For x0, if f(x) = _1^x(_et)/(1+t),dt, then f(e)+f(1e) is equal to

Question

For $x>0$, if $f(x) = \displaystyle\int_1^x\frac{\log_et}{1+t}\,dt$, then $f(e)+f\left(\dfrac1e\right)$ is equal to

Accepted Answer

$$f\left(\frac1e\right) = \int_1^{1/e}\frac{\log_et}{1+t}\,dt$$

Substituting $t=1/u$: $f(1/e) = \displaystyle\int_1^e\frac{\log_e u}{u(u+1)}\,du$.

$$f(e)+f\left(\frac1e\right) = \int_1^e\left[\frac{\log_et}{1+t}+\frac{\log_et}{t(1+t)}\right]dt = \int_1^e\frac{\log_et}{t}\,dt$$

With $v=\log_et$: $\displaystyle\int_0^1 v\,dv = \dfrac12$.

Solutions · 1