Menu Close

let-f-x-0-1-ln-1-xt-2-t-2-dt-with-x-R-1-find-a-explicit-form-of-f-x-2-calculate-0-1-ln-1-t-2-t-2-dt-3-calculate-0-1-ln-1-2t-2-t-2-dt-4-calculate-0-1-ln-1-t-2

Tinku Tara 0 Comments
Pin




Question Number 44201 by abdo.msup.com last updated on 23/Sep/18
let f(x)=∫_0 ^1 ((ln(1+xt^2 ))/t^2 )dt with x ∈R 1) find a explicit form of f(x) 2)calculate ∫_0 ^1 ((ln(1+t^2 ))/t^2 )dt 3)calculate ∫_0 ^1 ((ln(1+2t^2 ))/t^2 )dt 4) calculate ∫_0 ^1 ((ln(1−t^2 ))/t^2 )dt
$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{xt}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }{dt}\:\:{with}\:{x}\:\in{R} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right){calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }{dt} \\ $$
Commented by maxmathsup by imad last updated on 25/Sep/18
1) we have f^′ (x) = ∫_0 ^1 (t^2 /(t^2 (1+xt^2 )))dt = ∫_0 ^1 (dt/(1 +xt^2 )) case 1 x>0 changement t(√x)=u give f^′ (x)= ∫_0 ^(√x) (1/(1+u^2 )) (du/( (√x))) =(1/( (√x))) arctan((√x)) ⇒ f(x) =∫_0 ^x ((arctan((√t)))/( (√t))) dt +c but c=f(0)=0 ⇒ f(x) =∫_0 ^x ((actan((√t)))/( (√t)))dt =_((√t)=u) ∫_0 ^(√x) ((arctan(u))/u) (2u)du =2 ∫_0 ^(√x) arctanu du by parts f(x) =2{ [u arctanu]_0 ^(√x) −∫_0 ^(√x) (u/(1+u^2 ))du} =2{(√x)arctan((√x)) −[(1/2)ln(1+u^2 )]_0 ^(√x) } =2{ (√x)arctan((√x)) −(1/2)ln(1+x)} ⇒f(x)=2(√x)arctan((√x))−ln(1+x) case 2 x<0 ⇒f^′ (x)= ∫_0 ^1 (dt/(1−(−x)t^2 )) =_(t(√(−x))=u) ∫_0 ^(√(−x)) (1/(1−u^2 )) (du/( (√(−x)))) =(1/(2(√(−x)))){ ∫_0 ^(√(−x)) { (1/(1−u)) +(1/(1+u))}du} = (1/(2(√(−x))))[ln∣((1+u)/(1−u))∣]_0 ^(√(−x)) ⇒ f^′ (x) = (1/(2(√(−x))))ln∣((1+(√(−x)))/(1−(√(−x))))∣ ⇒f(x) = ∫_x ^0 (1/(2(√(−t))))ln∣((1+(√(−t)))/(1−(√(−t))))∣ +c c=f(0) =0 ⇒f(x)= ∫_x ^0 (1/(2(√(−t))))ln∣((1+(√(−t)))/(1−(√(−t))))∣ dt changement (√(−t))=u give f(x) = ∫_(√(−x)) ^0 (1/(2u))ln∣((1+u)/(1−u))∣(−2u)du = ∫_0 ^(√(−x)) ln∣1+u∣du−∫_0 ^(√(−x)) ln∣1−u∣ du =∫_0 ^(√(−x)) ln(1+u)du −∫_0 ^(√(−x)) ln(1−u)du but ∫_0 ^(√(−x)) ln(1+u)du =_(1+u =α) ∫_1 ^(1+(√(−x))) ln(α)dα [αln(α)−α]_1 ^(1+(√(−x))) =(1+(√(−x)))ln(1+(√(−x)))−(1+(√(−x))) +1 =(1+(√(−x)))ln(1+(√(−x))) −(√(−x)).also ∫_0 ^(√(−x)) ln(1−u)du =_(1−u =α) −∫_1 ^(1−(√(−x))) ln(α)dα =−[αln(α) −α]_1 ^(1−(√(−x))) =−{(1−(√(−x)))ln(1−(√(−x)))−(1−(√(−x))) +1} =−(1−(√(−x)))ln(1−(√(−x))) −(√(−x)) ⇒ f(x)=(1+(√(−x)))ln(1+(√(−x))) +(1−(√(−x)))ln(1−(√(−x))).
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} \left(\mathrm{1}+{xt}^{\mathrm{2}} \right)}{dt}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\mathrm{1}\:+{xt}^{\mathrm{2}} } \\ $$$${case}\:\mathrm{1}\:\:{x}>\mathrm{0}\:{changement}\:{t}\sqrt{{x}}={u}\:{give}\:\:{f}^{'} \left({x}\right)=\:\int_{\mathrm{0}} ^{\sqrt{{x}}} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{{du}}{\:\sqrt{{x}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{x}}}\:{arctan}\left(\sqrt{{x}}\right)\:\Rightarrow\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{arctan}\left(\sqrt{{t}}\right)}{\:\sqrt{{t}}}\:{dt}\:+{c}\:{but}\:{c}={f}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\int_{\mathrm{0}} ^{{x}} \:\:\:\frac{{actan}\left(\sqrt{{t}}\right)}{\:\sqrt{{t}}}{dt}\:=_{\sqrt{{t}}={u}} \:\:\:\:\int_{\mathrm{0}} ^{\sqrt{{x}}} \:\:\:\frac{{arctan}\left({u}\right)}{{u}}\:\left(\mathrm{2}{u}\right){du}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\sqrt{{x}}} \:{arctanu}\:{du} \\ $$$${by}\:{parts}\:{f}\left({x}\right)\:=\mathrm{2}\left\{\:\:\left[{u}\:{arctanu}\right]_{\mathrm{0}} ^{\sqrt{{x}}} \:−\int_{\mathrm{0}} ^{\sqrt{{x}}} \frac{{u}}{\mathrm{1}+{u}^{\mathrm{2}} }{du}\right\} \\ $$$$=\mathrm{2}\left\{\sqrt{{x}}{arctan}\left(\sqrt{{x}}\right)\:−\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\sqrt{{x}}} \right\} \\ $$$$=\mathrm{2}\left\{\:\sqrt{{x}}{arctan}\left(\sqrt{{x}}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}\right)\right\}\:\Rightarrow{f}\left({x}\right)=\mathrm{2}\sqrt{{x}}{arctan}\left(\sqrt{{x}}\right)−{ln}\left(\mathrm{1}+{x}\right) \\ $$$${case}\:\mathrm{2}\:{x}<\mathrm{0}\:\Rightarrow{f}^{'} \left({x}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}−\left(−{x}\right){t}^{\mathrm{2}} }\:=_{{t}\sqrt{−{x}}={u}} \:\:\:\int_{\mathrm{0}} ^{\sqrt{−{x}}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}−{u}^{\mathrm{2}} }\:\frac{{du}}{\:\sqrt{−{x}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{−{x}}}\left\{\:\int_{\mathrm{0}} ^{\sqrt{−{x}}} \:\left\{\:\frac{\mathrm{1}}{\mathrm{1}−{u}}\:+\frac{\mathrm{1}}{\mathrm{1}+{u}}\right\}{du}\right\}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{−{x}}}\left[{ln}\mid\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\mid\right]_{\mathrm{0}} ^{\sqrt{−{x}}} \:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{−{x}}}{ln}\mid\frac{\mathrm{1}+\sqrt{−{x}}}{\mathrm{1}−\sqrt{−{x}}}\mid\:\Rightarrow{f}\left({x}\right)\:=\:\int_{{x}} ^{\mathrm{0}} \:\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{−{t}}}{ln}\mid\frac{\mathrm{1}+\sqrt{−{t}}}{\mathrm{1}−\sqrt{−{t}}}\mid\:+{c} \\ $$$${c}={f}\left(\mathrm{0}\right)\:=\mathrm{0}\:\Rightarrow{f}\left({x}\right)=\:\int_{{x}} ^{\mathrm{0}} \:\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{−{t}}}{ln}\mid\frac{\mathrm{1}+\sqrt{−{t}}}{\mathrm{1}−\sqrt{−{t}}}\mid\:{dt} \\ $$$${changement}\:\sqrt{−{t}}={u}\:{give}\:{f}\left({x}\right)\:=\:\int_{\sqrt{−{x}}} ^{\mathrm{0}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}{u}}{ln}\mid\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\mid\left(−\mathrm{2}{u}\right){du} \\ $$$$=\:\int_{\mathrm{0}} ^{\sqrt{−{x}}} {ln}\mid\mathrm{1}+{u}\mid{du}−\int_{\mathrm{0}} ^{\sqrt{−{x}}} {ln}\mid\mathrm{1}−{u}\mid\:{du}\:\:=\int_{\mathrm{0}} ^{\sqrt{−{x}}} {ln}\left(\mathrm{1}+{u}\right){du}\:−\int_{\mathrm{0}} ^{\sqrt{−{x}}} {ln}\left(\mathrm{1}−{u}\right){du} \\ $$$${but}\:\:\int_{\mathrm{0}} ^{\sqrt{−{x}}} {ln}\left(\mathrm{1}+{u}\right){du}\:=_{\mathrm{1}+{u}\:=\alpha} \:\:\int_{\mathrm{1}} ^{\mathrm{1}+\sqrt{−{x}}} {ln}\left(\alpha\right){d}\alpha \\ $$$$\left[\alpha{ln}\left(\alpha\right)−\alpha\right]_{\mathrm{1}} ^{\mathrm{1}+\sqrt{−{x}}} \:=\left(\mathrm{1}+\sqrt{−{x}}\right){ln}\left(\mathrm{1}+\sqrt{−{x}}\right)−\left(\mathrm{1}+\sqrt{−{x}}\right)\:+\mathrm{1} \\ $$$$=\left(\mathrm{1}+\sqrt{−{x}}\right){ln}\left(\mathrm{1}+\sqrt{−{x}}\right)\:−\sqrt{−{x}}.{also} \\ $$$$\int_{\mathrm{0}} ^{\sqrt{−{x}}} {ln}\left(\mathrm{1}−{u}\right){du}\:=_{\mathrm{1}−{u}\:=\alpha} \:\:−\int_{\mathrm{1}} ^{\mathrm{1}−\sqrt{−{x}}} \:{ln}\left(\alpha\right){d}\alpha \\ $$$$=−\left[\alpha{ln}\left(\alpha\right)\:−\alpha\right]_{\mathrm{1}} ^{\mathrm{1}−\sqrt{−{x}}} \:\:=−\left\{\left(\mathrm{1}−\sqrt{−{x}}\right){ln}\left(\mathrm{1}−\sqrt{−{x}}\right)−\left(\mathrm{1}−\sqrt{−{x}}\right)\:+\mathrm{1}\right\} \\ $$$$=−\left(\mathrm{1}−\sqrt{−{x}}\right){ln}\left(\mathrm{1}−\sqrt{−{x}}\right)\:−\sqrt{−{x}}\:\Rightarrow \\ $$$${f}\left({x}\right)=\left(\mathrm{1}+\sqrt{−{x}}\right){ln}\left(\mathrm{1}+\sqrt{−{x}}\right)\:+\left(\mathrm{1}−\sqrt{−{x}}\right){ln}\left(\mathrm{1}−\sqrt{−{x}}\right). \\ $$
Commented by maxmathsup by imad last updated on 25/Sep/18
2) we have f(x) =∫_0 ^1 ((ln(1+xt^2 ))/t^2 ) dt ⇒ ∫_0 ^1 ((ln(1+t^2 ))/t^2 )dt =f(1) = 2 arctan(1)−ln(2) =((2π)/4) −ln(2) =(π/2) −ln(2) .
$$\left.\mathrm{2}\right)\:\:{we}\:{have}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{xt}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }\:{dt}\:\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }{dt} \\ $$$$={f}\left(\mathrm{1}\right)\:=\:\mathrm{2}\:{arctan}\left(\mathrm{1}\right)−{ln}\left(\mathrm{2}\right)\:=\frac{\mathrm{2}\pi}{\mathrm{4}}\:−{ln}\left(\mathrm{2}\right)\:=\frac{\pi}{\mathrm{2}}\:−{ln}\left(\mathrm{2}\right)\:. \\ $$
Commented by maxmathsup by imad last updated on 25/Sep/18
∫_0 ^1 ((ln(1+2t^2 ))/t^2 )dt =f(2) = 2 (√2)arctan((√2)) −ln(3).
$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{ln}\left(\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }{dt}\:={f}\left(\mathrm{2}\right)\:=\:\mathrm{2}\:\sqrt{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{2}}\right)\:−{ln}\left(\mathrm{3}\right). \\ $$
Commented by maxmathsup by imad last updated on 25/Sep/18
4) ∫_0 ^1 ((ln(1−t^2 ))/t^2 )dt =f(−1) =2ln(2) .
$$\left.\mathrm{4}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }{dt}\:={f}\left(−\mathrm{1}\right)\:=\mathrm{2}{ln}\left(\mathrm{2}\right)\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Sep/18
(df/dx)=∫_0 ^1 (∂/∂x)×{((ln(1+xt^2 ))/t^2 )}dt =∫_0 ^1 (1/t^2 )×((0+t^2 )/(1+xt^2 )) dt =∫_0 ^1 (dt/(x((1/x)+t^2 ))) =(1/x)×(1/(1/( (√x) )))∣tan^(−1) ((t/(1/( (√x_ )))))∣_0 ^1 =(1/( (√x) ))tan^(−1) ((√x) ) df(x)=((tan^(−1) ((√x) ))/( (√x) ))dx f(x)=∫((tan^(−1) ((√x) ))/( (√x) ))dx y^2 =x dx=2ydy ∫((tan^(−1) (y) 2ydy)/y) =2∫tan^(−1) (y)dy =2ytan^(−1) (y)−∫(2/(1+y^2 ))×ydy =2ytan^(−1) (y)−ln(1+y^2 )+c =2(√x) tan^(−1) ((√x) )−ln(1+x)+c
$$\frac{{df}}{{dx}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\partial}{\partial{x}}×\left\{\frac{{ln}\left(\mathrm{1}+{xt}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }\right\}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{t}^{\mathrm{2}} }×\frac{\mathrm{0}+{t}^{\mathrm{2}} }{\mathrm{1}+{xt}^{\mathrm{2}} }\:{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{{x}\left(\frac{\mathrm{1}}{{x}}+{t}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{{x}}×\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\sqrt{{x}}\:}}\mid{tan}^{−\mathrm{1}} \left(\frac{{t}}{\frac{\mathrm{1}}{\:\sqrt{{x}_{\:} }}}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{x}}\:}{tan}^{−\mathrm{1}} \left(\sqrt{{x}}\:\right) \\ $$$${df}\left({x}\right)=\frac{{tan}^{−\mathrm{1}} \left(\sqrt{{x}}\:\right)}{\:\sqrt{{x}}\:}{dx} \\ $$$${f}\left({x}\right)=\int\frac{{tan}^{−\mathrm{1}} \left(\sqrt{{x}}\:\right)}{\:\sqrt{{x}}\:}{dx} \\ $$$${y}^{\mathrm{2}} ={x} \\ $$$${dx}=\mathrm{2}{ydy} \\ $$$$\int\frac{{tan}^{−\mathrm{1}} \left({y}\right)\:\mathrm{2}{ydy}}{{y}} \\ $$$$=\mathrm{2}\int{tan}^{−\mathrm{1}} \left({y}\right){dy} \\ $$$$=\mathrm{2}{ytan}^{−\mathrm{1}} \left({y}\right)−\int\frac{\mathrm{2}}{\mathrm{1}+{y}^{\mathrm{2}} }×{ydy} \\ $$$$=\mathrm{2}{ytan}^{−\mathrm{1}} \left({y}\right)−{ln}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)+{c} \\ $$$$=\mathrm{2}\sqrt{{x}}\:{tan}^{−\mathrm{1}} \left(\sqrt{{x}}\:\right)−{ln}\left(\mathrm{1}+{x}\right)+{c} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *