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Question Number 109208 by Dwaipayan Shikari last updated on 21/Aug/20

∫_0 ^∞ ((x^n −1)/(x−1)).(x/e^x )dx

$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{n}} −\mathrm{1}}{{x}−\mathrm{1}}.\frac{{x}}{{e}^{{x}} }{dx} \\ $$

Answered by mathmax by abdo last updated on 22/Aug/20

if n integr   I =∫_0 ^∞ (Σ_(k=0) ^(n−1)  x^k )xe^(−x)  dx =Σ_(k=0) ^(n−1)  ∫_0 ^∞  x^(k+1)  e^(−x)  dx  =Σ_(k=0) ^(n−1)  Γ(k+2) =Σ_(k=0) ^(n−1)   (k+1)! =1! +2! +.....+n!

$$\mathrm{if}\:\mathrm{n}\:\mathrm{integr}\: \\ $$$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \left(\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\mathrm{x}^{\mathrm{k}} \right)\mathrm{xe}^{−\mathrm{x}} \:\mathrm{dx}\:=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{\mathrm{k}+\mathrm{1}} \:\mathrm{e}^{−\mathrm{x}} \:\mathrm{dx} \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\Gamma\left(\mathrm{k}+\mathrm{2}\right)\:=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \:\:\left(\mathrm{k}+\mathrm{1}\right)!\:=\mathrm{1}!\:+\mathrm{2}!\:+.....+\mathrm{n}! \\ $$

Commented by Dwaipayan Shikari last updated on 22/Aug/20

Great sir!  Σ_(n=1) ^n Γ(n+1)=Σ_(n=1) ^n ∫_0 ^∞ x^n e^(−x) dx=∫_0 ^∞ Σ_(n=1) ^n (x^n /e^x )=∫_0 ^∞ ((x^n −1)/(x−1)).(x^n /e^x )  Σ_(n=1) ^n Γ(n+1)=1!+2!+3!+...+n!

$${Great}\:{sir}! \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\Gamma\left({n}+\mathrm{1}\right)=\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\int_{\mathrm{0}} ^{\infty} {x}^{{n}} {e}^{−{x}} {dx}=\int_{\mathrm{0}} ^{\infty} \underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{x}^{{n}} }{{e}^{{x}} }=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{n}} −\mathrm{1}}{{x}−\mathrm{1}}.\frac{{x}^{{n}} }{{e}^{{x}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\Gamma\left({n}+\mathrm{1}\right)=\mathrm{1}!+\mathrm{2}!+\mathrm{3}!+...+{n}! \\ $$

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