Question Number 83565 by Jidda28 last updated on 03/Mar/20
Commented by abdomathmax last updated on 04/Mar/20
![I=∫_(−∞) ^(+∞) (e^(2x) /(e^(3x) +1))dx changement e^x =t give I =∫_0 ^(+∞) (t^2 /(t^3 +1))×(dt/t) =∫_0 ^∞ ((tdt)/(t^3 +1)) let decompose F(t)=(t/(t^3 +1)) ⇒F(t)=(t/((t+1)(t^2 −t+1))) =(a/(t+1)) +((bt +c)/(t^2 −t +1)) a=(t+1)F(t)∣_(t=−1) =((−1)/3) lim_(t→+∞) tF(t)=0=a+b ⇒b=(1/3) F(0)=0=a+c ⇒c=(1/3)⇒F(t)=((−1)/(3(t+1))) +(1/3)((t+1)/(t^2 −t+1)) ⇒∫ F(t)dt =−(1/3)∫ (dt/(t+1)) +(1/6)∫ ((2t+2)/(t^2 −t+1))dt =−(1/3)∫ (dt/(t+1)) +(1/6)∫ ((2t−1)/(t^2 −t+1))dt +(1/2)∫ (dt/(t^2 −t+1)) =−(1/3)ln∣t+1∣+(1/6)ln(t^2 −t+1) +(1/2)∫ (dt/((t−(1/2))^2 +(3/4))) (t−(1/2)=((√3)/2)u) =ln((((t^2 −t+1)^(1/6) )/(∣t+1∣^(1/3) )))+(1/2)×(4/3)∫ (1/(u^2 +1))×((√3)/2)du =ln(....)+(1/( (√3))) arctan(((2t−1)/( (√3)))) ⇒ I =∫_0 ^∞ F(t)dt =[ln((((t^2 −t+1)^(1/6) )/((t+1)^(1/3) )))]_0 ^(+∞) +(1/( (√3)))[arctan(((2t−1)/( (√3))))]_0 ^(+∞) =(1/( (√3))){(π/2) +arctan((1/( (√3))))} =(1/( (√3))){ (π/2)+(π/6)} =(1/( (√3))){((4π)/6)} =((2π)/(3(√3)))](https://www.tinkutara.com/question/Q83579.png)
$${I}=\int_{−\infty} ^{+\infty} \:\frac{{e}^{\mathrm{2}{x}} }{{e}^{\mathrm{3}{x}} \:+\mathrm{1}}{dx}\:{changement}\:{e}^{{x}} ={t}\:{give} \\ $$$${I}\:\:=\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{3}} \:+\mathrm{1}}×\frac{{dt}}{{t}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{tdt}}{{t}^{\mathrm{3}} +\mathrm{1}} \\ $$$${let}\:{decompose}\:{F}\left({t}\right)=\frac{{t}}{{t}^{\mathrm{3}} \:+\mathrm{1}}\:\Rightarrow{F}\left({t}\right)=\frac{{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)} \\ $$$$=\frac{{a}}{{t}+\mathrm{1}}\:+\frac{{bt}\:+{c}}{{t}^{\mathrm{2}} −{t}\:+\mathrm{1}} \\ $$$${a}=\left({t}+\mathrm{1}\right){F}\left({t}\right)\mid_{{t}=−\mathrm{1}} =\frac{−\mathrm{1}}{\mathrm{3}} \\ $$$${lim}_{{t}\rightarrow+\infty} \:{tF}\left({t}\right)=\mathrm{0}={a}+{b}\:\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{3}}\: \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{0}={a}+{c}\:\Rightarrow{c}=\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow{F}\left({t}\right)=\frac{−\mathrm{1}}{\mathrm{3}\left({t}+\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{3}}\frac{{t}+\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}} \\ $$$$\Rightarrow\int\:{F}\left({t}\right){dt}\:=−\frac{\mathrm{1}}{\mathrm{3}}\int\:\frac{{dt}}{{t}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{6}}\int\:\frac{\mathrm{2}{t}+\mathrm{2}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\int\:\frac{{dt}}{{t}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{6}}\int\:\frac{\mathrm{2}{t}−\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt}\:+\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{{dt}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{t}+\mathrm{1}\mid+\frac{\mathrm{1}}{\mathrm{6}}{ln}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right) \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{{dt}}{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}\:\:\:\left({t}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{u}\right) \\ $$$$={ln}\left(\frac{\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} }{\mid{t}+\mathrm{1}\mid^{\frac{\mathrm{1}}{\mathrm{3}}} }\right)+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{3}}\int\:\:\:\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:+\mathrm{1}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{du} \\ $$$$={ln}\left(….\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:\Rightarrow \\ $$$${I}\:=\int_{\mathrm{0}} ^{\infty} \:{F}\left({t}\right){dt}\:=\left[{ln}\left(\frac{\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} }{\left({t}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} }\right)\right]_{\mathrm{0}} ^{+\infty} \\ $$$$+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left[{arctan}\left(\frac{\mathrm{2}{t}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left\{\frac{\pi}{\mathrm{2}}\:+{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right\}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left\{\:\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{6}}\right\} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left\{\frac{\mathrm{4}\pi}{\mathrm{6}}\right\}\:=\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$$ \\ $$$$ \\ $$
Commented by abdomathmax last updated on 04/Mar/20
$$\Rightarrow{I}\:=\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{9}}\:\:\:\:\:\left(\:\:\:\:{error}\:{in}\:{the}\:{Q}!…\right) \\ $$
Commented by Jidda28 last updated on 04/Mar/20
$${thank}\:{you}\:{sir} \\ $$
Commented by msup trace by abdo last updated on 04/Mar/20
$${you}\:{are}\:{welcome} \\ $$
Answered by mind is power last updated on 03/Mar/20
![i found just this near to gamma Ψ(x)=((Γ′(x))/(Γ(x))) and ∫_0 ^(+∞) e^(−ax) dx=(1/a)∫_0 ^(+∞) x^(1−1) e^(−x) =((Γ(1))/a),lign (5→6)bellow =∫_0 ^(+∞) (e^(−2x) /(1+e^(−3x) ))dx=∫_0 ^(+∞) (e^x /(1+e^(3x) ))dx ⇒∫_(−∞) ^(+∞) (e^(2x) /(1+e^(3x) ))dx=∫_0 ^(+∞) (e^x /(1+e^(3x) ))dx+∫_0 ^(+∞) (e^(2x) /(1+e^(3x) )) =∫_0 ^(+∞) (e^(−2x) /(e^(−3x) +1))dx+∫_0 ^(+∞) (e^(−x) /(1+e^(−3x) ))dx =∫_0 ^(+∞) e^(−2x) {Σ_(k≥0) (−e^(−3x) )^k )dx+∫_0 ^(+∞) e^(−x) (Σ_(k≥0) (−e^(−3x) )^k )dx =Σ_(k≥0) (−1)^k ∫_0 ^(+∞) (e^(−(3k+2)x) +e^(−(3k+1)x) )dx =Σ_(k≥0) (((−1)^k )/(3k+2))+Σ_(k≥0) (((−1)^k )/(3k+1)) =Σ_(k≥0) ((1/(6k+2))−(1/(6k+5)))+Σ_(k≥0) ((1/(6k+1))−(1/(6k+4))) =Σ_(k≥0) ((3/((6k+2)(6k+5))))+Σ_(k≥0) ((3/((6k+1)(6k+4)))) =(1/(12))Σ_(k≥0) ((1/((k+(2/6))(k+(5/6)))))+(1/(12))Σ_(k≥0) (1/((k+(1/6))(k+(4/6)))) =(1/(12))((Ψ((5/6))−Ψ((2/6)))/((5−2)/6))+(1/(12))((Ψ((4/6))−Ψ((1/6)))/((4−1)/6)) =(1/6)[Ψ((5/6))−Ψ((1/6))+Ψ((4/6))−Ψ((2/6))] =(1/6)[−(π/2)cot(((5π)/6))+(π/2)cot((π/6))−(π/2)cot(((4π)/6))+(π/2)cot(((2π)/6))] =(1/6)[πcot((π/6))+πcot((π/3))] =(π/6)[(1/( (√3)))+(√3)]=(π/6).((4/( (√3))))=((2π)/(3(√3)))=((2π(√3))/9) 2 nd way =∫_0 ^(+∞) (e^x /(e^(3x) +1))dx+∫_0 ^(+∞) (e^(2x) /(1+e^(3x) ))dx=∫_0 ^(+∞) ((e^x (1+e^x )dx)/((e^x +1)(e^(2x) −e^x +1))) =∫_0 ^(+∞) (e^x /(e^(2x) −e^x +1))dx =∫_1 ^(+∞) (du/(u^2 −u+1))=∫_1 ^(+∞) (du/((u−(1/2))^2 +(3/4))) =(2/( (√3)))[arctan(((2u)/( (√3)))−(1/( (√3))))]_1 ^(+∞) (2/( (√3)))[(π/2)−(π/6)]=((4π)/(6(√3)))=((2π(√3))/9)](https://www.tinkutara.com/question/Q83572.png)
$$\:{i}\:{found}\:{just}\:{this}\:{near}\:{to}\:{gamma}\:\Psi\left({x}\right)=\frac{\Gamma'\left({x}\right)}{\Gamma\left({x}\right)}\:\:\: \\ $$$${and}\:\int_{\mathrm{0}} ^{+\infty} {e}^{−{ax}} {dx}=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{+\infty} {x}^{\mathrm{1}−\mathrm{1}} {e}^{−{x}} =\frac{\Gamma\left(\mathrm{1}\right)}{{a}},{lign}\:\left(\mathrm{5}\rightarrow\mathrm{6}\right){bellow} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{−\mathrm{2}{x}} }{\mathrm{1}+{e}^{−\mathrm{3}{x}} }{dx}=\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{{x}} }{\mathrm{1}+{e}^{\mathrm{3}{x}} }{dx} \\ $$$$\Rightarrow\int_{−\infty} ^{+\infty} \frac{{e}^{\mathrm{2}{x}} }{\mathrm{1}+{e}^{\mathrm{3}{x}} }{dx}=\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{{x}} }{\mathrm{1}+{e}^{\mathrm{3}{x}} }{dx}+\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{\mathrm{2}{x}} }{\mathrm{1}+{e}^{\mathrm{3}{x}} } \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{−\mathrm{2}{x}} }{{e}^{−\mathrm{3}{x}} +\mathrm{1}}{dx}+\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{−{x}} }{\mathrm{1}+{e}^{−\mathrm{3}{x}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} {e}^{−\mathrm{2}{x}} \left\{\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−{e}^{−\mathrm{3}{x}} \right)^{{k}} \right){dx}+\int_{\mathrm{0}} ^{+\infty} {e}^{−{x}} \left(\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−{e}^{−\mathrm{3}{x}} \right)^{{k}} \right){dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{k}} \int_{\mathrm{0}} ^{+\infty} \left({e}^{−\left(\mathrm{3}{k}+\mathrm{2}\right){x}} +{e}^{−\left(\mathrm{3}{k}+\mathrm{1}\right){x}} \right){dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{3}{k}+\mathrm{2}}+\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{3}{k}+\mathrm{1}} \\ $$$$ \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\mathrm{6}{k}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{6}{k}+\mathrm{5}}\right)+\underset{{k}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\mathrm{6}{k}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{6}{k}+\mathrm{4}}\right) \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{3}}{\left(\mathrm{6}{k}+\mathrm{2}\right)\left(\mathrm{6}{k}+\mathrm{5}\right)}\right)+\underset{{k}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{3}}{\left(\mathrm{6}{k}+\mathrm{1}\right)\left(\mathrm{6}{k}+\mathrm{4}\right)}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\underset{{k}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\left({k}+\frac{\mathrm{2}}{\mathrm{6}}\right)\left({k}+\frac{\mathrm{5}}{\mathrm{6}}\right)}\right)+\frac{\mathrm{1}}{\mathrm{12}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({k}+\frac{\mathrm{1}}{\mathrm{6}}\right)\left({k}+\frac{\mathrm{4}}{\mathrm{6}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\frac{\Psi\left(\frac{\mathrm{5}}{\mathrm{6}}\right)−\Psi\left(\frac{\mathrm{2}}{\mathrm{6}}\right)}{\frac{\mathrm{5}−\mathrm{2}}{\mathrm{6}}}+\frac{\mathrm{1}}{\mathrm{12}}\frac{\Psi\left(\frac{\mathrm{4}}{\mathrm{6}}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{6}}\right)}{\frac{\mathrm{4}−\mathrm{1}}{\mathrm{6}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left[\Psi\left(\frac{\mathrm{5}}{\mathrm{6}}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{6}}\right)+\Psi\left(\frac{\mathrm{4}}{\mathrm{6}}\right)−\Psi\left(\frac{\mathrm{2}}{\mathrm{6}}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left[−\frac{\pi}{\mathrm{2}}{cot}\left(\frac{\mathrm{5}\pi}{\mathrm{6}}\right)+\frac{\pi}{\mathrm{2}}{cot}\left(\frac{\pi}{\mathrm{6}}\right)−\frac{\pi}{\mathrm{2}}{cot}\left(\frac{\mathrm{4}\pi}{\mathrm{6}}\right)+\frac{\pi}{\mathrm{2}}{cot}\left(\frac{\mathrm{2}\pi}{\mathrm{6}}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left[\pi{cot}\left(\frac{\pi}{\mathrm{6}}\right)+\pi{cot}\left(\frac{\pi}{\mathrm{3}}\right)\right] \\ $$$$=\frac{\pi}{\mathrm{6}}\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\sqrt{\mathrm{3}}\right]=\frac{\pi}{\mathrm{6}}.\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\right)=\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}=\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{9}} \\ $$$$\mathrm{2}\:{nd}\:{way} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{{x}} }{{e}^{\mathrm{3}{x}} +\mathrm{1}}{dx}+\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{\mathrm{2}{x}} }{\mathrm{1}+{e}^{\mathrm{3}{x}} }{dx}=\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{{x}} \left(\mathrm{1}+{e}^{{x}} \right){dx}}{\left({e}^{{x}} +\mathrm{1}\right)\left({e}^{\mathrm{2}{x}} −{e}^{{x}} +\mathrm{1}\right)} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{{x}} }{{e}^{\mathrm{2}{x}} −{e}^{{x}} +\mathrm{1}}{dx} \\ $$$$=\int_{\mathrm{1}} ^{+\infty} \frac{{du}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}=\int_{\mathrm{1}} ^{+\infty} \frac{{du}}{\left({u}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left[{arctan}\left(\frac{\mathrm{2}{u}}{\:\sqrt{\mathrm{3}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right]_{\mathrm{1}} ^{+\infty} \\ $$$$\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left[\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{6}}\right]=\frac{\mathrm{4}\pi}{\mathrm{6}\sqrt{\mathrm{3}}}=\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{9}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Jidda28 last updated on 04/Mar/20
$${thank}\:{you}\:{sir} \\ $$