If-f-x-1-x-log-t-1-t-dt-then-f-x-f-1-x-1-2-log-x-2-

Question Number 52948 by gunawan last updated on 15/Jan/19

$$\mathrm{If}\:\:\:{f}\left({x}\right)\:=\underset{\:\mathrm{1}} {\overset{{x}} {\int}}\:\frac{\mathrm{log}\:{t}}{\mathrm{1}+{t}}\:{dt},\:\mathrm{then} \\ $$$$\:{f}\left({x}\right)+{f}\:\left(\frac{\mathrm{1}}{{x}}\:\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{log}\:{x}\right)^{\mathrm{2}} \\ $$

Commented by maxmathsup by imad last updated on 15/Jan/19

let ϕ(x)=f(x)+f((1/x)) ⇒ϕ^′ (x)=f^′ (x)−(1/x^2 )f^′ ((1/x)) =((logx)/(1+x)) −(1/x^2 ) ((log((1/x)))/(1+(1/x))) =((logx)/(1+x)) +((logx)/(x^2 +x)) =((xlogx +logx)/(x^2 +x)) =(((x+1)logx)/(x(x+1))) =((logx)/x) from another side we have (d/dx)((1/2)(logx)^2 )=(1/2) ((2logx)/x) =((logx)/x) ⇒ ϕ(x)=(1/2)(logx)^2 +c but ϕ(1)=0=c ⇒ϕ(x)=(1/2)(logx)^2 ⇒ f(x)+f((1/x))=(1/2)(logx)^2 .

$${let}\:\varphi\left({x}\right)={f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{{x}}\right)\:\:\:\Rightarrow\varphi^{'} \left({x}\right)={f}^{'} \left({x}\right)−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{f}^{'} \left(\frac{\mathrm{1}}{{x}}\right) \\ $$$$=\frac{{logx}}{\mathrm{1}+{x}}\:−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\frac{{log}\left(\frac{\mathrm{1}}{{x}}\right)}{\mathrm{1}+\frac{\mathrm{1}}{{x}}}\:=\frac{{logx}}{\mathrm{1}+{x}}\:+\frac{{logx}}{{x}^{\mathrm{2}} \:+{x}}\:=\frac{{xlogx}\:+{logx}}{{x}^{\mathrm{2}} \:+{x}}\:=\frac{\left({x}+\mathrm{1}\right){logx}}{{x}\left({x}+\mathrm{1}\right)}\:=\frac{{logx}}{{x}} \\ $$$${from}\:{another}\:{side}\:{we}\:{have}\:\frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{\mathrm{2}}\left({logx}\right)^{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{2}{logx}}{{x}}\:=\frac{{logx}}{{x}}\:\Rightarrow \\ $$$$\varphi\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({logx}\right)^{\mathrm{2}} \:+{c}\:\:\:\:{but}\:\varphi\left(\mathrm{1}\right)=\mathrm{0}={c}\:\:\Rightarrow\varphi\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({logx}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$${f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({logx}\right)^{\mathrm{2}} \:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

f(x)=∫_1 ^x ((lnt)/(1+t))dt f((1/x))=∫_1 ^(1/x) ((lnt)/(1+t))dt k=(1/t) t=(1/k) dt=−(1/k^2 )dk ∫_1 ^x ((ln((1/k)))/(1+(1/k)))×((−dk)/k^2 ) ∫_1 ^x ((−lnk)/(1+k))×(k/(−k^2 ))dk f((1/x))=∫_1 ^x ((lnk)/(1+k))×(dk/k) f(x)+f((1/x))=∫_1 ^x ((lnt)/(1+t))+((lnt)/(1+t))×(dt/t) =∫_1 ^x ((lnt)/(1+t))×(1+(1/t))dt =∫_1 ^x ((lnt)/t) =∣(((lnt)^2 )/2)∣_1 ^x =(((lnx)^2 )/2) answer pls see below... I=∫((lnt)/t)dt =lnt×lnt−∫(1/t)×lntdt 2I=(lnt)^2 so I=(((lnt)^2 )/2)

$${f}\left({x}\right)=\int_{\mathrm{1}} ^{{x}} \frac{{lnt}}{\mathrm{1}+{t}}{dt} \\ $$$${f}\left(\frac{\mathrm{1}}{{x}}\right)=\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{{x}}} \frac{{lnt}}{\mathrm{1}+{t}}{dt} \\ $$$${k}=\frac{\mathrm{1}}{{t}}\:\:{t}=\frac{\mathrm{1}}{{k}}\:\:\:{dt}=−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }{dk} \\ $$$$\int_{\mathrm{1}} ^{{x}} \frac{{ln}\left(\frac{\mathrm{1}}{{k}}\right)}{\mathrm{1}+\frac{\mathrm{1}}{{k}}}×\frac{−{dk}}{{k}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{1}} ^{{x}} \frac{−{lnk}}{\mathrm{1}+{k}}×\frac{{k}}{−{k}^{\mathrm{2}} }{dk} \\ $$$${f}\left(\frac{\mathrm{1}}{{x}}\right)=\int_{\mathrm{1}} ^{{x}} \frac{{lnk}}{\mathrm{1}+{k}}×\frac{{dk}}{{k}} \\ $$$${f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{{x}}\right)=\int_{\mathrm{1}} ^{{x}} \frac{{lnt}}{\mathrm{1}+{t}}+\frac{{lnt}}{\mathrm{1}+{t}}×\frac{{dt}}{{t}} \\ $$$$=\int_{\mathrm{1}} ^{{x}} \frac{{lnt}}{\mathrm{1}+{t}}×\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right){dt} \\ $$$$=\int_{\mathrm{1}} ^{{x}} \frac{{lnt}}{{t}} \\ $$$$=\mid\frac{\left({lnt}\right)^{\mathrm{2}} }{\mathrm{2}}\mid_{\mathrm{1}} ^{{x}} \\ $$$$=\frac{\left({lnx}\right)^{\mathrm{2}} }{\mathrm{2}}\:\:{answer} \\ $$$${pls}\:{see}\:{below}… \\ $$$${I}=\int\frac{{lnt}}{{t}}{dt} \\ $$$$={lnt}×{lnt}−\int\frac{\mathrm{1}}{{t}}×{lntdt} \\ $$$$\mathrm{2}{I}=\left({lnt}\right)^{\mathrm{2}} \:\:\:\:{so}\:{I}=\frac{\left({lnt}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$

Commented by gunawan last updated on 16/Jan/19