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Question Number 219355 by SdC355 last updated on 23/Apr/25
∫_0 ^( ∞) ((sin(z))/z)e^(−zt) dz=??
$$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{sin}\left({z}\right)}{{z}}{e}^{−{zt}} \mathrm{d}{z}=?? \\ $$
Answered by breniam last updated on 23/Apr/25
=I(t) I′(t)=lim_(t_0 →t) ∫_0 ^∞ ((sin(z))/z)×((e^(−zt) −e^(−zt_0 ) )/(t−t_0 ))dz By the mean value theorem, f.a. t∈R and f.a. t_0 ∈R\{t} there exists ξ∈(min({t,t_0 }),max({t,t_0 })) s.t. ((sin(z))/z)×((e^(−zt) −e^(−zt_0 ) )/(t−t_0 ))=−sin(z)e^(−zξ) Let ξ(t_0 ) be any function which for every t_0 ∈R\{t} gives one of the values guaranteed by theorem. Then, for each t_0 close enought to t we have −sin(z)e^(−2z) ≥((sin(z))/z)×((e^(−zt) −e^(−zt_0 ) )/(t−t_0 ))=−sin(z)e^(−zξ(t_0 )) ≥−sin(z) What more, lim_(t_0 →t) ((sin(z))/z)×((e^(−zt) −e^(−zt_0 ) )/(t−t_0 ))=−sin(z)e^(−tz) So, by Lebesgues diminated convergence theorem, for each (t_0 )_n convergent to t I′(t)=lim_(n→∞) ∫_0 ^∞ ((sin(z))/z)×((e^(−zt) −e^(−z(t_0 )_n ) )/(t−(t_0 )_n ))dz= I′(t)=∫_0 ^∞ lim_(n→∞) ((sin(z))/z)×((e^(−zt) −e^(−z(t_0 )_n ) )/(t−(t_0 )_n ))dz=−∫_0 ^∞ sin(z)e^(−tz) dz So, by Haine′s continous limit definition I′(t)=lim_(t_0 →t) ∫_0 ^∞ ((sin(z))/z)×((e^(−zt) −e^(−zt_0 ) )/(t−t_0 ))dz=−∫_0 ^∞ sin(z)e^(−tz) dz= −∫_0 ^∞ sin(z)e^(−tz) dz=∫_0 ^∞ (cos(z))′e^(−tz) dz=−1+t∫_0 ^∞ cos(z)e^(−tz) dz=−1+t∫_0 ^∞ (sin(z))′e^(−tz) dz=−1−t^2 ∫_0 ^∞ sin(z)e^(−tz) dz⇒ ⇒(t^2 +1)∫_0 ^∞ sin(z)e^(−tz) dz=−1⇒∫_0 ^∞ sin(z)e^(−tz) dz=((−1)/(t^2 +1)) I(t)∈−∫(1/(t^2 +1))dt=−arctan(t)+C lim_(t→∞) I(t)=0⇒C=(π/2)
$$={I}\left({t}\right) \\ $$$${I}'\left({t}\right)=\underset{{t}_{\mathrm{0}} \rightarrow{t}} {\mathrm{lim}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{sin}\left({z}\right)}{{z}}×\frac{{e}^{−{zt}} −{e}^{−{zt}_{\mathrm{0}} } }{{t}−{t}_{\mathrm{0}} }\mathrm{d}{z} \\ $$$$\mathrm{By}\:\mathrm{the}\:\mathrm{mean}\:\mathrm{value}\:\mathrm{theorem},\:\mathrm{f}.\mathrm{a}.\:{t}\in\mathbb{R}\:\mathrm{and}\:\mathrm{f}.\mathrm{a}.\:{t}_{\mathrm{0}} \in\mathbb{R}\backslash\left\{{t}\right\} \\ $$$$\mathrm{there}\:\mathrm{exists}\:\xi\in\left(\mathrm{min}\left(\left\{{t},{t}_{\mathrm{0}} \right\}\right),\mathrm{max}\left(\left\{{t},{t}_{\mathrm{0}} \right\}\right)\right)\:\mathrm{s}.\mathrm{t}. \\ $$$$\frac{\mathrm{sin}\left({z}\right)}{{z}}×\frac{{e}^{−{zt}} −{e}^{−{zt}_{\mathrm{0}} } }{{t}−{t}_{\mathrm{0}} }=−\mathrm{sin}\left({z}\right){e}^{−{z}\xi} \\ $$$$\mathrm{Let}\:\xi\left({t}_{\mathrm{0}} \right)\:\mathrm{be}\:\mathrm{any}\:\mathrm{function}\:\mathrm{which} \\ $$$$\mathrm{for}\:\mathrm{every}\:{t}_{\mathrm{0}} \in\mathbb{R}\backslash\left\{{t}\right\}\:\mathrm{gives}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{values}\:\mathrm{guaranteed}\:\mathrm{by}\:\mathrm{theorem}.\: \\ $$$$\mathrm{Then},\:\mathrm{for}\:\mathrm{each}\:{t}_{\mathrm{0}} \:\mathrm{close}\:\mathrm{enought}\:\mathrm{to}\:{t}\:\mathrm{we}\:\mathrm{have} \\ $$$$−\mathrm{sin}\left({z}\right){e}^{−\mathrm{2}{z}} \geqslant\frac{\mathrm{sin}\left({z}\right)}{{z}}×\frac{{e}^{−{zt}} −{e}^{−{zt}_{\mathrm{0}} } }{{t}−{t}_{\mathrm{0}} }=−\mathrm{sin}\left({z}\right){e}^{−{z}\xi\left({t}_{\mathrm{0}} \right)} \geqslant−\mathrm{sin}\left({z}\right) \\ $$$$\mathrm{What}\:\mathrm{more}, \\ $$$$\underset{{t}_{\mathrm{0}} \rightarrow{t}} {\mathrm{lim}}\frac{\mathrm{sin}\left({z}\right)}{{z}}×\frac{{e}^{−{zt}} −{e}^{−{zt}_{\mathrm{0}} } }{{t}−{t}_{\mathrm{0}} }=−\mathrm{sin}\left({z}\right){e}^{−{tz}} \\ $$$$\mathrm{So},\:\mathrm{by}\:\mathrm{Lebesgues}\:\mathrm{diminated}\:\mathrm{convergence}\:\mathrm{theorem},\:\mathrm{for}\:\mathrm{each}\:\left({t}_{\mathrm{0}} \right)_{{n}} \:\mathrm{convergent}\:\mathrm{to}\:{t} \\ $$$${I}'\left({t}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{sin}\left({z}\right)}{{z}}×\frac{{e}^{−{zt}} −{e}^{−{z}\left({t}_{\mathrm{0}} \right)_{{n}} } }{{t}−\left({t}_{\mathrm{0}} \right)_{{n}} }\mathrm{d}{z}= \\ $$$${I}'\left({t}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{sin}\left({z}\right)}{{z}}×\frac{{e}^{−{zt}} −{e}^{−{z}\left({t}_{\mathrm{0}} \right)_{{n}} } }{{t}−\left({t}_{\mathrm{0}} \right)_{{n}} }\mathrm{d}{z}=−\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{sin}\left({z}\right){e}^{−{tz}} \mathrm{d}{z} \\ $$$$\mathrm{So},\:\mathrm{by}\:\mathrm{Haine}'\mathrm{s}\:\mathrm{continous}\:\mathrm{limit}\:\mathrm{definition} \\ $$$${I}'\left({t}\right)=\underset{{t}_{\mathrm{0}} \rightarrow{t}} {\mathrm{lim}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{sin}\left({z}\right)}{{z}}×\frac{{e}^{−{zt}} −{e}^{−{zt}_{\mathrm{0}} } }{{t}−{t}_{\mathrm{0}} }\mathrm{d}{z}=−\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{sin}\left({z}\right){e}^{−{tz}} \mathrm{d}{z}= \\ $$$$−\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{sin}\left({z}\right){e}^{−{tz}} \mathrm{d}{z}=\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\mathrm{cos}\left({z}\right)\right)'{e}^{−{tz}} \mathrm{d}{z}=−\mathrm{1}+{t}\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{cos}\left({z}\right){e}^{−{tz}} \mathrm{d}{z}=−\mathrm{1}+{t}\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\mathrm{sin}\left({z}\right)\right)'{e}^{−{tz}} \mathrm{d}{z}=−\mathrm{1}−{t}^{\mathrm{2}} \underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{sin}\left({z}\right){e}^{−{tz}} \mathrm{d}{z}\Rightarrow \\ $$$$\Rightarrow\left({t}^{\mathrm{2}} +\mathrm{1}\right)\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{sin}\left({z}\right){e}^{−{tz}} \mathrm{d}{z}=−\mathrm{1}\Rightarrow\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{sin}\left({z}\right){e}^{−{tz}} \mathrm{d}{z}=\frac{−\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$${I}\left({t}\right)\in−\int\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{t}=−\mathrm{arctan}\left({t}\right)+{C} \\ $$$$\underset{{t}\rightarrow\infty} {\mathrm{lim}}{I}\left({t}\right)=\mathrm{0}\Rightarrow{C}=\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$

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