.Let `alpha in(0,1)` and `beta=log_(e)(1-alpha)`. Let `P_(n)(x)=x+(x^(2))/(2)+(x^(3))/(3)+ . . . +(x^(n))/(n) , x in (0,1)`.
Then the integral `int_(0)^(alpha) (t^(50))/(1-t) dt` is equal to

Options

A. `P_(50)(alpha)-beta`
B. `-(beta+P_(50)(alpha))`
C. `(beta+P_(50)(alpha))`
D. `beta-P_(50)(alpha)`

Answer

To solve the integral \( I = \int_{0}^{\alpha} \frac{t^{50}}{1-t} dt \), we can follow these steps:

### Step 1: Rewrite the integrand
We can rewrite the integrand \( \frac{1}{1-t} \) using its Taylor series expansion:
\[ \frac{1}{1-t} = 1 + t + t^2 + t^3 + \ldots = \sum_{k=0}^{\infty} t^k \quad \text{for } |t| < 1 \]
Thus, we can express the integral as:
\[ I = \int_{0}^{\alpha} t^{50} \left( \sum_{k=0}^{\infty} t^k \right) dt \]

### Step 2: Interchange the sum and the integral
We can interchange the sum and the integral (justified by uniform convergence on the interval):
\[ I = \sum_{k=0}^{\infty} \int_{0}^{\alpha} t^{50+k} dt \]

### Step 3: Evaluate the integral
The integral \( \int_{0}^{\alpha} t^{50+k} dt \) can be evaluated as:
\[ \int_{0}^{\alpha} t^{50+k} dt = \frac{\alpha^{51+k}}{51+k} \]
Thus, we have:
\[ I = \sum_{k=0}^{\infty} \frac{\alpha^{51+k}}{51+k} \]

### Step 4: Recognize the series
The series \( \sum_{k=0}^{\infty} \frac{x^{k}}{k+a} \) can be expressed in terms of the logarithm and the function \( P_n(x) \). Specifically, we can relate it to:
\[ I = \alpha^{51} \sum_{k=0}^{\infty} \frac{\alpha^{k}}{51+k} \]
This series can be recognized as \( -\ln(1-\alpha) - P_{50}(\alpha) \) where \( P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \ldots + \frac{x^n}{n} \).

### Step 5: Substitute \( \beta \)
Given that \( \beta = \ln(1 - \alpha) \), we can rewrite the integral:
\[ I = -\beta - P_{50}(\alpha) \]

### Final Result
Thus, the final result for the integral is:
\[ I = -\ln(1 - \alpha) - P_{50}(\alpha) \]