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let-u-n-0-cos-nx-2-dx-and-v-n-0-sin-nx-2-dx-with-n-gt-0-1-calculste-u-n-and-v-n-2-find-nsture-of-u-n-2v-n-and-u-n-2-4v-n-2-3-find-nature-of-u-n-2v-n-2-

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Question Number 48169 by Abdo msup. last updated on 20/Nov/18
let u_n =∫_0 ^∞ cos(nx^2 )dx and v_n =∫_0 ^∞ sin(nx^2 )dx with n >0 1) calculste u_n and v_n 2)find nsture of Σ(u_n +2v_n ) and Σ (u_n ^2 +4v_n ^2 ) 3)find nature of Σ(u_n +2v_n )^2
$${let}\:{u}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{cos}\left({nx}^{\mathrm{2}} \right){dx}\:{and}\:{v}_{{n}} =\int_{\mathrm{0}} ^{\infty} {sin}\left({nx}^{\mathrm{2}} \right){dx}\:{with}\:{n}\:>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculste}\:{u}_{{n}} {and}\:{v}_{{n}} \\ $$$$\left.\mathrm{2}\right){find}\:{nsture}\:{of}\:\Sigma\left({u}_{{n}} +\mathrm{2}{v}_{{n}} \right)\:{and}\:\Sigma\:\left({u}_{{n}} ^{\mathrm{2}} \:+\mathrm{4}{v}_{{n}} ^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{3}\right){find}\:{nature}\:{of}\:\Sigma\left({u}_{{n}} +\mathrm{2}{v}_{{n}} \right)^{\mathrm{2}} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 23/Nov/18
1) we have u_n −iv_n =∫_0 ^∞ e^(−inx^2 ) dx =_(t=(√(in))x) ∫_0 ^∞ e^(−t^2 ) (dt/( (√(in)))) = (1/( (√n)e^((iπ)/4) )) ∫_0 ^∞ e^(−t^2 ) dt =(e^(−((iπ)/4)) /( (√n))) ((√π)/2) =((√π)/(2(√n))) {cos((π/4))−isin((π/4))} ⇒u_n =((√π)/(2(√n))) .((√2)/2) =((√(2π))/(4(√n))) and v_n =((√π)/(2(√n))) ((√2)/2) =((√(2π))/(4(√n))) . 2)due to u_(n ) =v_n we have Σ_(n=1) ^∞ (u_n +2v_n )=3 Σ_(n=1) ^∞ ((√(2π))/(4(√n))) =3 ((√(2π))/4) Σ_(n=1) ^∞ (1/( (√n))) and this serie diverges also Σ_(n=1) ^∞ (u_n ^2 +4v_n ^2 )=5 Σ_(n=1) ^n ((2π)/(16n)) =((5π)/8) Σ_(n=1) ^∞ (1/n) diverges 3) Σ(u_n +2v_n )^2 =9 Σ u_n ^2 and this serie diverges.
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{u}_{{n}} −{iv}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{e}^{−{inx}^{\mathrm{2}} } {dx}\:=_{{t}=\sqrt{{in}}{x}} \:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}^{\mathrm{2}} } \frac{{dt}}{\:\sqrt{{in}}} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{{n}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}^{\mathrm{2}} } {dt}\:=\frac{{e}^{−\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{n}}}\:\frac{\sqrt{\pi}}{\mathrm{2}}\:=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{{n}}}\:\left\{{cos}\left(\frac{\pi}{\mathrm{4}}\right)−{isin}\left(\frac{\pi}{\mathrm{4}}\right)\right\} \\ $$$$\Rightarrow{u}_{{n}} =\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{{n}}}\:.\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{4}\sqrt{{n}}}\:\:{and}\:\:{v}_{{n}} =\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{{n}}}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{4}\sqrt{{n}}}\:. \\ $$$$\left.\mathrm{2}\right){due}\:{to}\:{u}_{{n}\:} ={v}_{{n}} \:\:{we}\:{have}\:\sum_{{n}=\mathrm{1}} ^{\infty} \left({u}_{{n}} +\mathrm{2}{v}_{{n}} \right)=\mathrm{3}\:\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\sqrt{\mathrm{2}\pi}}{\mathrm{4}\sqrt{{n}}} \\ $$$$=\mathrm{3}\:\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{4}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\:\sqrt{{n}}}\:\:{and}\:{this}\:{serie}\:{diverges}\:\:{also} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \left({u}_{{n}} ^{\mathrm{2}} \:+\mathrm{4}{v}_{{n}} ^{\mathrm{2}} \right)=\mathrm{5}\:\sum_{{n}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{2}\pi}{\mathrm{16}{n}}\:=\frac{\mathrm{5}\pi}{\mathrm{8}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}}\:\:\:{diverges} \\ $$$$\left.\mathrm{3}\right)\:\Sigma\left({u}_{{n}} +\mathrm{2}{v}_{{n}} \right)^{\mathrm{2}} \:=\mathrm{9}\:\Sigma\:{u}_{{n}} ^{\mathrm{2}} \:\:{and}\:{this}\:\:{serie}\:{diverges}. \\ $$

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