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Question Number 39035 by maxmathsup by imad last updated on 01/Jul/18
find f(t) =∫_0 ^∞ sin(x)e^(−t [x]) dx with t>0
$${find}\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:{sin}\left({x}\right){e}^{−{t}\:\left[{x}\right]} {dx}\:\:\:{with}\:{t}>\mathrm{0} \\ $$
Commented by maxmathsup by imad last updated on 02/Jul/18
we have f(t)=Σ_(n=0) ^∞ ∫_n ^(n+1) sin(x) e^(−nt) dx =Σ_(n=0) ^∞ e^(−nt) {cos(n) −cos(n+1)} =Σ_(n=0) ^∞ e^(−nt) cos(n) −Σ_(n=0) ^∞ e^(−nt) cos(n+1) =Σ_(n=0) ^∞ e^(−nt) cos(n) −Σ_(n=1) ^∞ e^(−(n−1)t) cos(n) = 1+(1−e^t )Σ_(n=1) ^∞ e^(−nt) cos(n) but Σ_(n=1) ^∞ e^(−nt) cos(n) =Σ_(n=0) ^∞ e^(−nt) cos(n) −1 =Re( Σ_(n=0) ^∞ e^(−nt +in) ) =Re( Σ_(n=0) ^∞ (e^(−t+i) )^n ) but Σ_(n=0) ^∞ (e^(−t+i) )^n = (1/(1−e^(−t+i) )) = (1/(1−e^(−t) (cos1 +isin(1)))) = (1/(1−e^(−t) cos(1)−i e^(−t) sin(1))) = ((1−e^(−t) cos(1) +i e^(−t) sin(1))/((1−e^(−t) cos(1))^2 +e^(−2t) sin^2 (1))) ⇒ f(t) =1+(1−e^t ){ ((1−e^(−t) cos(1))/((1−e^(−t) cos(1))^2 +e^(−2t) sin^2 (1))) −1}
$${we}\:{have}\:{f}\left({t}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\int_{{n}} ^{{n}+\mathrm{1}} \:{sin}\left({x}\right)\:{e}^{−{nt}} \:{dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:{e}^{−{nt}} \:\left\{{cos}\left({n}\right)\:−{cos}\left({n}+\mathrm{1}\right)\right\} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{nt}} \:{cos}\left({n}\right)\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{nt}} \:{cos}\left({n}+\mathrm{1}\right) \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{nt}} {cos}\left({n}\right)\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{−\left({n}−\mathrm{1}\right){t}} {cos}\left({n}\right) \\ $$$$=\:\mathrm{1}+\left(\mathrm{1}−{e}^{{t}} \right)\sum_{{n}=\mathrm{1}} ^{\infty} \:\:{e}^{−{nt}} \:{cos}\left({n}\right)\:{but} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{−{nt}} \:{cos}\left({n}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{nt}} {cos}\left({n}\right)\:−\mathrm{1} \\ $$$$={Re}\left(\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{nt}\:+{in}} \right)\:={Re}\left(\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left({e}^{−{t}+{i}} \right)^{{n}} \right)\:{but} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\left({e}^{−{t}+{i}} \right)^{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{t}+{i}} }\:=\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{t}} \left({cos}\mathrm{1}\:+{isin}\left(\mathrm{1}\right)\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−{t}} \:{cos}\left(\mathrm{1}\right)−{i}\:{e}^{−{t}} {sin}\left(\mathrm{1}\right)}\:=\:\frac{\mathrm{1}−{e}^{−{t}} {cos}\left(\mathrm{1}\right)\:+{i}\:{e}^{−{t}} {sin}\left(\mathrm{1}\right)}{\left(\mathrm{1}−{e}^{−{t}} {cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:+{e}^{−\mathrm{2}{t}} \:{sin}^{\mathrm{2}} \left(\mathrm{1}\right)}\:\Rightarrow \\ $$$${f}\left({t}\right)\:=\mathrm{1}+\left(\mathrm{1}−{e}^{{t}} \right)\left\{\:\frac{\mathrm{1}−{e}^{−{t}} {cos}\left(\mathrm{1}\right)}{\left(\mathrm{1}−{e}^{−{t}} {cos}\left(\mathrm{1}\right)\right)^{\mathrm{2}} \:+{e}^{−\mathrm{2}{t}} {sin}^{\mathrm{2}} \left(\mathrm{1}\right)}\:−\mathrm{1}\right\} \\ $$$$ \\ $$

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