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Question Number 74017 by mathmax by abdo last updated on 17/Nov/19
let f(x)=∫_x ^(x^2 +3) e^(−xt) ln(1+e^(−xt) )dt with x>0 1) calculate f(x) 2)find lim_(x→+∞) f(x).
letf(x)=∫xx2+3e−xtln(1+e−xt)dtwithx>01)calculatef(x)2)findlimx→+∞f(x).
Commented by mathmax by abdo last updated on 17/Nov/19
1) calculate f^′ (x)
1)calculatef′(x)
Commented by mind 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
u=xt⇒du=xdtf(x)=1x∫x2x3+3xe−uln(1+e−u)du∫e−uln(1+e−u)du=−e−uln(1+e−u)−∫e−2u1+e−udu=−e−uln(1+e−u)−∫(e−u+1)(e−u−1)(1+e−u)−∫eueu+1du=−e−uln(1+e−u)+∫(−e−u+1)du−ln(eu+1)=−e−uln(1+e−u)+e−u+u−ln(eu+1)+cf(x)=1x{[−e−(x3+3x)ln(1+e−x3−3x)+e−(x3+3x)+(x3+3x)−ln(ex3+3x+1)]+e−x2ln(1+e−x2)−e−x2−x2+ln(ex2+1))]limf(x)=0
Commented by mathmax 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
1)wehavef(x)=∫xx2+3e−xtln(1+e−xt)dt(x>0)changementxt=ugivef(x)=1x∫x2x3+3xe−uln(1+e−u)du⇒xf(x)=∫x2x3+3xe−uln(1+e−u)dy=bypsrts[−e−uln(1+e−u)]x2x3+3x−∫x2x3+3x(−e−u)×−e−u1+e−udu=e−x2ln(1+e−x2)−e−(x3+3x)ln(1+e−(x3+3x))+∫x2x3+3xe−2u1+e−uduwehave∫x2x3+3xe−2u1+e−udu=∫x2x3+3x1e2u+eudu=eu=z∫ex2ex3+3x1z2+zdzz=∫ex2ex3+3xdzz2(z+1)decompositionofF(z)=1z2(z+1)F(z)=az+bz2+cz+1b=1,c=1⇒F(z)=az+1z2+1z+1limz→+∞zF(z)=0=a+c⇒a=−c=−1⇒F(z)=−1z+1z2+1z+1∫ex2ex3+3xdzz2(z+1)=[ln∣z+1z∣]ex2ex3+3x+[−1z]ex2ex3+3x=ln(1+ex3+3xex3+3x)−ln(1+ex2ex2)+(e−x2−e−(x3+3x))⇒f(x)=1x{e−x2ln(1+e−x2)−e−(x3+3x)ln(1+e−(x3+3x))ln(1+ex3+3xex3+3x)−ln(1+ex2ex2)+e−x2−e−(x3+3x)}2)limx→+∞f(x)=0

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