Question Number 35992 by abdo mathsup 649 cc last updated on 26/May/18
![let f(x)= ((sin(2x))/x) χ_(]−a,a[) (x) with a>0 calculate the fourier trsnsform of f .](Q35992.png)
$${let}\:{f}\left({x}\right)=\:\frac{{sin}\left(\mathrm{2}{x}\right)}{{x}}\:\chi_{\left.\right]−{a},{a}\left[\right.} \left({x}\right)\:\:{with}\:{a}>\mathrm{0} \\ $$ $${calculate}\:{the}\:{fourier}\:{trsnsform}\:{of}\:{f}\:. \\ $$
Commented byabdo mathsup 649 cc last updated on 27/May/18
![F(f(x))= (1/(√(2π))) ∫_(−∞) ^(+∞) f(t) e^(−ixt) dt = (1/(√(2π))) ∫_(−a) ^a ((sin(2t))/t) e^(−ixt) dt = (1/(√(2π))) ∫_(−a) ^a ((sin(2t))/t) cos(xt)dt =(√((2/π) )) ∫_0 ^a ((sin(2t)cos(xt))/t)dt=(√(2/π)) w(x) w(x)= ∫_0 ^a ((sin(2t)cos(xt))/t)dt w^′ (x) = −∫_0 ^a sin(2t)sin(xt)dt but sin(2t)sin(xt)=(1/2){ cos(x−2)t−cos(x+2)t}⇒ w^′ (x) =−(1/2) { ∫_0 ^a cos(x−2)tdt −∫_0 ^a cos(x+2)t dt} = (1/2)[ (1/(x+2)) sin(x+2)t]_0 ^a −(1/2)[(1/(x−2))sin(x−2)t]_0 ^a = (1/(2x+4)) sin{(x+2)a} −(1/(2x−4))sin{(x−2)a}⇒ w(x) = ∫_0 ^x ((sin{(t+2)a})/(2t+4))dt −∫_0 ^x ((sin{(t−2)a})/(2t−4))dt +c ( c=w(0)= ∫_0 ^a ((sin(2t))/t)dt)...be continued...](Q36048.png)
$${F}\left({f}\left({x}\right)\right)=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}\pi}}\:\int_{−\infty} ^{+\infty} \:\:{f}\left({t}\right)\:{e}^{−{ixt}} \:{dt}\: \\ $$ $$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}\pi}}\:\int_{−{a}} ^{{a}} \:\frac{{sin}\left(\mathrm{2}{t}\right)}{{t}}\:{e}^{−{ixt}} \:{dt} \\ $$ $$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}\pi}}\:\int_{−{a}} ^{{a}} \:\frac{{sin}\left(\mathrm{2}{t}\right)}{{t}}\:{cos}\left({xt}\right){dt} \\ $$ $$=\sqrt{\frac{\mathrm{2}}{\pi}\:}\:\:\int_{\mathrm{0}} ^{{a}} \:\:\:\frac{{sin}\left(\mathrm{2}{t}\right){cos}\left({xt}\right)}{{t}}{dt}=\sqrt{\frac{\mathrm{2}}{\pi}}\:{w}\left({x}\right) \\ $$ $${w}\left({x}\right)=\:\int_{\mathrm{0}} ^{{a}} \:\:\frac{{sin}\left(\mathrm{2}{t}\right){cos}\left({xt}\right)}{{t}}{dt} \\ $$ $${w}^{'} \left({x}\right)\:=\:−\int_{\mathrm{0}} ^{{a}} \:{sin}\left(\mathrm{2}{t}\right){sin}\left({xt}\right){dt}\:\:{but} \\ $$ $${sin}\left(\mathrm{2}{t}\right){sin}\left({xt}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{cos}\left({x}−\mathrm{2}\right){t}−{cos}\left({x}+\mathrm{2}\right){t}\right\}\Rightarrow \\ $$ $${w}^{'} \left({x}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\:\:\int_{\mathrm{0}} ^{{a}} \:{cos}\left({x}−\mathrm{2}\right){tdt}\:−\int_{\mathrm{0}} ^{{a}} \:{cos}\left({x}+\mathrm{2}\right){t}\:{dt}\right\} \\ $$ $$=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\:\frac{\mathrm{1}}{{x}+\mathrm{2}}\:{sin}\left({x}+\mathrm{2}\right){t}\right]_{\mathrm{0}} ^{{a}} \:−\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{{x}−\mathrm{2}}{sin}\left({x}−\mathrm{2}\right){t}\right]_{\mathrm{0}} ^{{a}} \\ $$ $$=\:\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{4}}\:{sin}\left\{\left({x}+\mathrm{2}\right){a}\right\}\:−\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{4}}{sin}\left\{\left({x}−\mathrm{2}\right){a}\right\}\Rightarrow \\ $$ $${w}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{{x}} \:\:\frac{{sin}\left\{\left({t}+\mathrm{2}\right){a}\right\}}{\mathrm{2}{t}+\mathrm{4}}{dt}\:−\int_{\mathrm{0}} ^{{x}} \:\frac{{sin}\left\{\left({t}−\mathrm{2}\right){a}\right\}}{\mathrm{2}{t}−\mathrm{4}}{dt}\:+{c} \\ $$ $$\left(\:{c}={w}\left(\mathrm{0}\right)=\:\int_{\mathrm{0}} ^{{a}} \:\:\frac{{sin}\left(\mathrm{2}{t}\right)}{{t}}{dt}\right)...{be}\:{continued}... \\ $$ $$ \\ $$