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Question Number 141931 by mathmax by abdo last updated on 24/May/21

calculate ∫_0 ^∞ e^(−2x) ln(1+e^(3x) )dx

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{2x}} \mathrm{ln}\left(\mathrm{1}+\mathrm{e}^{\mathrm{3x}} \right)\mathrm{dx} \\ $$

Answered by mindispower last updated on 24/May/21

ln(1+e^(3x) )=3x+ln(1+e^(−3x) ) =∫_0 ^∞ 3xe^(−2x) dx+∫_0 ^∞ e^(−2x) .Σ(((−1)^k )/(k+1))e^(−3(1+k)x) dx =∫_0 ^∞ (3/4)ye^(−y) dy+Σ_(k≥0) (((−1)^k )/(k+1))∫_0 ^∞ e^(−(5+3k)x) dx =(3/4)Γ(2)+Σ_(k≥0) (((−1)^k )/(k+1)).(1/(5+3k)) =(3/4)−Σ_(k≥0) (((−1)^k )/(2(5+3k)))+(3/2)Σ_(k≥0) (((−1)^k )/(k+1)) =(3/4)+(3/2)ln(2)+(1/2)(−Σ_(k≥0) (1/(6k+5))+(1/(6k+8))) =(3/2)ln(2e^(1/2) )+(1/(12))(Σ_(k≥0) (1/((k+(5/8))(k+(8/6))))) =(3/2)ln(2e^(1/2) )+(1/(12))((Ψ((8/6))−Ψ((5/6)))/(1/6)) =(3/2)ln(2e^(1/2) )+(1/2)Ψ((4/3))−(1/2)Ψ((5/6)) Ψ((4/3))=Ψ((1/3))+3 Ψ((1/3))=−γ−ln(6)−(π/2)cot((π/3))+2cos(((2π)/3))ln(sin((π/3))) =−γ−ln(6)−(π/(2(√3)))−ln(((√3)/2)) Ψ((5/6))=−γ−ln(12)−(π/2)cot(((5π)/6))+2cos(((2π)/6))ln(sin((π/6))) +2cos(((4π)/6))ln(sin(((2π)/6))) =−γ−ln(12)+(π/2)(√3)+ln((1/2))−ln(((√3)/2))

$${ln}\left(\mathrm{1}+{e}^{\mathrm{3}{x}} \right)=\mathrm{3}{x}+{ln}\left(\mathrm{1}+{e}^{−\mathrm{3}{x}} \right) \\ $$$$=\int_{\mathrm{0}} ^{\infty} \mathrm{3}{xe}^{−\mathrm{2}\boldsymbol{{x}}} \boldsymbol{{dx}}+\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{2}{x}} .\Sigma\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}+\mathrm{1}}{e}^{−\mathrm{3}\left(\mathrm{1}+{k}\right){x}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{3}}{\mathrm{4}}{ye}^{−{y}} {dy}+\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} {e}^{−\left(\mathrm{5}+\mathrm{3}{k}\right){x}} {dx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\Gamma\left(\mathrm{2}\right)+\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}+\mathrm{1}}.\frac{\mathrm{1}}{\mathrm{5}+\mathrm{3}{k}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}−\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}\left(\mathrm{5}+\mathrm{3}{k}\right)}+\frac{\mathrm{3}}{\mathrm{2}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}+\mathrm{1}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(−\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{6}{k}+\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}{k}+\mathrm{8}}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\mathrm{2}{e}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)+\frac{\mathrm{1}}{\mathrm{12}}\left(\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({k}+\frac{\mathrm{5}}{\mathrm{8}}\right)\left({k}+\frac{\mathrm{8}}{\mathrm{6}}\right)}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\mathrm{2}{e}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)+\frac{\mathrm{1}}{\mathrm{12}}\frac{\Psi\left(\frac{\mathrm{8}}{\mathrm{6}}\right)−\Psi\left(\frac{\mathrm{5}}{\mathrm{6}}\right)}{\frac{\mathrm{1}}{\mathrm{6}}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\mathrm{2}{e}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\Psi\left(\frac{\mathrm{4}}{\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\Psi\left(\frac{\mathrm{5}}{\mathrm{6}}\right) \\ $$$$\Psi\left(\frac{\mathrm{4}}{\mathrm{3}}\right)=\Psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{3} \\ $$$$\Psi\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=−\gamma−{ln}\left(\mathrm{6}\right)−\frac{\pi}{\mathrm{2}}{cot}\left(\frac{\pi}{\mathrm{3}}\right)+\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right){ln}\left({sin}\left(\frac{\pi}{\mathrm{3}}\right)\right) \\ $$$$=−\gamma−{ln}\left(\mathrm{6}\right)−\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}−{ln}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\Psi\left(\frac{\mathrm{5}}{\mathrm{6}}\right)=−\gamma−{ln}\left(\mathrm{12}\right)−\frac{\pi}{\mathrm{2}}{cot}\left(\frac{\mathrm{5}\pi}{\mathrm{6}}\right)+\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{6}}\right){ln}\left({sin}\left(\frac{\pi}{\mathrm{6}}\right)\right) \\ $$$$+\mathrm{2}{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{6}}\right){ln}\left({sin}\left(\frac{\mathrm{2}\pi}{\mathrm{6}}\right)\right) \\ $$$$=−\gamma−{ln}\left(\mathrm{12}\right)+\frac{\pi}{\mathrm{2}}\sqrt{\mathrm{3}}+{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−{ln}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$ \\ $$

Commented by Mathspace last updated on 24/May/21

thsnks sir

$${thsnks}\:{sir} \\ $$

Answered by mathmax by abdo last updated on 26/May/21

Φ=∫_0 ^∞ e^(−2x) log(1+e^(3x) )dx by parts Φ=[−(1/2)e^(−2x) log(1+e^(3x) )]_0 ^∞ +(1/2)∫_0 ^∞ e^(−2x) ×((3e^(3x) )/(1+e^(3x) ))dx =((log2)/2) +(3/2)∫_0 ^∞ (e^x /(1+e^(3x) ))dx but ∫_0 ^∞ (e^x /(1+e^(3x) ))dx =_(e^x =t) ∫_0 ^∞ (t/(1+t^3 ))×(dt/t)=∫_1 ^∞ (dt/(t^3 +1)) F(t)=(1/(t^3 +1))=(1/((t+1)(t^2 −t +1)))=(a/(t+1))+((bt +c)/(t^2 −t+1)) a=(1/3),lim_(t→+∞) tF(t)=0 =a+b ⇒b=−(1/3) F(0)=1=a+c ⇒c=1−(1/3)=(2/3) ⇒ F(t)=(1/(3(t+1)))+((−(1/3)t+(2/3))/(t^2 −t+1))=(1/(3(t+1)))−(1/3)((t−2)/(t^2 −t +1)) ⇒ ∫_1 ^∞ F(t)dt =[(1/3)log∣((t+1)/( (√(t^2 −t+1))))∣]_1 ^∞ (→0) +(2/3)∫_1 ^∞ (dt/(t^2 −t +1)) ∫_1 ^∞ (dt/(t^2 −t +1))=∫_1 ^∞ (dt/((t−(1/2))^2 +(3/4)))=_(t−(1/2)=((√3)/2)y →y=((2t−1)/( (√3)))) =(4/3)∫_(1/( (√3))) ^∞ (1/(y^2 +1))((√3)/2)dy =(2/( (√3)))[arctany]_(1/( (√3))) ^∞ =(2/( (√3)))((π/2)−(π/6))=(2/( (√3))).(π/3) =((2π)/(3(√3))) ⇒Φ=((log2)/2)+(3/2).((2π)/(3(√3))) ⇒Φ=((log2)/2) +(π/( (√3)))

$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{2x}} \mathrm{log}\left(\mathrm{1}+\mathrm{e}^{\mathrm{3x}} \right)\mathrm{dx}\:\:\mathrm{by}\:\mathrm{parts} \\ $$$$\Phi=\left[−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−\mathrm{2x}} \mathrm{log}\left(\mathrm{1}+\mathrm{e}^{\mathrm{3x}} \right)\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{2x}} ×\frac{\mathrm{3e}^{\mathrm{3x}} }{\mathrm{1}+\mathrm{e}^{\mathrm{3x}} }\mathrm{dx} \\ $$$$=\frac{\mathrm{log2}}{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{1}+\mathrm{e}^{\mathrm{3x}} }\mathrm{dx}\:\:\mathrm{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{\mathrm{x}} }{\mathrm{1}+\mathrm{e}^{\mathrm{3x}} }\mathrm{dx}\:=_{\mathrm{e}^{\mathrm{x}} =\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{t}}{\mathrm{1}+\mathrm{t}^{\mathrm{3}} }×\frac{\mathrm{dt}}{\mathrm{t}}=\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{3}} +\mathrm{1}}\: \\ $$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{3}} \:+\mathrm{1}}=\frac{\mathrm{1}}{\left(\mathrm{t}+\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{2}} −\mathrm{t}\:+\mathrm{1}\right)}=\frac{\mathrm{a}}{\mathrm{t}+\mathrm{1}}+\frac{\mathrm{bt}\:+\mathrm{c}}{\mathrm{t}^{\mathrm{2}} −\mathrm{t}+\mathrm{1}} \\ $$$$\mathrm{a}=\frac{\mathrm{1}}{\mathrm{3}},\mathrm{lim}_{\mathrm{t}\rightarrow+\infty} \mathrm{tF}\left(\mathrm{t}\right)=\mathrm{0}\:=\mathrm{a}+\mathrm{b}\:\Rightarrow\mathrm{b}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{F}\left(\mathrm{0}\right)=\mathrm{1}=\mathrm{a}+\mathrm{c}\:\Rightarrow\mathrm{c}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{t}+\mathrm{1}\right)}+\frac{−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{t}+\frac{\mathrm{2}}{\mathrm{3}}}{\mathrm{t}^{\mathrm{2}} −\mathrm{t}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{t}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{3}}\frac{\mathrm{t}−\mathrm{2}}{\mathrm{t}^{\mathrm{2}} −\mathrm{t}\:+\mathrm{1}}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{\infty} \mathrm{F}\left(\mathrm{t}\right)\mathrm{dt}\:=\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{log}\mid\frac{\mathrm{t}+\mathrm{1}}{\:\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{t}+\mathrm{1}}}\mid\right]_{\mathrm{1}} ^{\infty} \left(\rightarrow\mathrm{0}\right)\:+\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{t}\:+\mathrm{1}} \\ $$$$\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{t}\:+\mathrm{1}}=\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{dt}}{\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}=_{\mathrm{t}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{y}\:\rightarrow\mathrm{y}=\frac{\mathrm{2t}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} +\mathrm{1}}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{dy}\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left[\mathrm{arctany}\right]_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\infty} \:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{6}}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}.\frac{\pi}{\mathrm{3}} \\ $$$$=\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:\Rightarrow\Phi=\frac{\mathrm{log2}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}.\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:\Rightarrow\Phi=\frac{\mathrm{log2}}{\mathrm{2}}\:+\frac{\pi}{\:\sqrt{\mathrm{3}}} \\ $$

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