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

Prove that; ∫_(−∞) ^0 e^(−∣t∣) dt = 1

$$\boldsymbol{{Prove}}\:\boldsymbol{{that}}; \\ $$$$\:\:\int_{−\infty} ^{\mathrm{0}} \:\boldsymbol{{e}}^{−\mid\boldsymbol{{t}}\mid} \:\boldsymbol{{dt}}\:=\:\mathrm{1} \\ $$

Commented by alephzero last updated on 08/Jan/22

∫_(−∞) ^0 e^(−∣t∣) dt = 1 lim_(q→−∞) (∫_q ^0 e^(−∣t∣) dt) = 1 ∫e^(−∣t∣) dt = e^t + C ⇔ t ≤ 0 e^(−t) + C ⇔ t ≥ 0 ⇒ ∫_q ^0 e^(−∣t∣) dt = e^0 − e^q = 1 − e^q ⇔ q ≤ 0 ⇒ lim_(q→−∞) (∫_q ^0 e^(−∣t∣) dt) = = lim_(q→−∞) (1 − e^q ) = = lim_(q→−∞) (1) −lim_(q→−∞) (e^q ) ∀a {a ∈ R ∣ a^(−∞) := 0} ⇒ 1 − lim_(q→−∞) (e^q ) = 1 − 0 = 1 ⇒ ∫_(−∞) ^0 e^(−∣t∣) dt = 1 ■ Q.E.D.

$$\int_{−\infty} ^{\mathrm{0}} {e}^{−\mid{t}\mid} {dt}\:=\:\mathrm{1} \\ $$$$\underset{{q}\rightarrow−\infty} {\mathrm{lim}}\left(\int_{{q}} ^{\mathrm{0}} {e}^{−\mid{t}\mid} {dt}\right)\:=\:\mathrm{1} \\ $$$$\int{e}^{−\mid{t}\mid} {dt}\:= \\ $$$${e}^{{t}} \:+\:{C}\:\:\:\:\Leftrightarrow\:{t}\:\leqslant\:\mathrm{0} \\ $$$${e}^{−{t}} \:+\:{C}\:\Leftrightarrow\:{t}\:\geqslant\:\mathrm{0} \\ $$$$\Rightarrow\:\int_{{q}} ^{\mathrm{0}} {e}^{−\mid{t}\mid} {dt}\:=\:{e}^{\mathrm{0}} \:−\:{e}^{{q}} \:=\:\mathrm{1}\:−\:{e}^{{q}} \:\Leftrightarrow\:{q}\:\leqslant\:\mathrm{0} \\ $$$$\Rightarrow\:\underset{{q}\rightarrow−\infty} {\mathrm{lim}}\left(\int_{{q}} ^{\mathrm{0}} {e}^{−\mid{t}\mid} {dt}\right)\:= \\ $$$$=\:\underset{{q}\rightarrow−\infty} {\mathrm{lim}}\left(\mathrm{1}\:−\:{e}^{{q}} \right)\:= \\ $$$$=\:\underset{{q}\rightarrow−\infty} {\mathrm{lim}}\left(\mathrm{1}\right)\:−\underset{{q}\rightarrow−\infty} {\mathrm{lim}}\left({e}^{{q}} \right) \\ $$$$\forall{a}\:\left\{{a}\:\in\:\mathbb{R}\:\mid\:{a}^{−\infty} \::=\:\mathrm{0}\right\} \\ $$$$\Rightarrow\:\mathrm{1}\:−\:\underset{{q}\rightarrow−\infty} {\mathrm{lim}}\left({e}^{{q}} \right)\:=\:\mathrm{1}\:−\:\mathrm{0}\:=\:\mathrm{1} \\ $$$$\Rightarrow\:\int_{−\infty} ^{\mathrm{0}} {e}^{−\mid{t}\mid} {dt}\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare\:\mathrm{Q}.\mathrm{E}.\mathrm{D}. \\ $$

Commented by Zaynal last updated on 09/Jan/22

thank you sir. perpect

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

Answered by MJS_new last updated on 08/Jan/22

−∞<t≤0 ⇒ ∣t∣=−t ∫_(−∞) ^0 e^(−∣t∣) dt=∫_(−∞) ^0 e^t dt=[e^t ]_(−∞) ^0 =0−(−1)=1

$$−\infty<{t}\leqslant\mathrm{0}\:\Rightarrow\:\mid{t}\mid=−{t} \\ $$$$\underset{−\infty} {\overset{\mathrm{0}} {\int}}\mathrm{e}^{−\mid{t}\mid} {dt}=\underset{−\infty} {\overset{\mathrm{0}} {\int}}\mathrm{e}^{{t}} {dt}=\left[\mathrm{e}^{{t}} \right]_{−\infty} ^{\mathrm{0}} =\mathrm{0}−\left(−\mathrm{1}\right)=\mathrm{1} \\ $$

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