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Question Number 137708 by mohammad17 last updated on 05/Apr/21
Commented by mohammad17 last updated on 05/Apr/21
prove
$${prove} \\ $$
Answered by mr W last updated on 05/Apr/21
∫_(−∞) ^∞ ((e^(2θ) dθ)/(e^(3θ) +1)) =(1/2)∫_(−∞) ^∞ (de^(2θ) /(e^(3θ) +1)) =(1/2)∫_0 ^∞ (dx/(x^(3/2) +1)) =[(1/6)ln ((x−(√x)+1)/(x+2(√x)+1))+(1/( (√3)))tan^(−1) ((2(√x)−1)/( (√3)))]_0 ^∞ =(1/( (√3)))((π/2)+(π/6)) =((2π)/(3(√3)))
$$\int_{−\infty} ^{\infty} \frac{{e}^{\mathrm{2}\theta} {d}\theta}{{e}^{\mathrm{3}\theta} +\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{\infty} \frac{{de}^{\mathrm{2}\theta} }{{e}^{\mathrm{3}\theta} +\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{x}^{\frac{\mathrm{3}}{\mathrm{2}}} +\mathrm{1}} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\frac{{x}−\sqrt{{x}}+\mathrm{1}}{{x}+\mathrm{2}\sqrt{{x}}+\mathrm{1}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}\sqrt{{x}}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{6}}\right) \\ $$$$=\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$
Commented by Dwaipayan Shikari last updated on 05/Apr/21
e^θ =t⇒e^θ =(dt/dθ) =∫_(−∞) ^∞ (t/(t^3 +1))dt=(1/3)∫_(−∞) ^∞ (u^((2/3)−1) /(1+u))du=(2/3).((Γ((2/3))Γ((1/3)))/(Γ(1)))=((2π)/(3(√3)))
$${e}^{\theta} ={t}\Rightarrow{e}^{\theta} =\frac{{dt}}{{d}\theta} \\ $$$$=\int_{−\infty} ^{\infty} \frac{{t}}{{t}^{\mathrm{3}} +\mathrm{1}}{dt}=\frac{\mathrm{1}}{\mathrm{3}}\int_{−\infty} ^{\infty} \frac{{u}^{\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{1}} }{\mathrm{1}+{u}}{du}=\frac{\mathrm{2}}{\mathrm{3}}.\frac{\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)}{\Gamma\left(\mathrm{1}\right)}=\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$
Answered by Dwaipayan Shikari last updated on 05/Apr/21
∫_0 ^∞ (e^(2θ) /(e^(3θ) +1))dθ =Σ_(n=1) ^∞ (−1)^(n+1) ∫_0 ^∞ e^(−3nθ+2θ) dθ =Σ_(n=1) ^∞ (((−1)^(n+1) )/(3n−2))=Σ_(n=0) ^∞ (((−1)^n )/(3n+1)) =((1/1)−(1/4)+(1/7)−(1/(10))+(1/(13))−(1/(16))+..) =(1+(1/7)+(1/(13))+..)−((1/4)+(1/(10))+(1/(16))+..) =(1/6)Σ_(n=0) ^∞ ((1/(n+(1/6)))−(1/(n+(2/3))))=(1/6)(ψ((2/3))−ψ((1/6)))
$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{\mathrm{2}\theta} }{{e}^{\mathrm{3}\theta} +\mathrm{1}}{d}\theta \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{3}{n}\theta+\mathrm{2}\theta} {d}\theta \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{\mathrm{3}{n}−\mathrm{2}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}} \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{13}}−\frac{\mathrm{1}}{\mathrm{16}}+..\right) \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{13}}+..\right)−\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{16}}+..\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}+\frac{\mathrm{1}}{\mathrm{6}}}−\frac{\mathrm{1}}{{n}+\frac{\mathrm{2}}{\mathrm{3}}}\right)=\frac{\mathrm{1}}{\mathrm{6}}\left(\psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{6}}\right)\right) \\ $$
Commented by mohammad17 last updated on 05/Apr/21
put sir the result is ((2π)/(3(√3)))
$${put}\:{sir}\:{the}\:{result}\:{is}\:\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$
Answered by EnterUsername last updated on 06/Apr/21
∫_(−∞) ^∞ (e^(2θ) /(e^(3θ) +1))dθ=∫_(−∞) ^∞ (e^(−2θ) /(e^(−3θ) +1))dθ=∫_(−∞) ^∞ (e^θ /(e^(3θ) +1))dθ =(1/2)∫_(−∞) ^∞ ((e^(2θ) +e^θ )/(e^(3θ) +1))dθ=(1/2)∫_(−∞) ^∞ ((e^θ (e^θ +1))/((e^θ +1)(e^(2θ) −e^θ +1)))dθ =(1/2)∫_(−∞) ^∞ (e^θ /(e^(2θ) −e^θ +1))dθ=(1/2)∫_0 ^∞ (du/(u^2 −u+1))=(1/2)∫_0 ^∞ (du/((u−(1/2))^2 +(3/4))) =(1/2)[(2/( (√3)))tan^(−1) (((2u−1)/( (√3))))]_0 ^∞ =(1/2)[(2/( (√3)))((π/2)+(π/6))]=((2π)/(3(√3)))
$$\int_{−\infty} ^{\infty} \frac{{e}^{\mathrm{2}\theta} }{{e}^{\mathrm{3}\theta} +\mathrm{1}}{d}\theta=\int_{−\infty} ^{\infty} \frac{{e}^{−\mathrm{2}\theta} }{{e}^{−\mathrm{3}\theta} +\mathrm{1}}{d}\theta=\int_{−\infty} ^{\infty} \frac{{e}^{\theta} }{{e}^{\mathrm{3}\theta} +\mathrm{1}}{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{\infty} \frac{{e}^{\mathrm{2}\theta} +{e}^{\theta} }{{e}^{\mathrm{3}\theta} +\mathrm{1}}{d}\theta=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{\infty} \frac{{e}^{\theta} \left({e}^{\theta} +\mathrm{1}\right)}{\left({e}^{\theta} +\mathrm{1}\right)\left({e}^{\mathrm{2}\theta} −{e}^{\theta} +\mathrm{1}\right)}{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{\infty} \frac{{e}^{\theta} }{{e}^{\mathrm{2}\theta} −{e}^{\theta} +\mathrm{1}}{d}\theta=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{du}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{du}}{\left({u}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{u}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right]_{\mathrm{0}} ^{\infty} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{6}}\right)\right]=\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$

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