Question Number 91603 by mathmax by abdo last updated on 01/May/20
![calculate ∫_0 ^∞ xe^(−x^2 −[x]) dx](https://www.tinkutara.com/question/Q91603.png)
$${calculate}\:\int_{\mathrm{0}} ^{\infty} {xe}^{−{x}^{\mathrm{2}} −\left[{x}\right]} \:{dx} \\ $$
Commented by mathmax by abdo last updated on 02/May/20
![I =∫_0 ^∞ x e^(−x^2 −[x]) dx ⇒I =Σ_(n=0) ^∞ ∫_n ^(n+1) xe^(−x^2 −n) dx =Σ_(n=0) ^∞ e^(−n) ∫_n ^(n+1) xe^(−x^2 ) dx =Σ_(n=0) ^∞ e^(−n) [−(1/2)e^(−x^2 ) ]_n ^(n+1) =(1/2)Σ_(n=0) ^∞ e^(−n) (e^(−n^2 ) −e^((n+1)^2 ) ) =(1/2)Σ_(n=0) ^∞ e^(−n−n^2 ) −(1/2)Σ_(n=0) ^∞ e^(−n−(n+1)^2 ) ...be continued...](https://www.tinkutara.com/question/Q91709.png)
$${I}\:=\int_{\mathrm{0}} ^{\infty} {x}\:{e}^{−{x}^{\mathrm{2}} −\left[{x}\right]} \:{dx}\:\Rightarrow{I}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\int_{{n}} ^{{n}+\mathrm{1}} \:{xe}^{−{x}^{\mathrm{2}} −{n}} {dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{n}} \:\int_{{n}} ^{{n}+\mathrm{1}} \:{xe}^{−{x}^{\mathrm{2}} } {dx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{n}} \left[−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{x}^{\mathrm{2}} } \right]_{{n}} ^{{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{n}} \left({e}^{−{n}^{\mathrm{2}} } −{e}^{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \right)\:=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{n}−{n}^{\mathrm{2}} } −\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{n}−\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$…{be}\:{continued}… \\ $$