Question Number 30441 by abdo imad last updated on 22/Feb/18
![prove that ∫_0 ^∞ e^(−[x]^2 ) = Σ_(n≥0) e^(−n^2 ) .](https://www.tinkutara.com/question/Q30441.png)
$${prove}\:{that}\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left[{x}\right]^{\mathrm{2}} } =\:\sum_{{n}\geqslant\mathrm{0}} \:{e}^{−{n}^{\mathrm{2}} } . \\ $$
Answered by alex041103 last updated on 22/Feb/18
![∫_0 ^∞ e^(−[x]^2 ) dx=Σ_(n=0) ^∞ ∫_n ^(n+1) e^(−[x]^2 ) dx= =Σ_(n=0) ^∞ ∫_n ^(n+1) e^(−n^2 ) dx= =Σ_(n=0) ^∞ e^(−n^2 ) ∫_n ^(n+1) dx= =Σ_(n=0) ^∞ e^(−n^2 ) x∣_n ^(n+1) = =Σ_(n=0) ^∞ e^(−n^2 ) (n+1−n)=Σ_(n≥0) e^(−n^2 )](https://www.tinkutara.com/question/Q30474.png)
$$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left[{x}\right]^{\mathrm{2}} } {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}} {\overset{{n}+\mathrm{1}} {\int}}{e}^{−\left[{x}\right]^{\mathrm{2}} } {dx}= \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}} {\overset{{n}+\mathrm{1}} {\int}}{e}^{−{n}^{\mathrm{2}} } {dx}= \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{n}^{\mathrm{2}} } \underset{{n}} {\overset{{n}+\mathrm{1}} {\int}}{dx}= \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{n}^{\mathrm{2}} } {x}\mid_{{n}} ^{{n}+\mathrm{1}} = \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{n}^{\mathrm{2}} } \left({n}+\mathrm{1}−{n}\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}{e}^{−{n}^{\mathrm{2}} } \\ $$