Menu Close

Question-136950




Question Number 136950 by BHOOPENDRA last updated on 28/Mar/21
Answered by Olaf last updated on 28/Mar/21
f^∧ ^c (ν) = ∫_(−∞) ^(+∞) f(s)cos(2πνs)ds = Ref^∧ (ν)  F((1/s)) = −iπsign(ν)  F(g(s)sin(as)) = (1/2)[g^∧ (ν−(a/(2π)))+g^∧ (ν+(a/(2π)))]  F(((sin(as))/s)) = (1/2)[−iπsign(ν−(a/(2π)))−iπsign(ν+(a/(2π)))]  F(((sin(as))/s)) = −((iπ)/2)[sign(ν−(a/(2π)))+sign(ν+(a/(2π)))]  f^∧ ^c (ν) = Ref^∧ (ν) = 0 ????
$$\overset{\wedge} {{f}}\:^{{c}} \left(\nu\right)\:=\:\int_{−\infty} ^{+\infty} {f}\left({s}\right)\mathrm{cos}\left(\mathrm{2}\pi\nu{s}\right){ds}\:=\:\mathrm{Re}\overset{\wedge} {{f}}\left(\nu\right) \\ $$$$\mathcal{F}\left(\frac{\mathrm{1}}{{s}}\right)\:=\:−{i}\pi\mathrm{sign}\left(\nu\right) \\ $$$$\mathcal{F}\left({g}\left({s}\right)\mathrm{sin}\left({as}\right)\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\overset{\wedge} {{g}}\left(\nu−\frac{{a}}{\mathrm{2}\pi}\right)+\overset{\wedge} {{g}}\left(\nu+\frac{{a}}{\mathrm{2}\pi}\right)\right] \\ $$$$\mathcal{F}\left(\frac{\mathrm{sin}\left({as}\right)}{{s}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[−{i}\pi\mathrm{sign}\left(\nu−\frac{{a}}{\mathrm{2}\pi}\right)−{i}\pi\mathrm{sign}\left(\nu+\frac{{a}}{\mathrm{2}\pi}\right)\right] \\ $$$$\mathcal{F}\left(\frac{\mathrm{sin}\left({as}\right)}{{s}}\right)\:=\:−\frac{{i}\pi}{\mathrm{2}}\left[\mathrm{sign}\left(\nu−\frac{{a}}{\mathrm{2}\pi}\right)+\mathrm{sign}\left(\nu+\frac{{a}}{\mathrm{2}\pi}\right)\right] \\ $$$$\overset{\wedge} {{f}}\:^{{c}} \left(\nu\right)\:=\:\mathrm{Re}\overset{\wedge} {{f}}\left(\nu\right)\:=\:\mathrm{0}\:???? \\ $$
Answered by mathmax by abdo last updated on 28/Mar/21
f^★ (x)=(1/( (√(2π))))∫_R f(t)e^(−ixt)  dt  =(1/( (√(2π))))∫_(−∞) ^∞  f(t)e^(−ixt)  dt  =(√(2/π))∫_0 ^∞ f(t)cos(xt)dt(if f is even) ⇒ F^∧ (((sin(ax))/x))=(√(2/π))∫_0 ^∞ ((sin(at))/t)cos(xt)dt  we have cosp sinq =cosp.cos((π/2)−q)  =(1/2){cos(p−q+(π/2))+cos(p+q−(π/2)))  =(1/2){−sin(p−q)+sin(p+q)} ⇒cos(xt)sin(at)  =(1/2){sin(x+a)t−sin(x−a)t} ⇒ case 1   x>a  ∫_0 ^∞  ((cos(xt)sin(at))/t)dt =(1/2)∫_0 ^∞  ((sin(x+a)t)/t)dt(→(x+a)t=z)−(1/2)∫_0 ^∞ ((sin(x−a)t)/t)dt(→(x−a)t=z)  =(1/2)∫_0 ^∞  ((sinz)/(z/(x+a)))(dz/(x+a))  −(1/2)∫_0 ^∞  ((sinz)/(z/(x−a)))(dz/(x−a))=0 ⇒f^∧ (x)=0  case 2 −a<x<a ⇒∫_0 ^∞  ((cos(xt)sin(at))/t)dt  =(1/2)∫_0 ^∞  ((sinz)/z)dz +(1/2)∫_0 ^∞  ((sin(a−x)t)/t)dt =∫_0 ^∞  ((sinz)/z)dz=(π/2)  ⇒f^∧ (x)=(√(2/π)).(π/2)=((π(√2))/(2(√π))) =(√(π/2))
$$\mathrm{f}^{\bigstar} \left(\mathrm{x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\pi}}\int_{\mathrm{R}} \mathrm{f}\left(\mathrm{t}\right)\mathrm{e}^{−\mathrm{ixt}} \:\mathrm{dt}\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\pi}}\int_{−\infty} ^{\infty} \:\mathrm{f}\left(\mathrm{t}\right)\mathrm{e}^{−\mathrm{ixt}} \:\mathrm{dt} \\ $$$$=\sqrt{\frac{\mathrm{2}}{\pi}}\int_{\mathrm{0}} ^{\infty} \mathrm{f}\left(\mathrm{t}\right)\mathrm{cos}\left(\mathrm{xt}\right)\mathrm{dt}\left(\mathrm{if}\:\mathrm{f}\:\mathrm{is}\:\mathrm{even}\right)\:\Rightarrow\:\mathrm{F}^{\wedge} \left(\frac{\mathrm{sin}\left(\mathrm{ax}\right)}{\mathrm{x}}\right)=\sqrt{\frac{\mathrm{2}}{\pi}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left(\mathrm{at}\right)}{\mathrm{t}}\mathrm{cos}\left(\mathrm{xt}\right)\mathrm{dt} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{cosp}\:\mathrm{sinq}\:=\mathrm{cosp}.\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{q}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{cos}\left(\mathrm{p}−\mathrm{q}+\frac{\pi}{\mathrm{2}}\right)+\mathrm{cos}\left(\mathrm{p}+\mathrm{q}−\frac{\pi}{\mathrm{2}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{−\mathrm{sin}\left(\mathrm{p}−\mathrm{q}\right)+\mathrm{sin}\left(\mathrm{p}+\mathrm{q}\right)\right\}\:\Rightarrow\mathrm{cos}\left(\mathrm{xt}\right)\mathrm{sin}\left(\mathrm{at}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{sin}\left(\mathrm{x}+\mathrm{a}\right)\mathrm{t}−\mathrm{sin}\left(\mathrm{x}−\mathrm{a}\right)\mathrm{t}\right\}\:\Rightarrow\:\mathrm{case}\:\mathrm{1}\:\:\:\mathrm{x}>\mathrm{a} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{xt}\right)\mathrm{sin}\left(\mathrm{at}\right)}{\mathrm{t}}\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left(\mathrm{x}+\mathrm{a}\right)\mathrm{t}}{\mathrm{t}}\mathrm{dt}\left(\rightarrow\left(\mathrm{x}+\mathrm{a}\right)\mathrm{t}=\mathrm{z}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left(\mathrm{x}−\mathrm{a}\right)\mathrm{t}}{\mathrm{t}}\mathrm{dt}\left(\rightarrow\left(\mathrm{x}−\mathrm{a}\right)\mathrm{t}=\mathrm{z}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sinz}}{\frac{\mathrm{z}}{\mathrm{x}+\mathrm{a}}}\frac{\mathrm{dz}}{\mathrm{x}+\mathrm{a}}\:\:−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sinz}}{\frac{\mathrm{z}}{\mathrm{x}−\mathrm{a}}}\frac{\mathrm{dz}}{\mathrm{x}−\mathrm{a}}=\mathrm{0}\:\Rightarrow\mathrm{f}^{\wedge} \left(\mathrm{x}\right)=\mathrm{0} \\ $$$$\mathrm{case}\:\mathrm{2}\:−\mathrm{a}<\mathrm{x}<\mathrm{a}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{xt}\right)\mathrm{sin}\left(\mathrm{at}\right)}{\mathrm{t}}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sinz}}{\mathrm{z}}\mathrm{dz}\:+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left(\mathrm{a}−\mathrm{x}\right)\mathrm{t}}{\mathrm{t}}\mathrm{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sinz}}{\mathrm{z}}\mathrm{dz}=\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{f}^{\wedge} \left(\mathrm{x}\right)=\sqrt{\frac{\mathrm{2}}{\pi}}.\frac{\pi}{\mathrm{2}}=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\pi}}\:=\sqrt{\frac{\pi}{\mathrm{2}}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *