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


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Question Number 48169 by Abdo msup. last updated on 20/Nov/18

$l e t u_{n} = \int_{0}^{\infty} c o s ({n x}^{2}) d x a n d v_{n} = \int_{0}^{\infty} s i n ({n x}^{2}) d x w i t h n > 0$ $1) c a l c u l s t e u_{n} a n d v_{n}$ $2) f i n d n s t u r e o f Σ (u_{n} + 2 v_{n}) a n d Σ (u_{n}^{2} + 4 v_{n}^{2})$ $3) f i n d n a t u r e o f Σ {(u_{n} + 2 v_{n})}^{2}$
Commented bymaxmathsup by imad last updated on 23/Nov/18
$1) we have u_n −iv_n =∫_0 ^∞ e^(−inx^2 ) dx =_(t=(√(in))x) ∫_0 ^∞ e^(−t^2 ) (dt/(√(in))) = (1/((√n)e^((iπ)/4) )) ∫_0 ^∞ e^(−t^2 ) dt =(e^(−((iπ)/4)) /(√n)) ((√π)/2) =((√π)/(2(√n))) {cos((π/4))−isin((π/4))} ⇒u_n =((√π)/(2(√n))) .((√2)/2) =((√(2π))/(4(√n))) and v_n =((√π)/(2(√n))) ((√2)/2) =((√(2π))/(4(√n))) . 2)due to u_(n ) =v_n we have Σ_(n=1) ^∞ (u_n +2v_n )=3 Σ_(n=1) ^∞ ((√(2π))/(4(√n))) =3 ((√(2π))/4) Σ_(n=1) ^∞ (1/(√n)) and this serie diverges also Σ_(n=1) ^∞ (u_n ^2 +4v_n ^2 )=5 Σ_(n=1) ^n ((2π)/(16n)) =((5π)/8) Σ_(n=1) ^∞ (1/n) diverges 3) Σ(u_n +2v_n )^2 =9 Σ u_n ^2 and this serie diverges.$
$1) w e h a v e u_{n} - {i v}_{n} = \int_{0}^{\infty} e^{- {i n x}^{2}} d x =_{t = \sqrt{i n} x} \int_{0}^{\infty} e^{- t^{2}} \frac{d t}{\sqrt{i n}}$ $= \frac{1}{\sqrt{n} e^{\frac{i π}{4}}} \int_{0}^{\infty} e^{- t^{2}} d t = \frac{e^{- \frac{i π}{4}}}{\sqrt{n}} \frac{\sqrt{π}}{2} = \frac{\sqrt{π}}{2 \sqrt{n}} {c o s (\frac{π}{4}) - i s i n (\frac{π}{4})}$ $\Rightarrow u_{n} = \frac{\sqrt{π}}{2 \sqrt{n}} . \frac{\sqrt{2}}{2} = \frac{\sqrt{2 π}}{4 \sqrt{n}} a n d v_{n} = \frac{\sqrt{π}}{2 \sqrt{n}} \frac{\sqrt{2}}{2} = \frac{\sqrt{2 π}}{4 \sqrt{n}} .$ $2) d u e t o u_{n} = v_{n} w e h a v e \sum_{n = 1}^{\infty} (u_{n} + 2 v_{n}) = 3 \sum_{n = 1}^{\infty} \frac{\sqrt{2 π}}{4 \sqrt{n}}$ $= 3 \frac{\sqrt{2 π}}{4} \sum_{n = 1}^{\infty} \frac{1}{\sqrt{n}} a n d t h i s s e r i e d i v e r g e s a l s o$ $\sum_{n = 1}^{\infty} (u_{n}^{2} + 4 v_{n}^{2}) = 5 \sum_{n = 1}^{n} \frac{2 π}{16 n} = \frac{5 π}{8} \sum_{n = 1}^{\infty} \frac{1}{n} d i v e r g e s$ $3) Σ {(u_{n} + 2 v_{n})}^{2} = 9 Σ u_{n}^{2} a n d t h i s s e r i e d i v e r g e s .$
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