Question

For x > 0, let f(x) =$\int_1^x \frac{ln \, t}{1+t} dt .$ Find the function
f (x) + f (1 / x) and show that f (e) + f (1/e) = 1/2, where
$ln \, t =log_e t.$

Answer 1

f(x) =$\int_1^x \frac{ln \, t}{1+t} dt .$ for x > 0 $ \, \, \, \, \, \, \, \, \, \, \, \, $ [given]
Now, $ \, \, \, \, \, \, \, f(1/x)=\int_1^{1/x} \frac{ln \, t}{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{ln \, u}{u(u+1)}du =\int_1^x \frac{ln \, t}{t(1+t)}dt$
Now, $f(x)+f\bigg(\frac{1}{x}\bigg)=\int_1^x \frac{ln \, t}{(1+t)}dt +\int_1^x \frac{ln \, t}{t(1+t)}dt$
$=\int_1^x \frac{(1+t)ln t}{t(1+t)}dt =\int_1^x \frac{ln \, t}{t} dt=\frac{1}{2}[(ln \, t)^2]_1^x=\frac{1}{2}(ln \, x)^2$
Put x = e,
$ \, \, \, \, \, \, \, \, \, \, \, \, f(e)+f\bigg(\frac{1}{e}\bigg)=\frac{1}{2}(ln \, e)^2=\frac{1}{2}$ Hence proved