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let-x-1-a-2-x-2-if-x-lt-a-and-x-0-if-x-a-find-the-fourier-transform-of-




Question Number 35632 by abdo mathsup 649 cc last updated on 21/May/18
let ϕ(x)= (1/( (√(a^2 −x^2 ))))  if ∣x∣<a  and ϕ(x)=0 if ∣x∣≥a  find the fourier transform of ϕ .
$${let}\:\varphi\left({x}\right)=\:\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }}\:\:{if}\:\mid{x}\mid<{a}\:\:{and}\:\varphi\left({x}\right)=\mathrm{0}\:{if}\:\mid{x}\mid\geqslant{a} \\ $$$${find}\:{the}\:{fourier}\:{transform}\:{of}\:\varphi\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 21/May/18
F(f(x)) = (1/( (√(2π))))∫_(−∞) ^(+∞)   f(t)e^(−ixt)  dt .
$${F}\left({f}\left({x}\right)\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\pi}}\int_{−\infty} ^{+\infty} \:\:{f}\left({t}\right){e}^{−{ixt}} \:{dt}\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 24/May/18
we have F(ϕ(x))= (1/( (√(2π)))) ∫_(−∞) ^(+∞)  ϕ(t) e^(−ixt)  dt and ϕ is even  F(ϕ(x))o= (1/( (√(2π)))) ∫_(−∞) ^(+∞)  ϕ(t) e^(−ixt)  dt  =(1/( (√(2π)))) ∫_(−a) ^a    (e^(−ixt) /( (√(a^2  −t^2 )))) dt  =(√(2/π))  ∫_0 ^a    ((cos(xt))/( (√(a^2  −t^2 ))))dt   let find   w(x) =∫_0 ^a       ((cos(xt))/( (√(a^2  −t^2 ))))dt we hsve  w^′ (x) = ∫_0 ^a   ((−t sin(xt))/( (√(a^2  −t^2 ))))   and by parts  w^′ (x)  = [ (√(a^2  −t^2 )) sin(xt)]_0 ^a  −∫_0 ^a  (√(a^2  −t^2 )) x cos(xt)dt  = −x ∫_0 ^a   (√(a^2  −t^2  ))  cos(xt)dt   chsngement t =asinα  give w^′ (x) = ∫_0 ^(π/2)   a cos(α)cos(ax sinα) a cosα dα  = a^2   ∫_0 ^(π/2)   cos^2 (α) cos(ax sinα)dα  =a^2   ∫_0 ^(π/2)  cos(α) (cos(α) cos(ax sinα))dα  =a^2   { (1/(ax)) sin(ax sin(α))cos(α)]_0 ^(π/2)   − ∫_0 ^(π/2)   −sin(α) sin(ax sinα) dα}  = ∫_0 ^(π/2)   sin(α) sin(ax sinα)dα ....be continued...
$${we}\:{have}\:{F}\left(\varphi\left({x}\right)\right)=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\pi}}\:\int_{−\infty} ^{+\infty} \:\varphi\left({t}\right)\:{e}^{−{ixt}} \:{dt}\:{and}\:\varphi\:{is}\:{even} \\ $$$${F}\left(\varphi\left({x}\right)\right){o}=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\pi}}\:\int_{−\infty} ^{+\infty} \:\varphi\left({t}\right)\:{e}^{−{ixt}} \:{dt} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\pi}}\:\int_{−{a}} ^{{a}} \:\:\:\frac{{e}^{−{ixt}} }{\:\sqrt{{a}^{\mathrm{2}} \:−{t}^{\mathrm{2}} }}\:{dt} \\ $$$$=\sqrt{\frac{\mathrm{2}}{\pi}}\:\:\int_{\mathrm{0}} ^{{a}} \:\:\:\frac{{cos}\left({xt}\right)}{\:\sqrt{{a}^{\mathrm{2}} \:−{t}^{\mathrm{2}} }}{dt}\:\:\:{let}\:{find}\: \\ $$$${w}\left({x}\right)\:=\int_{\mathrm{0}} ^{{a}} \:\:\:\:\:\:\frac{{cos}\left({xt}\right)}{\:\sqrt{{a}^{\mathrm{2}} \:−{t}^{\mathrm{2}} }}{dt}\:{we}\:{hsve} \\ $$$${w}^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{{a}} \:\:\frac{−{t}\:{sin}\left({xt}\right)}{\:\sqrt{{a}^{\mathrm{2}} \:−{t}^{\mathrm{2}} }}\:\:\:{and}\:{by}\:{parts} \\ $$$${w}^{'} \left({x}\right)\:\:=\:\left[\:\sqrt{{a}^{\mathrm{2}} \:−{t}^{\mathrm{2}} }\:{sin}\left({xt}\right)\right]_{\mathrm{0}} ^{{a}} \:−\int_{\mathrm{0}} ^{{a}} \:\sqrt{{a}^{\mathrm{2}} \:−{t}^{\mathrm{2}} }\:{x}\:{cos}\left({xt}\right){dt} \\ $$$$=\:−{x}\:\int_{\mathrm{0}} ^{{a}} \:\:\sqrt{{a}^{\mathrm{2}} \:−{t}^{\mathrm{2}} \:}\:\:{cos}\left({xt}\right){dt}\:\:\:{chsngement}\:{t}\:={asin}\alpha \\ $$$${give}\:{w}^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{a}\:{cos}\left(\alpha\right){cos}\left({ax}\:{sin}\alpha\right)\:{a}\:{cos}\alpha\:{d}\alpha \\ $$$$=\:{a}^{\mathrm{2}} \:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{cos}^{\mathrm{2}} \left(\alpha\right)\:{cos}\left({ax}\:{sin}\alpha\right){d}\alpha \\ $$$$={a}^{\mathrm{2}} \:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}\left(\alpha\right)\:\left({cos}\left(\alpha\right)\:{cos}\left({ax}\:{sin}\alpha\right)\right){d}\alpha \\ $$$$={a}^{\mathrm{2}} \:\:\left\{\:\frac{\mathrm{1}}{{ax}}\:{sin}\left({ax}\:{sin}\left(\alpha\right)\right){cos}\left(\alpha\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\left.−\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:−{sin}\left(\alpha\right)\:{sin}\left({ax}\:{sin}\alpha\right)\:{d}\alpha\right\} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{sin}\left(\alpha\right)\:{sin}\left({ax}\:{sin}\alpha\right){d}\alpha\:….{be}\:{continued}… \\ $$

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