Let F(x) = f(x)+f(1x), where f(x) = _1^x t1+t,dt.

Question

Let $F(x) = f(x)+f\left(\dfrac1x\right)$, where $f(x) = \displaystyle\int_1^x \dfrac{\log t}{1+t}\,dt$. Then $F(e)$ equals:

Accepted Answer

$F(e) = f(e)+f\left(\dfrac1e\right) = \int_1^e \dfrac{\log t}{1+t}\,dt + \int_1^{1/e} \dfrac{\log t}{1+t}\,dt$ (1)

Let $I = \int_1^{1/e} \dfrac{\log t}{1+t}\,dt$. Put $\dfrac1t=z \Rightarrow dt=-\dfrac{dz}{z^2}$; when $t=1,z=1$; when $t=\dfrac1e,z=e$

$I = \int_1^e \dfrac{\log z}{z(z+1)}\,dz$

From (1): $F(e) = \int_1^e \dfrac{\log t}{1+t}\,dt+\int_1^e \dfrac{\log t}{t(1+t)}\,dt = \int_1^e \dfrac{\log t\,(t+1)}{t(1+t)}\,dt = \int_1^e \dfrac{\log t}{t}\,dt$

Let $\log t = x \Rightarrow \dfrac1t\,dt=dx$; limits $t=1\Rightarrow x=0$, $t=e\Rightarrow x=1$

$F(e) = \int_0^1 x\,dx = \left[\dfrac{x^2}{2}\right]_0^1 = \dfrac12$

Solutions · 1