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Question Number 74345 by mathmax by abdo last updated on 22/Nov/19
1) calculate f(x)=∫_(x+1) ^(x^2 +1)    e^(−xt) arctan(t)dt  2) find lim_(x→0)    f(x)
$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({x}\right)=\int_{{x}+\mathrm{1}} ^{{x}^{\mathrm{2}} +\mathrm{1}} \:\:\:{e}^{−{xt}} {arctan}\left({t}\right){dt} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:{f}\left({x}\right) \\ $$
Commented by mathmax by abdo last updated on 24/Nov/19
1)f(x)=∫_(x+1) ^(x^2 +1)  e^(−xt)  arctan(t)dt  ⇒f(x)=_(xt=u)  (1/x)∫_(x^2 +x) ^(x^3 +x)  e^(−u)  arctan((u/x))du  ⇒xf(x)=∫_(x^2 +x) ^(x^3 +x)  e^(−u)  arctan((u/x))dx =_(by parts)  [−e^(−u)  arctan((u/x))]_(x^2 +x) ^(x^3 +x)   +∫_(x^2 +x) ^(x^3 +x)  e^(−u)   ×(1/(x(1+(u^2 /x^2 ))))du  =e^(−(x^2 +x))  arctan(x+1)−e^(−(x^3 +x))  arctan(x^2 +1)  +x ∫_(x^2 +x) ^(x^3 +x)    (e^(−u) /(x^2 +u^2 ))du ⇒  f(x)=(1/x){ e^(−(x^2 +x))  arctan(x+1)−e^(−(x^3 +x))  arctan(x^2 +1)}  +∫_(x^2 +x) ^(x^3 +x)   (e^(−u) /(x^2  +u^2 ))du  ...be continued...
$$\left.\mathrm{1}\right){f}\left({x}\right)=\int_{{x}+\mathrm{1}} ^{{x}^{\mathrm{2}} +\mathrm{1}} \:{e}^{−{xt}} \:{arctan}\left({t}\right){dt}\:\:\Rightarrow{f}\left({x}\right)=_{{xt}={u}} \:\frac{\mathrm{1}}{{x}}\int_{{x}^{\mathrm{2}} +{x}} ^{{x}^{\mathrm{3}} +{x}} \:{e}^{−{u}} \:{arctan}\left(\frac{{u}}{{x}}\right){du} \\ $$$$\Rightarrow{xf}\left({x}\right)=\int_{{x}^{\mathrm{2}} +{x}} ^{{x}^{\mathrm{3}} +{x}} \:{e}^{−{u}} \:{arctan}\left(\frac{{u}}{{x}}\right){dx}\:=_{{by}\:{parts}} \:\left[−{e}^{−{u}} \:{arctan}\left(\frac{{u}}{{x}}\right)\right]_{{x}^{\mathrm{2}} +{x}} ^{{x}^{\mathrm{3}} +{x}} \\ $$$$+\int_{{x}^{\mathrm{2}} +{x}} ^{{x}^{\mathrm{3}} +{x}} \:{e}^{−{u}} \:\:×\frac{\mathrm{1}}{{x}\left(\mathrm{1}+\frac{{u}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)}{du} \\ $$$$={e}^{−\left({x}^{\mathrm{2}} +{x}\right)} \:{arctan}\left({x}+\mathrm{1}\right)−{e}^{−\left({x}^{\mathrm{3}} +{x}\right)} \:{arctan}\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$+{x}\:\int_{{x}^{\mathrm{2}} +{x}} ^{{x}^{\mathrm{3}} +{x}} \:\:\:\frac{{e}^{−{u}} }{{x}^{\mathrm{2}} +{u}^{\mathrm{2}} }{du}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{{x}}\left\{\:{e}^{−\left({x}^{\mathrm{2}} +{x}\right)} \:{arctan}\left({x}+\mathrm{1}\right)−{e}^{−\left({x}^{\mathrm{3}} +{x}\right)} \:{arctan}\left({x}^{\mathrm{2}} +\mathrm{1}\right)\right\} \\ $$$$+\int_{{x}^{\mathrm{2}} +{x}} ^{{x}^{\mathrm{3}} +{x}} \:\:\frac{{e}^{−{u}} }{{x}^{\mathrm{2}} \:+{u}^{\mathrm{2}} }{du}\:\:…{be}\:{continued}… \\ $$
Commented by mathmax by abdo last updated on 24/Nov/19
2) ∃ c_x ∈]x+1,x^2 +1[ /f(x)=arctanc_x  ∫_(x+1) ^(x^2 +1)  e^(−xt)  dt  =_(xt=u)     arctan(c_x ) ∫_(x^2 +x) ^(x^3  +x)  e^(−u) (du/x) =(1/x) arctan(c_x )[−e^(−u) ]_(x^2 +x) ^(x^3 +x)   =arctan(c_x )×((e^(−(x^2 +x)) −e^(−(x^3 +x)) )/x)  we have lim_(x→0) c_x =(π/4)  also we have e^(−(x^2 +x)) ∼1−(x^2 +x)  and e^(−(x^3 +x)) ∼1−(x^3 +x) ⇒  e^(−(x^2 +x)) −e^(−(x^3 +x)) =1−x^2 −x−1+x^3 +x =x^3 −x^2  ⇒  ((e^(−(x^2 +x)) −e^(−(x^3 +x)) )/x) ∼ x^2 −x →0  (x→0) ⇒lim_(x→0)    f(x)=0
$$\left.\mathrm{2}\left.\right)\:\exists\:{c}_{{x}} \in\right]{x}+\mathrm{1},{x}^{\mathrm{2}} +\mathrm{1}\left[\:/{f}\left({x}\right)={arctanc}_{{x}} \:\int_{{x}+\mathrm{1}} ^{{x}^{\mathrm{2}} +\mathrm{1}} \:{e}^{−{xt}} \:{dt}\right. \\ $$$$=_{{xt}={u}} \:\:\:\:{arctan}\left({c}_{{x}} \right)\:\int_{{x}^{\mathrm{2}} +{x}} ^{{x}^{\mathrm{3}} \:+{x}} \:{e}^{−{u}} \frac{{du}}{{x}}\:=\frac{\mathrm{1}}{{x}}\:{arctan}\left({c}_{{x}} \right)\left[−{e}^{−{u}} \right]_{{x}^{\mathrm{2}} +{x}} ^{{x}^{\mathrm{3}} +{x}} \\ $$$$={arctan}\left({c}_{{x}} \right)×\frac{{e}^{−\left({x}^{\mathrm{2}} +{x}\right)} −{e}^{−\left({x}^{\mathrm{3}} +{x}\right)} }{{x}} \\ $$$${we}\:{have}\:{lim}_{{x}\rightarrow\mathrm{0}} {c}_{{x}} =\frac{\pi}{\mathrm{4}}\:\:{also}\:{we}\:{have}\:{e}^{−\left({x}^{\mathrm{2}} +{x}\right)} \sim\mathrm{1}−\left({x}^{\mathrm{2}} +{x}\right) \\ $$$${and}\:{e}^{−\left({x}^{\mathrm{3}} +{x}\right)} \sim\mathrm{1}−\left({x}^{\mathrm{3}} +{x}\right)\:\Rightarrow \\ $$$${e}^{−\left({x}^{\mathrm{2}} +{x}\right)} −{e}^{−\left({x}^{\mathrm{3}} +{x}\right)} =\mathrm{1}−{x}^{\mathrm{2}} −{x}−\mathrm{1}+{x}^{\mathrm{3}} +{x}\:={x}^{\mathrm{3}} −{x}^{\mathrm{2}} \:\Rightarrow \\ $$$$\frac{{e}^{−\left({x}^{\mathrm{2}} +{x}\right)} −{e}^{−\left({x}^{\mathrm{3}} +{x}\right)} }{{x}}\:\sim\:{x}^{\mathrm{2}} −{x}\:\rightarrow\mathrm{0}\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:{f}\left({x}\right)=\mathrm{0} \\ $$

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