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Question Number 136950 by BHOOPENDRA last updated on 28/Mar/21

	Answered by Olaf last updated on 28/Mar/21

	$\overset{\land}{f}^{c} (ν) = \int_{- \infty}^{+ \infty} f (s) \cos (2 π ν s) d s = Re \overset{\land}{f} (ν)$ $F (\frac{1}{s}) = - i π sign (ν)$ $F (g (s) \sin (a s)) = \frac{1}{2} [\overset{\land}{g} (ν - \frac{a}{2 π}) + \overset{\land}{g} (ν + \frac{a}{2 π})]$ $F (\frac{\sin (a s)}{s}) = \frac{1}{2} [- i π sign (ν - \frac{a}{2 π}) - i π sign (ν + \frac{a}{2 π})]$ $F (\frac{\sin (a s)}{s}) = - \frac{i π}{2} [sign (ν - \frac{a}{2 π}) + sign (ν + \frac{a}{2 π})]$ $\overset{\land}{f}^{c} (ν) = Re \overset{\land}{f} (ν) = 0 ? ? ? ?$
	Answered by mathmax by abdo last updated on 28/Mar/21
	$f^★ (x)=(1/( (√(2π))))∫_R f(t)e^(−ixt) dt =(1/( (√(2π))))∫_(−∞) ^∞ f(t)e^(−ixt) dt =(√(2/π))∫_0 ^∞ f(t)cos(xt)dt(if f is even) ⇒ F^∧ (((sin(ax))/x))=(√(2/π))∫_0 ^∞ ((sin(at))/t)cos(xt)dt we have cosp sinq =cosp.cos((π/2)−q) =(1/2){cos(p−q+(π/2))+cos(p+q−(π/2))) =(1/2){−sin(p−q)+sin(p+q)} ⇒cos(xt)sin(at) =(1/2){sin(x+a)t−sin(x−a)t} ⇒ case 1 x>a ∫_0 ^∞ ((cos(xt)sin(at))/t)dt =(1/2)∫_0 ^∞ ((sin(x+a)t)/t)dt(→(x+a)t=z)−(1/2)∫_0 ^∞ ((sin(x−a)t)/t)dt(→(x−a)t=z) =(1/2)∫_0 ^∞ ((sinz)/(z/(x+a)))(dz/(x+a)) −(1/2)∫_0 ^∞ ((sinz)/(z/(x−a)))(dz/(x−a))=0 ⇒f^∧ (x)=0 case 2 −a<x<a ⇒∫_0 ^∞ ((cos(xt)sin(at))/t)dt =(1/2)∫_0 ^∞ ((sinz)/z)dz +(1/2)∫_0 ^∞ ((sin(a−x)t)/t)dt =∫_0 ^∞ ((sinz)/z)dz=(π/2) ⇒f^∧ (x)=(√(2/π)).(π/2)=((π(√2))/(2(√π))) =(√(π/2))$
	$f^{★} (x) = \frac{1}{\sqrt{2 π}} \int_{R} f (t) e^{- ixt} dt = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} f (t) e^{- ixt} dt$ $= \sqrt{\frac{2}{π}} \int_{0}^{\infty} f (t) \cos (xt) dt (if f is even) \Rightarrow F^{\land} (\frac{\sin (ax)}{x}) = \sqrt{\frac{2}{π}} \int_{0}^{\infty} \frac{\sin (at)}{t} \cos (xt) dt$ $we have cosp sinq = cosp . \cos (\frac{π}{2} - q)$ $= \frac{1}{2} {\cos (p - q + \frac{π}{2}) + \cos (p + q - \frac{π}{2}))$ $= \frac{1}{2} {- \sin (p - q) + \sin (p + q)} \Rightarrow \cos (xt) \sin (at)$ $= \frac{1}{2} {\sin (x + a) t - \sin (x - a) t} \Rightarrow case 1 x > a$ $\int_{0}^{\infty} \frac{\cos (xt) \sin (at)}{t} dt = \frac{1}{2} \int_{0}^{\infty} \frac{\sin (x + a) t}{t} dt (\to (x + a) t = z) - \frac{1}{2} \int_{0}^{\infty} \frac{\sin (x - a) t}{t} dt (\to (x - a) t = z)$ $= \frac{1}{2} \int_{0}^{\infty} \frac{sinz}{\frac{z}{x + a}} \frac{dz}{x + a} - \frac{1}{2} \int_{0}^{\infty} \frac{sinz}{\frac{z}{x - a}} \frac{dz}{x - a} = 0 \Rightarrow f^{\land} (x) = 0$ $case 2 - a < x < a \Rightarrow \int_{0}^{\infty} \frac{\cos (xt) \sin (at)}{t} dt$ $= \frac{1}{2} \int_{0}^{\infty} \frac{sinz}{z} dz + \frac{1}{2} \int_{0}^{\infty} \frac{\sin (a - x) t}{t} dt = \int_{0}^{\infty} \frac{sinz}{z} dz = \frac{π}{2}$ $\Rightarrow f^{\land} (x) = \sqrt{\frac{2}{π}} . \frac{π}{2} = \frac{π \sqrt{2}}{2 \sqrt{π}} = \sqrt{\frac{π}{2}}$
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