Question Number 86480 by M±th+et£s last updated on 28/Mar/20
$$\int_{\mathrm{0}} ^{\infty} \left({x}\:{e}^{\mathrm{1}−{x}} \:−\lfloor{x}\rfloor{e}^{\mathrm{1}−\lfloor{x}\rfloor} \right){dx} \\ $$
Commented by abdomathmax last updated on 28/Mar/20
![I =∫_0 ^∞ (xe^(1−x) −[x]e^(1−[x]) )dx =e∫_0 ^∞ xe^(−x) dx−e∫_0 ^∞ [x]e^(−[x]) dx by parts ∫_0 ^∞ xe^(−x) dx =[−xe^(−x) ]_0 ^(+∞) +∫_0 ^∞ e^(−x) dx =[−e^(−x) ]_0 ^(+∞) =1 ∫_0 ^∞ [x] e^(−[x]) dx =Σ_(n=0) ^∞ ∫_n ^(n+1) n e^(−n) dx =Σ_(n=0) ^∞ n(e^(−1) )^n =w(e^(−1) ) with w(x)=Σ_(n=0) ^∞ nx^n we hsve Σ_(n=0) ^∞ x^n =(1/(1−x)) ⇒ Σ_(n=1) ^∞ nx^(n−1) =(1/((1−x)^2 )) ⇒ Σ_(n=1) ^∞ nx^n =(x/((1−x)^2 ))=w(x) (∣x∣<1) ⇒w(e^(−1) ) =(e^(−1) /((1−e^(−1) )^2 )) =(e^(−1) /((1−(1/e))^2 )) =(e/((e−1)^2 )) ⇒I =e(1−(e/((e−1)^2 ))) =e−(e^2 /((e−1)^2 ))](https://www.tinkutara.com/question/Q86481.png)
$${I}\:=\int_{\mathrm{0}} ^{\infty} \left({xe}^{\mathrm{1}−{x}} −\left[{x}\right]{e}^{\mathrm{1}−\left[{x}\right]} \right){dx} \\ $$$$={e}\int_{\mathrm{0}} ^{\infty} \:{xe}^{−{x}} {dx}−{e}\int_{\mathrm{0}} ^{\infty} \left[{x}\right]{e}^{−\left[{x}\right]} {dx} \\ $$$${by}\:{parts}\:\int_{\mathrm{0}} ^{\infty} \:{xe}^{−{x}} \:{dx}\:=\left[−{xe}^{−{x}} \right]_{\mathrm{0}} ^{+\infty} +\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}} {dx} \\ $$$$=\left[−{e}^{−{x}} \right]_{\mathrm{0}} ^{+\infty} =\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\left[{x}\right]\:{e}^{−\left[{x}\right]} {dx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\int_{{n}} ^{{n}+\mathrm{1}} {n}\:{e}^{−{n}} {dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:{n}\left({e}^{−\mathrm{1}} \right)^{{n}} \:={w}\left({e}^{−\mathrm{1}} \right)\:{with}\:{w}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:{nx}^{{n}} \\ $$$${we}\:{hsve}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}−\mathrm{1}} \:=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}} \:=\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }={w}\left({x}\right) \\ $$$$\left(\mid{x}\mid<\mathrm{1}\right)\:\Rightarrow{w}\left({e}^{−\mathrm{1}} \right)\:=\frac{{e}^{−\mathrm{1}} }{\left(\mathrm{1}−{e}^{−\mathrm{1}} \right)^{\mathrm{2}} }\:=\frac{{e}^{−\mathrm{1}} }{\left(\mathrm{1}−\frac{\mathrm{1}}{{e}}\right)^{\mathrm{2}} } \\ $$$$=\frac{{e}}{\left({e}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{I}\:={e}\left(\mathrm{1}−\frac{{e}}{\left({e}−\mathrm{1}\right)^{\mathrm{2}} }\right)\:={e}−\frac{{e}^{\mathrm{2}} }{\left({e}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$ \\ $$