If f() = _1^_10t1+t,dt, 0, then f(e^3)+f(e^-3) is equal to:

Question

If $f(\alpha) = \displaystyle\int_1^\alpha\frac{\log_{10}t}{1+t}\,dt$, $\alpha>0$, then $f(e^3)+f(e^{-3})$ is equal to:

Accepted Answer

$$f(\alpha ) = \int_1^\alpha  {{{{{\log }_{10}}t} \over {1 + t}}dt} $$ ...... (i)

$$f\left( {{1 \over \alpha }} \right) = \int_1^{{1 \over \alpha }} {{{{{\log }_{10}}t} \over {1 + t}}dt} $$

Substituting $$t \to {1 \over p}$$

$$f\left( {{1 \over \alpha }} \right) = \int_1^\alpha  {{{{{\log }_{10}}\left( {{1 \over p}} \right)} \over {1 + {1 \over p}}}\left( {{{ - 1} \over {{p^2}}}} \right)dp} $$

$$ = \int_1^\alpha  {{{{{\log }_{10}}p} \over {p(p + 1)}}dp = \int_1^\alpha  {\left( {{{{{\log }_{10}}t} \over t} - {{{{\log }_{10}}t} \over {t + 1}}} \right)dt} } $$ ....... (ii)

By (i) + (ii)

$$f(\alpha ) + f\left( {{1 \over \alpha }} \right) = \int_1^\alpha  {{{{{\log }_{10}}t} \over t}dt = \int_1^\alpha  {{{\ln t} \over t}\,.\,{{\log }_{10}}e\,dt} } $$

$$ = {{{{(\ln \alpha )}^2}} \over {2{{\log }_e}10}}$$

$$\alpha  = {e^3} \Rightarrow f({e^3}) + f({e^{ - 3}}) = {9 \over {2{{\log }_e}10}}$$

Solutions · 1