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Question Number 74017 by mathmax by abdo last updated on 17/Nov/19
$${let}\:{f}\left({x}\right)=\int_{{x}} ^{{x}^{\mathrm{2}} +\mathrm{3}} \:{e}^{−{xt}} \:{ln}\left(\mathrm{1}+{e}^{−{xt}} \right){dt}\:\:\:\:{with}\:{x}>\mathrm{0} \\ $$ $$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({x}\right) \\ $$ $$\left.\mathrm{2}\right){find}\:\:{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right). \\ $$
Commented bymathmax by abdo last updated on 17/Nov/19
$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{'} \left({x}\right) \\ $$
Commented bymind is power last updated on 17/Nov/19
![u=xt⇒du=xdt f(x)=(1/x)∫_x^2 ^(x^3 +3x) e^(−u) ln(1+e^(−u) )du ∫e^(−u) ln(1+e^(−u) )du =−e^(−u) ln(1+e^(−u) )−∫(e^(−2u) /(1+e^(−u) ))du =−e^(−u) ln(1+e^(−u) )−∫(((e^(−u) +1)(e^(−u) −1))/((1+e^(−u) )))−∫(e^u /(e^u +1))du =−e^(−u) ln(1+e^(−u) )+∫(−e^(−u) +1)du−ln(e^u +1) =−e^(−u) ln(1+e^(−u) )+e^(−u) +u−ln(e^u +1)+c f(x)=(1/x){[−e^(−(x^3 +3x)) ln(1+e^(−x^3 −3x) )+e^(−(x^3 +3x)) +(x^3 +3x)−ln(e^(x^3 +3x) +1)] +e^(−x^2 ) ln(1+e^(−x^2 ) )−e^(−x^2 ) −x^2 +ln(e^x^2 +1))] lim f(x)=0](Q74028.png)
$$\mathrm{u}=\mathrm{xt}\Rightarrow\mathrm{du}=\mathrm{xdt} \\ $$ $$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}}\int_{\mathrm{x}^{\mathrm{2}} } ^{\mathrm{x}^{\mathrm{3}} +\mathrm{3x}} \mathrm{e}^{−\mathrm{u}} \mathrm{ln}\left(\mathrm{1}+\mathrm{e}^{−\mathrm{u}} \right)\mathrm{du} \\ $$ $$\int\mathrm{e}^{−\mathrm{u}} \mathrm{ln}\left(\mathrm{1}+\mathrm{e}^{−\mathrm{u}} \right)\mathrm{du} \\ $$ $$=−\mathrm{e}^{−\mathrm{u}} \mathrm{ln}\left(\mathrm{1}+\mathrm{e}^{−\mathrm{u}} \right)−\int\frac{\mathrm{e}^{−\mathrm{2u}} }{\mathrm{1}+\mathrm{e}^{−\mathrm{u}} }\mathrm{du} \\ $$ $$=−\mathrm{e}^{−\mathrm{u}} \mathrm{ln}\left(\mathrm{1}+\mathrm{e}^{−\mathrm{u}} \right)−\int\frac{\left(\mathrm{e}^{−\mathrm{u}} +\mathrm{1}\right)\left(\mathrm{e}^{−\mathrm{u}} −\mathrm{1}\right)}{\left(\mathrm{1}+\mathrm{e}^{−\mathrm{u}} \right)}−\int\frac{\mathrm{e}^{\mathrm{u}} }{\mathrm{e}^{\mathrm{u}} +\mathrm{1}}\mathrm{du} \\ $$ $$=−\mathrm{e}^{−\mathrm{u}} \mathrm{ln}\left(\mathrm{1}+\mathrm{e}^{−\mathrm{u}} \right)+\int\left(−\mathrm{e}^{−\mathrm{u}} +\mathrm{1}\right)\mathrm{du}−\mathrm{ln}\left(\mathrm{e}^{\mathrm{u}} +\mathrm{1}\right) \\ $$ $$=−\mathrm{e}^{−\mathrm{u}} \mathrm{ln}\left(\mathrm{1}+\mathrm{e}^{−\mathrm{u}} \right)+\mathrm{e}^{−\mathrm{u}} +\mathrm{u}−\mathrm{ln}\left(\mathrm{e}^{\mathrm{u}} +\mathrm{1}\right)+\mathrm{c} \\ $$ $$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}}\left\{\left[−\mathrm{e}^{−\left(\mathrm{x}^{\mathrm{3}} +\mathrm{3x}\right)} \mathrm{ln}\left(\mathrm{1}+\mathrm{e}^{−\mathrm{x}^{\mathrm{3}} −\mathrm{3x}} \right)+\mathrm{e}^{−\left(\mathrm{x}^{\mathrm{3}} +\mathrm{3x}\right)} +\left(\mathrm{x}^{\mathrm{3}} +\mathrm{3x}\right)−\mathrm{ln}\left(\mathrm{e}^{\mathrm{x}^{\mathrm{3}} +\mathrm{3x}} +\mathrm{1}\right)\right]\right. \\ $$ $$\left.+\left.\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{ln}\left(\mathrm{1}+\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \right)−\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } −\mathrm{x}^{\mathrm{2}} +\mathrm{ln}\left(\mathrm{e}^{\mathrm{x}^{\mathrm{2}} } +\mathrm{1}\right)\right)\right] \\ $$ $$\mathrm{lim}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{0} \\ $$
Commented bymathmax by abdo last updated on 21/Nov/19
![1) we have f(x)=∫_x ^(x^2 +3) e^(−xt) ln(1+e^(−xt) )dt (x>0) changement xt=u give f(x)=(1/x)∫_x^2 ^(x^3 +3x) e^(−u) ln(1+e^(−u) )du ⇒ xf(x)=∫_x^2 ^(x^3 +3x) e^(−u) ln(1+e^(−u) )dy =_(bypsrts) [−e^(−u) ln(1+e^(−u) )]_x^2 ^(x^3 +3x) −∫_x^2 ^(x^3 +3x) (−e^(−u) )×((−e^(−u) )/(1+e^(−u) )) du =e^(−x^2 ) ln(1+e^(−x^2 ) ) −e^(−(x^3 +3x)) ln(1+e^(−(x^3 +3x)) ) +∫_x^2 ^(x^3 +3x) (e^(−2u) /(1+e^(−u) ))du we have ∫_x^2 ^(x^3 +3x) (e^(−2u) /(1+e^(−u) ))du =∫_x^2 ^(x^3 +3x) (1/(e^(2u) +e^u ))du =_(e^u =z) ∫_e^x^2 ^e^(x^3 +3x) (1/(z^2 +z))(dz/z) = ∫_e^x^2 ^e^(x^3 +3x) (dz/(z^2 (z+1))) decomposition of F(z)=(1/(z^2 (z+1))) F(z)=(a/z) +(b/z^2 ) +(c/(z+1)) b=1 , c=1 ⇒F(z)=(a/z) +(1/z^2 ) +(1/(z+1)) lim_(z→+∞) zF(z)=0=a+c ⇒a=−c=−1 ⇒F(z)=−(1/z)+(1/z^2 ) +(1/(z+1)) ∫_e^x^2 ^e^(x^3 +3x) (dz/(z^2 (z+1))) =[ln∣((z+1)/z)∣]_e^x^2 ^e^(x^3 +3x) +[−(1/z)]_e^x^2 ^e^(x^3 +3x) =ln(((1+e^(x^3 +3x) )/e^(x^3 +3x) ))−ln(((1+e^x^2 )/e^x^2 ))+(e^(−x^2 ) −e^(−(x^3 +3x)) ) ⇒ f(x)=(1/x){e^(−x^2 ) ln(1+e^(−x^2 ) )−e^(−(x^3 +3x)) ln(1+e^(−(x^3 +3x)) ) ln(((1+e^(x^3 +3x) )/e^(x^3 +3x) ))−ln(((1+e^x^2 )/e^x^2 )) +e^(−x^2 ) −e^(−(x^3 +3x)) } 2) lim_(x→+∞) f(x)=0](Q74316.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)=\int_{{x}} ^{{x}^{\mathrm{2}} +\mathrm{3}} \:{e}^{−{xt}} {ln}\left(\mathrm{1}+{e}^{−{xt}} \right){dt}\:\:\:\left({x}>\mathrm{0}\right)\:{changement} \\ $$ $${xt}={u}\:{give}\:{f}\left({x}\right)=\frac{\mathrm{1}}{{x}}\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{3}} +\mathrm{3}{x}} \:{e}^{−{u}} {ln}\left(\mathrm{1}+{e}^{−{u}} \right){du}\:\Rightarrow \\ $$ $${xf}\left({x}\right)=\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{3}} +\mathrm{3}{x}} \:{e}^{−{u}} {ln}\left(\mathrm{1}+{e}^{−{u}} \right){dy}\:=_{{bypsrts}} \:\:\left[−{e}^{−{u}} {ln}\left(\mathrm{1}+{e}^{−{u}} \right)\right]_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{3}} +\mathrm{3}{x}} \\ $$ $$−\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{3}} +\mathrm{3}{x}} \:\left(−{e}^{−{u}} \right)×\frac{−{e}^{−{u}} }{\mathrm{1}+{e}^{−{u}} }\:{du}\:={e}^{−{x}^{\mathrm{2}} } {ln}\left(\mathrm{1}+{e}^{−{x}^{\mathrm{2}} } \right) \\ $$ $$−{e}^{−\left({x}^{\mathrm{3}} +\mathrm{3}{x}\right)} {ln}\left(\mathrm{1}+{e}^{−\left({x}^{\mathrm{3}} +\mathrm{3}{x}\right)} \right)\:\:\:+\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{3}} +\mathrm{3}{x}} \:\:\frac{{e}^{−\mathrm{2}{u}} }{\mathrm{1}+{e}^{−{u}} }{du}\:\:{we}\:{have} \\ $$ $$\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{3}} +\mathrm{3}{x}} \:\:\frac{{e}^{−\mathrm{2}{u}} }{\mathrm{1}+{e}^{−{u}} }{du}\:=\int_{{x}^{\mathrm{2}} } ^{{x}^{\mathrm{3}} +\mathrm{3}{x}} \:\:\frac{\mathrm{1}}{{e}^{\mathrm{2}{u}} \:+{e}^{{u}} }{du}\:=_{{e}^{{u}} ={z}} \:\:\:\int_{{e}^{{x}^{\mathrm{2}} } } ^{{e}^{{x}^{\mathrm{3}} \:+\mathrm{3}{x}} } \:\frac{\mathrm{1}}{{z}^{\mathrm{2}} +{z}}\frac{{dz}}{{z}} \\ $$ $$=\:\int_{{e}^{{x}^{\mathrm{2}} } } ^{{e}^{{x}^{\mathrm{3}} +\mathrm{3}{x}} } \:\:\:\frac{{dz}}{{z}^{\mathrm{2}} \left({z}+\mathrm{1}\right)}\:\:{decomposition}\:{of}\:{F}\left({z}\right)=\frac{\mathrm{1}}{{z}^{\mathrm{2}} \left({z}+\mathrm{1}\right)} \\ $$ $${F}\left({z}\right)=\frac{{a}}{{z}}\:+\frac{{b}}{{z}^{\mathrm{2}} }\:+\frac{{c}}{{z}+\mathrm{1}} \\ $$ $${b}=\mathrm{1}\:\:\:\:,\:\:{c}=\mathrm{1}\:\Rightarrow{F}\left({z}\right)=\frac{{a}}{{z}}\:+\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{z}+\mathrm{1}} \\ $$ $${lim}_{{z}\rightarrow+\infty} {zF}\left({z}\right)=\mathrm{0}={a}+{c}\:\:\Rightarrow{a}=−{c}=−\mathrm{1}\:\Rightarrow{F}\left({z}\right)=−\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{z}+\mathrm{1}} \\ $$ $$\int_{{e}^{{x}^{\mathrm{2}} } } ^{{e}^{{x}^{\mathrm{3}} +\mathrm{3}{x}} } \:\:\:\:\frac{{dz}}{{z}^{\mathrm{2}} \left({z}+\mathrm{1}\right)}\:=\left[{ln}\mid\frac{{z}+\mathrm{1}}{{z}}\mid\right]_{{e}^{{x}^{\mathrm{2}} } } ^{{e}^{{x}^{\mathrm{3}} +\mathrm{3}{x}} } \:+\left[−\frac{\mathrm{1}}{{z}}\right]_{{e}^{{x}^{\mathrm{2}} } } ^{{e}^{{x}^{\mathrm{3}} +\mathrm{3}{x}} } \\ $$ $$={ln}\left(\frac{\mathrm{1}+{e}^{{x}^{\mathrm{3}} +\mathrm{3}{x}} }{{e}^{{x}^{\mathrm{3}} +\mathrm{3}{x}} }\right)−{ln}\left(\frac{\mathrm{1}+{e}^{{x}^{\mathrm{2}} } }{{e}^{{x}^{\mathrm{2}} } }\right)+\left({e}^{−{x}^{\mathrm{2}} } −{e}^{−\left({x}^{\mathrm{3}} +\mathrm{3}{x}\right)} \right)\:\Rightarrow \\ $$ $${f}\left({x}\right)=\frac{\mathrm{1}}{{x}}\left\{{e}^{−{x}^{\mathrm{2}} } {ln}\left(\mathrm{1}+{e}^{−{x}^{\mathrm{2}} } \right)−{e}^{−\left({x}^{\mathrm{3}} +\mathrm{3}{x}\right)} {ln}\left(\mathrm{1}+{e}^{−\left({x}^{\mathrm{3}} +\mathrm{3}{x}\right)} \right)\right. \\ $$ $$\left.{ln}\left(\frac{\mathrm{1}+{e}^{{x}^{\mathrm{3}} +\mathrm{3}{x}} }{{e}^{{x}^{\mathrm{3}} +\mathrm{3}{x}} }\right)−{ln}\left(\frac{\mathrm{1}+{e}^{{x}^{\mathrm{2}} } }{{e}^{{x}^{\mathrm{2}} } }\right)\:+{e}^{−{x}^{\mathrm{2}} } −{e}^{−\left({x}^{\mathrm{3}} +\mathrm{3}{x}\right)} \right\} \\ $$ $$\left.\mathrm{2}\right)\:{lim}_{{x}\rightarrow+\infty} \:\:\:\:{f}\left({x}\right)=\mathrm{0} \\ $$