2} + \frac{{{t^3}}}{3} + \dots + \frac{{{t^{50}}}}{{50}}} \right]_0^\alpha$ Applying the limits of integration: $= \left( {\alpha + \frac{{{\alpha ^2}}}{2} + \frac{{{\alpha ^3}}}{3} + \dots + \frac{{{\alpha ^{50}}}}{{50}}} \right) - \left( {0 + \frac{{{0^2}}}{2} + \dots + \frac{{{0^{50}}}}{{50}}} \right)$ $= \alpha + \frac{{{\alpha ^2}}}{2} + \frac{{{\alpha ^3}}}{3} + \dots + \frac{{{\alpha ^{50}}}}{{50}}$ This sum is precisely the definition of $P_{50}(\alpha)$.

For the second integral: $\int\limits_0^\alpha {\frac{1}{{1 - t}}dt} = \left[ { - \ln |1 - t|} \right]_0^\alpha$ Applying the limits of integration: $= \left( { - \ln |1 - \alpha |} \right) - \left( { - \ln |1 - 0|} \right)$ Given that $\alpha \in (0,1)$, $1-\alpha > 0$, so $|1-\alpha| = 1-\alpha$. Also, $\ln(1) = 0$. $= - \ln(1 - \alpha) - ( - \ln(1)) = - \ln(1 - \alpha) - 0 = - \ln(1 - \alpha)$ Explanation: We have evaluated both integrals using the power rule for integration and the integral of $\frac{1}{1-t}$. The definitions of $P_{50}(\alpha)$ and $\beta$ are now ready to be applied.

Step 5: Combining Results and Substituting Definitions Substitute the evaluated integrals back into the expression for $I$: $I = - \left( {\alpha + \frac{{{\alpha ^2}}}{2} + \frac{{{\alpha ^3}}}{3} + \dots + \frac{{{\alpha ^{50}}}}{{50}}} \right) + \left( { - \ln(1 - \alpha )} \right)$ Using the given definition $P_{50}(\alpha) = \alpha + \frac{{\alpha ^2}}{2} + \frac{{\alpha ^3}}{3} + \dots + \frac{{\alpha ^{50}}}{50}$: $I = - P_{50}(\alpha) - \ln(1 - \alpha)$ Finally, substitute the given definition $\beta = {\log _e}(1 - \alpha ) = \ln(1-\alpha)$: $I = - P_{50}(\alpha) - \beta$ This can be rewritten as: $I = - (\beta + P_{50}(\alpha))$ Explanation: We have combined the results of the two integrals and substituted the given definitions for $P_{50}(\alpha)$ and $\beta$ to arrive at the final answer in the required format.

Common Mistakes and Tips

• Sign Errors: Be meticulous with signs when manipulating fractions and integrating terms like $\frac{1}{1-t}$.
• Geometric Series Formula: Ensure the correct formula and exponent ($n$ for the numerator, $n-1$ for the last term of the sum) are used.
• Logarithm Domain: Remember that for $\ln|1-t|$, the absolute value is important, but for the definite integral, the sign of $1-t$ within the limits of integration determines the evaluation. Here, $1-\alpha > 0$.
• Recognizing $P_n(x)$: The definition of $P_n(x)$ is key to simplifying the final expression.

Summary

The integral was evaluated by first manipulating the integrand algebraically to express it as a sum of a geometric series and a simple logarithmic term. Both parts were then integrated separately. The results were combined, and the given definitions of $P_{50}(\alpha)$ and $\beta$ were substituted to obtain the final answer. This problem demonstrates the power of algebraic simplification in making complex integrals tractable.

The final answer is $\boxed{-(\beta + P_{50}(\alpha))}$. This corresponds to option (A).