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Question Number 163619 by Zaynal last updated on 08/Jan/22

$$\mathit{P}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathit{t}\mathit{h}\mathit{a}\mathit{t};$$$${\int }_{-\mathrm{\infty }}^0{\mathit{e}}^{-\mid \mathit{t}\mid }\mathit{d}\mathit{t}=1$$

Commented by alephzero last updated on 08/Jan/22

$${\int }_{-\mathrm{\infty }}^0{e}^{-\mid t\mid }dt=1$$$$\underset{q\to -\mathrm{\infty }}{\lim }\left({\int }_q^0{e}^{-\mid t\mid }dt\right)=1$$$$\int e^{-\mid t\mid }dt=$$$$e^t+C\iff t⩽0$$$$e^{-t}+C\iff t⩾0$$$$\Rightarrow {\int }_q^0{e}^{-\mid t\mid }dt=e^0-e^q=1-e^q\iff q⩽0$$$$\Rightarrow \underset{q\to -\mathrm{\infty }}{\lim }\left({\int }_q^0{e}^{-\mid t\mid }dt\right)=$$$$=\underset{q\to -\mathrm{\infty }}{\lim }(1-e^q)=$$$$=\underset{q\to -\mathrm{\infty }}{\lim }\left(1\right)-\underset{q\to -\mathrm{\infty }}{\lim }\left(e^q\right)$$$$\mathrm{\forall }a\{a\in \mathbb{R}\mid a^{-\mathrm{\infty }}:=0\}$$$$\Rightarrow 1-\underset{q\to -\mathrm{\infty }}{\lim }\left(e^q\right)=1-0=1$$$$\Rightarrow {\int }_{-\mathrm{\infty }}^0{e}^{-\mid t\mid }dt=1◼\mathrm{Q}.\mathrm{E}.\mathrm{D}.$$

Commented by Zaynal last updated on 09/Jan/22

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.\mathrm{perpect}$$

Answered by MJS_new last updated on 08/Jan/22

$$-\mathrm{\infty }<t⩽0\Rightarrow \mid t\mid =-t$$$$\underset{-\mathrm{\infty }}{\stackrel{0}{\int }}{\mathrm{e}}^{-\mid t\mid }dt=\underset{-\mathrm{\infty }}{\stackrel{0}{\int }}{\mathrm{e}}^{t}dt={\left[{\mathrm{e}}^t\right]}_{-\mathrm{\infty }}^0=0-(-1)=1$$

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