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