Question

Consider a function $f\left(x\right)$ and a series $S$ are defined as
$$f\left(x\right)=\underset{0}{\overset{x}{\int }}{x}^2\ln (1-x^2)dx$$ $$S=\frac{1}{1\cdot 5}+\frac{1}{2\cdot 7}+\frac{1}{3\cdot 9}+\cdots \infty .$$
​​​​​​​If $S$ can be expressed as $S=A-B\ln 2.$ Then

• A. $f\left(1\right)$ is not a finite number.
• B. $f\left(a\right)=kS$
  (where $a$ and $k$ are some positive real constant)
• C. $A=\frac{8}{9}$
• D. $B=\frac{2}{3}$

Answer 1

Answer: (C), (D)

The correct options are
C $A=\frac{8}{9}$
D $B=\frac{2}{3}$

We know that,
$\ln (1-x)=-[x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots \infty ]$
Putting $x\to x^2$,
$\ln (1-x^2)=-[x^2+\frac{x^4}{2}+\frac{x^6}{3}+\cdots \infty ]x^2\ln (1-x^2)=-[x^4+\frac{x^6}{2}+\frac{x^8}{3}+\cdots \infty ]\underset{0}{\overset{1}{\int }}{x}^2\ln (1-x^2)dx=-\underset{0}{\overset{1}{\int }}[x^4+\frac{x^6}{2}+\frac{x^8}{3}+\cdots \infty ]dx\Rightarrow f\left(1\right)=-S\cdots \left(1\right)$
$f\left(a\right)=kS\Rightarrow a=1,k=-1$

Now,
$f\left(1\right)=\underset{0}{\overset{1}{\int }}{x}^2\ln (1-x^2)dx\text{Using integration by parts}\Rightarrow f\left(1\right)={\left[\frac{x^3}{3}\ln \right(1-x^2\left)\right]}_0^1-\underset{0}{\overset{1}{\int }}\frac{x^3(-2x)}{3(1-x^2)}dx\Rightarrow f\left(1\right)={\left[\frac{x^3}{3}\ln \right(1-x^2\left)\right]}_0^1-\frac{2}{3}\underset{0}{\overset{1}{\int }}\frac{-x^4}{1-x^2}dx$
Let,
$I=\underset{0}{\overset{1}{\int }}\frac{-x^4}{1-x^2}dx\Rightarrow I=\underset{0}{\overset{1}{\int }}\frac{1-x^4-1}{1-x^2}dx\Rightarrow I=\underset{0}{\overset{1}{\int }}(1+x^2)dx-\underset{0}{\overset{1}{\int }}\frac{1}{1-x^2}dx\Rightarrow I=\frac{4}{3}-\frac{1}{2}\underset{0}{\overset{1}{\int }}\frac{1}{1-x}+\frac{1}{1+x}dx\Rightarrow I=\frac{4}{3}-\frac{1}{2}{[\ln |1+x|-\ln |1-x|]}_0^1$
Therefore,
$f\left(1\right)=-\frac{8}{9}+{\left[\frac{x^3}{3}\ln \right(1+x)+\frac{x^3}{3}\ln (1-x)}^{}{+\frac{1}{3}\ln (1+x)-\frac{1}{3}\ln (1-x)]}_0^1=-\frac{8}{9}+{\left[\frac{x^3+1}{3}\ln \right(1+x)+\frac{x^3-1}{3}\ln (1-x\left)\right]}_0^1=-\frac{8}{9}+\frac{2}{3}\ln 2+\underset{x\to 1}{lim}\frac{x^3-1}{3}\ln (1-x)$
Let,
$L=\underset{x\to 1}{lim}\frac{x^3-1}{3}\ln (1-x)\Rightarrow L=\underset{x\to 1}{lim}\frac{x^2+x+1}{3}\frac{\ln (1-x)}{\frac{1}{x-1}}\Rightarrow L=\underset{x\to 1}{lim}\frac{\ln (1-x)}{\frac{1}{x-1}}$
Using L'hospital's Rule,
$\Rightarrow L=\underset{x\to 1}{lim}\frac{-\frac{1}{1-x}}{-\frac{1}{(x-1{)}^2}}=0$
So,
$f\left(1\right)=-\frac{8}{9}+\frac{2}{3}\ln 2=-S\Rightarrow S=\frac{8}{9}-\frac{2}{3}\ln 2\Rightarrow S=\frac{8-6\ln 2}{9}\approx 0.5$
Hence, $f\left(1\right)$ is finite.
$A=\frac{8}{9},B=\frac{2}{3}$