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Question Number 145941 by Mathspace last updated on 09/Jul/21
find ∫_0 ^∞ e^(−3x) log(1+x^3 )dx
$${find}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\mathrm{3}{x}} {log}\left(\mathrm{1}+{x}^{\mathrm{3}} \right){dx} \\ $$
Answered by ArielVyny last updated on 24/Jul/21
((3x^2 )/(1+x^3 ))=3Σ(−1)^n x^(3n+2) ln(1+x^3 )=Σ3(−1)^n (1/(3n+3))x^(3n+3) =Σ(((−1)^n )/(n+1))x^(3n+3) ∫_0 ^∞ e^(−3x) ln(1+x^3 )=Σ(((−1)^n )/(n+1))∫_0 ^∞ e^(−3x) x^(3n+3) dx 3x=t→3dx=dt ∫_0 ^∞ e^(−t) ((t/3))^(3n+3) (1/3)dt=((1/3))^(3n+4) ∫_0 ^∞ e^(−t) t^(3n+3) dt =((1/3))^(3n+4) Γ(3n+4)=((1/3))^(3n+4) (3n+3)! ∫_0 ^∞ e^(−3x) ln(1+x^3 )dx=Σ(((−1)^n )/(n+1))((1/3))^(3n+4) (3n+3)! =Σ(((−1)^n (3n+3))/(n+1))((1/3))^(3n+4) (3n+2)! =Σ(−1)^n 3×3^(−(3n+4)) (3n+2)! =Σ(−1)^n ((1/3))^(3n+3) (3n+2)! =(1/(27))Σ(−1)^n ((1/(27)))^n (3n+2)!
$$\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{3}} }=\mathrm{3}\Sigma\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{3}{n}+\mathrm{2}} \\ $$$${ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)=\Sigma\mathrm{3}\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{1}}{\mathrm{3}{n}+\mathrm{3}}{x}^{\mathrm{3}{n}+\mathrm{3}} =\Sigma\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}{x}^{\mathrm{3}{n}+\mathrm{3}} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{3}{x}} {ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)=\Sigma\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{3}{x}} {x}^{\mathrm{3}{n}+\mathrm{3}} {dx} \\ $$$$\mathrm{3}{x}={t}\rightarrow\mathrm{3}{dx}={dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} \left(\frac{{t}}{\mathrm{3}}\right)^{\mathrm{3}{n}+\mathrm{3}} \frac{\mathrm{1}}{\mathrm{3}}{dt}=\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}{n}+\mathrm{4}} \int_{\mathrm{0}} ^{\infty} {e}^{−{t}} {t}^{\mathrm{3}{n}+\mathrm{3}} {dt} \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}{n}+\mathrm{4}} \Gamma\left(\mathrm{3}{n}+\mathrm{4}\right)=\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}{n}+\mathrm{4}} \left(\mathrm{3}{n}+\mathrm{3}\right)! \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−\mathrm{3}{x}} {ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right){dx}=\Sigma\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}{n}+\mathrm{4}} \left(\mathrm{3}{n}+\mathrm{3}\right)! \\ $$$$=\Sigma\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{3}{n}+\mathrm{3}\right)}{{n}+\mathrm{1}}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}{n}+\mathrm{4}} \left(\mathrm{3}{n}+\mathrm{2}\right)! \\ $$$$=\Sigma\left(−\mathrm{1}\right)^{{n}} \mathrm{3}×\mathrm{3}^{−\left(\mathrm{3}{n}+\mathrm{4}\right)} \left(\mathrm{3}{n}+\mathrm{2}\right)! \\ $$$$=\Sigma\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}{n}+\mathrm{3}} \left(\mathrm{3}{n}+\mathrm{2}\right)! \\ $$$$=\frac{\mathrm{1}}{\mathrm{27}}\Sigma\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{\mathrm{27}}\right)^{{n}} \left(\mathrm{3}{n}+\mathrm{2}\right)! \\ $$

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