Question

If $f\left(x\right)=\stackrel{x}{\underset{1}{\int }}\frac{logt}{t}dt$ find the value of $f\left(e\right)+f\left(e^{-1}\right)$

Answer 1

Given,

$f\left(x\right)=\stackrel{x}{\underset{1}{\int }}\frac{\log t}{t}dt$

or, $f\left(x\right)=\stackrel{x}{\underset{1}{\int }}\left(\log t\right)d\left(\log t\right)$

or, $f\left(x\right)=\frac{(\log x{)}^2}{2}$.

Now,

$f\left(e\right)+f\left(e^{-1}\right)$

$=\frac{(1{)}^2}{2}+\frac{(-1{)}^2}{2}$

$=1$.