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Question Number 37888 by abdo mathsup 649 cc last updated on 19/Jun/18
find f(α) = ∫_0 ^1   arctan(e^(−αx) )dx with α≥0
$${find}\:{f}\left(\alpha\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{arctan}\left({e}^{−\alpha{x}} \right){dx}\:{with}\:\alpha\geqslant\mathrm{0}\: \\ $$
Commented by math khazana by abdo last updated on 21/Jun/18
we have f^′ (α)=∫_0 ^1  ((−x e^(−αx) )/(1+e^(−2αx) ))dx  = ∫_0 ^1  −x e^(−αx) { Σ_(n=0) ^∞  (−1)^n  e^(−2nαx) }  =Σ_(n=0) ^∞  (−1)^(n+1)   ∫_0 ^1  x e^(−(2n+1)αx) dx but  A_n =∫_0 ^1  x e^(−(2n+1)αx) dx=_((2n+1)αx=t)  ∫_0 ^((2n+1)α)  (t/((2n+1)α)) e^(−t)  (dt/((2n+1)α))  = (1/((2n+1)^2 α^2 )) ∫_0 ^((2n+1)α)   t .e^(−t) dt  and by parts  ∫_0 ^((2n+1)α)  t.e^(−t) dt =[−t e^(−t) ]_0 ^((2n+1)α)  +∫_0 ^((2n+1)α)  e^(−t) dt  =−(2n+1)α e^(−(2n+1)α)   +[−e^(−t) ]_0 ^((2n+1)α)   =−(2n+1)α e^(−(2n+1)α)  +(1−e^(−(2n+1)α) ) ⇒  A_n = (1/((2n+1)^2 α^2 )){ −(2n+1)α e^(−(2n+1)α)  +(1−e^(−(2n+1)α) )  =−(1/((2n+1)α)) e^(−(2n+1)α)    +((1−e^(−(2n+1)α) )/((2n+1)^2  α^2 ))  I =Σ_(n=0) ^∞   (((−1)^n )/((2n+1)α)) e^(−(2n+1)α)   +Σ_(n=0) ^∞ (−1)^n ((e^(−(2n+1)α)  −1)/((2n+1)^2  α^2 ))  ...be continued...
$${we}\:{have}\:{f}^{'} \left(\alpha\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{−{x}\:{e}^{−\alpha{x}} }{\mathrm{1}+{e}^{−\mathrm{2}\alpha{x}} }{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:−{x}\:{e}^{−\alpha{x}} \left\{\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\mathrm{2}{n}\alpha{x}} \right\} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha{x}} {dx}\:{but} \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha{x}} {dx}=_{\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha{x}={t}} \:\int_{\mathrm{0}} ^{\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha} \:\frac{{t}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha}\:{e}^{−{t}} \:\frac{{dt}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha} \\ $$$$=\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \alpha^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha} \:\:{t}\:.{e}^{−{t}} {dt}\:\:{and}\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha} \:{t}.{e}^{−{t}} {dt}\:=\left[−{t}\:{e}^{−{t}} \right]_{\mathrm{0}} ^{\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha} \:+\int_{\mathrm{0}} ^{\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha} \:{e}^{−{t}} {dt} \\ $$$$=−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha} \:\:+\left[−{e}^{−{t}} \right]_{\mathrm{0}} ^{\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha} \\ $$$$=−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha} \:+\left(\mathrm{1}−{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha} \right)\:\Rightarrow \\ $$$${A}_{{n}} =\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \alpha^{\mathrm{2}} }\left\{\:−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha} \:+\left(\mathrm{1}−{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha} \right)\right. \\ $$$$=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha}\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha} \:\:\:+\frac{\mathrm{1}−{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \:\alpha^{\mathrm{2}} } \\ $$$${I}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha}\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha} \:\:+\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \frac{{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha} \:−\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \:\alpha^{\mathrm{2}} } \\ $$$$…{be}\:{continued}… \\ $$
Commented by math khazana by abdo last updated on 21/Jun/18
we have artan^′ (u)= (1/(1+u^2 )) =Σ_(n=0) ^∞  (−1)^n  u^(2n)  ⇒  arctan(u) = Σ_(n=0) ^∞   (((−1)^n )/(2n+1)) u^(2n+1)  +c  (c=0)⇒  arctan(e^(−αx) ) =Σ_(n=0) ^∞  (((−1)^n )/(2n+1)) e^(−(2n+1)αx)   and  f(α)= ∫_0 ^1  arctan(e^(−αx) )dx  =Σ_(n=0) ^∞   (((−1)^n )/(2n+1))  ∫_0 ^1    e^(−(2n+1)αx) dx  =Σ_(n=0) ^∞     (((−1)^(n+1) )/((2n+1)^2 α))[ e^(−(2n+1)αx) ]_0 ^1   =(1/α) Σ_(n=0) ^∞    (((−1)^(n+1) )/((2n+1)^2 )){ e^(−(2n+1)α)  −1}  = (1/α) Σ_(n=0) ^∞    (((−1)^(n+1) )/((2n+1)^2 )) e^(−(2n+1)α)    +(1/α) Σ_(n=0) ^∞   (((−1)^n )/((2n+1)^2 ))  ...be continued...
$${we}\:{have}\:{artan}^{'} \left({u}\right)=\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{u}^{\mathrm{2}{n}} \:\Rightarrow \\ $$$${arctan}\left({u}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:{u}^{\mathrm{2}{n}+\mathrm{1}} \:+{c}\:\:\left({c}=\mathrm{0}\right)\Rightarrow \\ $$$${arctan}\left({e}^{−\alpha{x}} \right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha{x}} \:\:{and} \\ $$$${f}\left(\alpha\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left({e}^{−\alpha{x}} \right){dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha{x}} {dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \alpha}\left[\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha{x}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\alpha}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\left\{\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha} \:−\mathrm{1}\right\} \\ $$$$=\:\frac{\mathrm{1}}{\alpha}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha} \:\:\:+\frac{\mathrm{1}}{\alpha}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$…{be}\:{continued}… \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jun/18
  let t=e^(−αx   ) dt=e^(−αx) ×−α dx  dx=(dt/(−αt))  ∫_1 ^e^(−α)   tan^− (t)×(dt/(−αt))  =(1/(−α))∫_1 ^e^(−α)  (1/t)(t−(t^3 /3)+(t^5 /5)...)dt  ((−1)/α)∫_1 ^e^(−α)  (1−(t^2 /3)+(t^4 /5)..)dt  =((−1)/α)×∣t−(t^3 /9)+(t^5 /(25))...)∣_1 ^e^(−α)
$$ \\ $$$${let}\:{t}={e}^{−\alpha{x}\:\:\:} {dt}={e}^{−\alpha{x}} ×−\alpha\:{dx} \\ $$$${dx}=\frac{{dt}}{−\alpha{t}} \\ $$$$\int_{\mathrm{1}} ^{{e}^{−\alpha} } \:{tan}^{−} \left({t}\right)×\frac{{dt}}{−\alpha{t}} \\ $$$$=\frac{\mathrm{1}}{−\alpha}\int_{\mathrm{1}} ^{{e}^{−\alpha} } \frac{\mathrm{1}}{{t}}\left({t}−\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+\frac{{t}^{\mathrm{5}} }{\mathrm{5}}…\right){dt} \\ $$$$\frac{−\mathrm{1}}{\alpha}\int_{\mathrm{1}} ^{{e}^{−\alpha} } \left(\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{3}}+\frac{{t}^{\mathrm{4}} }{\mathrm{5}}..\right){dt} \\ $$$$\left.=\frac{−\mathrm{1}}{\alpha}×\mid{t}−\frac{{t}^{\mathrm{3}} }{\mathrm{9}}+\frac{{t}^{\mathrm{5}} }{\mathrm{25}}…\right)\mid_{\mathrm{1}} ^{{e}^{−\alpha} } \\ $$$$ \\ $$

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