Integration Questions

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

$${I}\:=\int_{\mathrm{0}} ^{\infty} \left({xe}^{\mathrm{1}−{x}} −\left[{x}\right]{e}^{\mathrm{1}−\left[{x}\right]} \right){dx} \\$$$$={e}\int_{\mathrm{0}} ^{\infty} \:{xe}^{−{x}} {dx}−{e}\int_{\mathrm{0}} ^{\infty} \left[{x}\right]{e}^{−\left[{x}\right]} {dx} \\$$$${by}\:{parts}\:\int_{\mathrm{0}} ^{\infty} \:{xe}^{−{x}} \:{dx}\:=\left[−{xe}^{−{x}} \right]_{\mathrm{0}} ^{+\infty} +\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}} {dx} \\$$$$=\left[−{e}^{−{x}} \right]_{\mathrm{0}} ^{+\infty} =\mathrm{1} \\$$$$\int_{\mathrm{0}} ^{\infty} \:\left[{x}\right]\:{e}^{−\left[{x}\right]} {dx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\int_{{n}} ^{{n}+\mathrm{1}} {n}\:{e}^{−{n}} {dx} \\$$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:{n}\left({e}^{−\mathrm{1}} \right)^{{n}} \:={w}\left({e}^{−\mathrm{1}} \right)\:{with}\:{w}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:{nx}^{{n}} \\$$$${we}\:{hsve}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow \\$$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}−\mathrm{1}} \:=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}} \:=\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }={w}\left({x}\right) \\$$$$\left(\mid{x}\mid<\mathrm{1}\right)\:\Rightarrow{w}\left({e}^{−\mathrm{1}} \right)\:=\frac{{e}^{−\mathrm{1}} }{\left(\mathrm{1}−{e}^{−\mathrm{1}} \right)^{\mathrm{2}} }\:=\frac{{e}^{−\mathrm{1}} }{\left(\mathrm{1}−\frac{\mathrm{1}}{{e}}\right)^{\mathrm{2}} } \\$$$$=\frac{{e}}{\left({e}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow{I}\:={e}\left(\mathrm{1}−\frac{{e}}{\left({e}−\mathrm{1}\right)^{\mathrm{2}} }\right)\:={e}−\frac{{e}^{\mathrm{2}} }{\left({e}−\mathrm{1}\right)^{\mathrm{2}} } \\$$$$\\$$