Question

For $-1\le x\le1$, if f (x) is the sum of the convergent power series $x+\frac{x^2}{2^2}+\frac{x^3}{3^2}+...+\frac{x^n}{n^2}+...$ then $f(\frac{1}{2})$ is equal to

• A. $\displaystyle\int\limits^{\frac{1}{2}}_{0}\frac{In(1-t)}{t}dt.$
• B. $-\displaystyle\int\limits^{\frac{1}{2}}_{0}\frac{In(1-t)}{t}dt.$
• C. $\displaystyle\int\limits^{\frac{1}{2}}_{0}tIn(1+t)dt.$
• D. $\displaystyle\int\limits^{\frac{1}{2}}_{0}tIn(1-t)dt.$

Answer 1

Answer: (B)

$-\displaystyle\int\limits^{\frac{1}{2}}_{0}\frac{In(1-t)}{t}dt.$

Answer 2

The correct option is (B): $-\displaystyle\int\limits^{\frac{1}{2}}_{0}\frac{In(1-t)}{t}dt.$