For x 0, if [f(x) = _1^x _et (1 + t)dt ], then [f(e) + f( 1 e )] is equal to :

Question

For x > 0, if $$f(x) = \int\limits_1^x {{{{{\log }_e}t} \over {(1 + t)}}dt} $$, then $$f(e) + f\left( {{1 \over e}} \right)$$ is equal to :

Accepted Answer

$$f(x) = \int_1^x {{{\ln t} \over {1 + t}}dt} $$

then $$f\left( {{1 \over x}} \right) = \int_1^{1/x} {{{\ln t} \over {1 + t}}dt} $$

Let $$t = {1 \over u} \Rightarrow dt =  - {1 \over {{u^2}}}du$$

$$ \Rightarrow f\left( {{1 \over x}} \right) = \int_1^x {{{\ln {1 \over u}} \over {1 + {1 \over u}}}\left( { - {1 \over {{u^2}}}} \right)dx} $$

$$f\left( {{1 \over x}} \right) = \int_1^x {{{\ln u} \over {u(1 + u)}}} du = \int_1^x {{{\ln t} \over {t(1 + t)}}dt} $$

$$ \therefore $$ $$f(x) + f\left( {{1 \over x}} \right) = \int_1^x {\ln t\left( {{1 \over {1 + t}} + {1 \over {t(1 + t)}}} \right)} dt$$

$$ = \int_1^x {\ln t\left( {{1 \over {1 + t}} + {1 \over t} - {1 \over {t + 1}}} \right)} dt$$

$$ = \int_1^x {{{\ln t} \over t}dt = {1 \over 2}{{(\ln x)}^2}} $$

$$ \therefore $$ $$f(e) + f\left( {{1 \over e}} \right) = {1 \over 2}{(\ln e)^2} = {1 \over 2}$$

Solutions · 1