Question

For x > 0, let f(x) =${\int }_1^x\frac{lnt}{1+t}dt.$ Find the function
f (x) + f (1 / x) and show that f (e) + f (1/e) = 1/2, where
$lnt=lo{g}_{e}t.$

Answer 1

f(x) =${\int }_1^x\frac{lnt}{1+t}dt.$ for x > 0 [given]
Now, $f(1/x)={\int }_1^{1/x}\frac{lnt}{1+t}dt$
Put t = 1/u $\Rightarrow dt=(-1/u^2)du$
$\therefore f(1/x)={\int }_1^x\frac{ln(1/u}{1+1/u}.\frac{(-1)}{u^2}du$
$={\int }_1^x\frac{lnu}{u(u+1)}du={\int }_1^x\frac{lnt}{t(1+t)}dt$
Now, $f(x)+f\left(\frac{1}{x}\right)={\int }_1^x\frac{lnt}{(1+t)}dt+{\int }_1^x\frac{lnt}{t(1+t)}dt$
$={\int }_1^x\frac{(1+t)lnt}{t(1+t)}dt={\int }_1^x\frac{lnt}{t}dt=\frac{1}{2}[(lnt{)}^2{]}_1^x=\frac{1}{2}(lnx{)}^2$
Put x = e,
$f(e)+f\left(\frac{1}{e}\right)=\frac{1}{2}(lne{)}^2=\frac{1}{2}$ Hence proved