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Question Number 41678 by math khazana by abdo last updated on 11/Aug/18
prove that ∫_0 ^∞ cos(x^2 )dx=∫_0 ^∞ sin(x^2 )dx by using only series.
$${prove}\:{that}\:\:\int_{\mathrm{0}} ^{\infty} \:{cos}\left({x}^{\mathrm{2}} \right){dx}=\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}^{\mathrm{2}} \right){dx}\:{by}\:{using} \\ $$$${only}\:{series}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 11/Aug/18
p+iq=∫_0 ^∞ e^(ix^2 ) dx −t=ix^2 i^2 t=ix^2 it=x^2 2xdx=idt dx=((idt)/(2x))=((idt)/(2(√(it)) ))=(((√i) t^((−1)/2) )/2)dt ∫_0 ^∞ e^(−t) .(((√i) t^((−1)/2) )/2)dt (((√i) )/2)∫_0 ^∞ e^(−t) .t^((1/2)−1) dt =(((√i) )/2)⌈((1/2))=(((√Π) )/2)×(((√i) )/) now calculating the value of (√i) (√i) =(1/( (√2) ))×(√(1−1+2i)) =(1/( (√2) ))×(√(1^2 +i^2 +2×1×i)) =(1/( (√2) ))×(√((1+i)^2 )) =(1/( (√2)))×(1+i) =(1/( (√2) ))+i×(1/( (√2) )) so p+iq=∫_0 ^∞ e^(ix^2 ) dx=(((√Π) )/2)×(√i) =(((√Π) )/2)((1/( (√2) ))+i(1/( (√2) ))) p=(((√Π) )/(2(√2) )) q=(((√Π) )/(2(√2)))
$${p}+{iq}=\int_{\mathrm{0}} ^{\infty} {e}^{{ix}^{\mathrm{2}} } {dx} \\ $$$$−{t}={ix}^{\mathrm{2}} \:\:\:\:{i}^{\mathrm{2}} {t}={ix}^{\mathrm{2}} \\ $$$${it}={x}^{\mathrm{2}} \\ $$$$\mathrm{2}{xdx}={idt} \\ $$$${dx}=\frac{{idt}}{\mathrm{2}{x}}=\frac{{idt}}{\mathrm{2}\sqrt{{it}}\:}=\frac{\sqrt{{i}}\:{t}^{\frac{−\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}}{dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} .\frac{\sqrt{{i}}\:{t}^{\frac{−\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}}{dt} \\ $$$$\frac{\sqrt{{i}}\:}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} .{t}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {dt} \\ $$$$=\frac{\sqrt{{i}}\:}{\mathrm{2}}\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\Pi}\:}{\mathrm{2}}×\frac{\sqrt{{i}}\:}{} \\ $$$${now}\:{calculating}\:{the}\:{value}\:{of}\:\sqrt{{i}}\: \\ $$$$\sqrt{{i}}\: \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}×\sqrt{\mathrm{1}−\mathrm{1}+\mathrm{2}{i}}\: \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}×\sqrt{\mathrm{1}^{\mathrm{2}} +{i}^{\mathrm{2}} +\mathrm{2}×\mathrm{1}×{i}}\: \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}×\sqrt{\left(\mathrm{1}+{i}\right)^{\mathrm{2}} \:} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}×\left(\mathrm{1}+{i}\right)\: \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}+{i}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:} \\ $$$${so}\:{p}+{iq}=\int_{\mathrm{0}} ^{\infty} {e}^{{ix}^{\mathrm{2}} } {dx}=\frac{\sqrt{\Pi}\:}{\mathrm{2}}×\sqrt{{i}}\:=\frac{\sqrt{\Pi}\:}{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}+{i}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\right) \\ $$$${p}=\frac{\sqrt{\Pi}\:}{\mathrm{2}\sqrt{\mathrm{2}}\:}\:\:\:\:{q}=\frac{\sqrt{\Pi}\:}{\mathrm{2}\sqrt{\mathrm{2}}}\:\: \\ $$$$ \\ $$

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