let A_n = ∫_0 ^∞ e^(−nx^2 ) cos(x^2 ) dx and B_n =∫_0 ^∞ e^(−nx^2 ) sin(x^2 )dx (n∈ N^★ ) 1) calculate A_n and B_n 2) find lim_(n→+∞) (A_n /B


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Question Number 41513 by maxmathsup by imad last updated on 08/Aug/18

$l e t A_{n} = \int_{0}^{\infty} e^{- {n x}^{2}} c o s (x^{2}) d x a n d B_{n} = \int_{0}^{\infty} e^{- {n x}^{2}} s i n (x^{2}) d x (n \in N^{★})$ $1) c a l c u l a t e A_{n} a n d B_{n}$ $2) f i n d {l i m}_{n \to + \infty} \frac{A_{n}}{B_{n}}$
	Answered by alex041103 last updated on 09/Aug/18

	$L e t A_{n} = 2 A_{n} = \int_{- \infty}^{\infty} e^{- {n x}^{2}} c o s (x^{2}) d x a n d$ $B_{n} = 2 B_{n} = \int_{- \infty}^{\infty} e^{- {n x}^{2}} s i n (x^{2}) d x$ $L e t Z_{n} = A_{n} + i B_{n} = \int_{- \infty}^{\infty} e^{- (n - i) x^{2}} d x$ $T h e n Z_{n} = \sqrt{\int_{- \infty}^{\infty} \int_{- \infty}^{\infty} e^{- (n - i) (x^{2} + y^{2})} d x d y}$ $W e c h a n g e t o p o l a r c o o r d i n a t e s$ $Z_{n} = \sqrt{\int_{0}^{2 π} \int_{0}^{\infty} {r e}^{- (n - i) r^{2}} d r d θ}$ $Z_{n}^{2} = - \frac{e^{- (n - i) r^{2}} ∣_{r = 0}^{r = \infty}}{2 n - 2 i} \int_{0}^{2 π} d θ =$ $= \frac{π}{n - i} = Z_{n}^{2} = \frac{π (n + i)}{n^{2} + 1}$ $\Rightarrow Z_{n} = \sqrt{\frac{π}{n^{2} + 1}} \sqrt{n + i}$ $\Rightarrow A_{n} = \frac{1}{2} \sqrt{\frac{π}{n^{2} + 1}} R e (\sqrt{n + i})$ $B_{n} = \frac{1}{2} \sqrt{\frac{π}{n^{2} + 1}} I m (\sqrt{n + i})$
	Commented by alex041103 last updated on 09/Aug/18

	$F o r 2)$ $\frac{A_{n}}{B_{n}} = \frac{R e (\sqrt{n + i})}{I m (\sqrt{n + i})} = c o t (a r g (\sqrt{n + i})) =$ $= c o t (\frac{1}{2} a r c c o t (n))$ $\Rightarrow \underset{n \to \infty}{l i m c o t} (\frac{1}{2} a r c c o t (n)) =$ $= c o t (\frac{1}{2} a r c c o t (+ \infty)) =$ $= c o t (\frac{1}{2} \times 0^{+}) = c o t (0^{+}) = + \infty$ $\Rightarrow \underset{n \to \infty}{l i m} \frac{A_{n}}{B_{n}} = + \infty$
	Answered by math khazana by abdo last updated on 09/Aug/18
	$we have A_n +iB_n = ∫_0 ^∞ e^(−nx^2 +ix^2 ) dx = ∫_0 ^∞ e^(−(n−i)x^2 ) dx ⇒2A_n +2i B_n =∫_(−∞) ^(+∞) e^(−(n−i)x^2 ) dx =_((√(n−i))x=t) ∫_(−∞) ^(+∞) e^(−t^2 ) (dt/(√(n−i))) =((√π)/(√(n−i))) but n−i =(√(n^2 +1)){ (n/(√(n^2 +1))) −(i/(√(n^2 +1)))}=r e^(iθ) ⇒ r =(√(n^2 +1)) and tanθ =−(1/n) ⇒θ=−arctan((1/n))⇒ (√(n−i))=(n^2 +1)^(1/4) e^((iθ)/2) =(n^2 +1)^(1/4) e^(−i((arctan((1/n)))/2)) =^4 (√(n^2 +1)){ cos((1/2) arctan((1/n)) −i sin((1/2)arctan((1/n))) =^4 (√(n^2 +1)){ cos((π/4) −((arctan(n))/2))−isin((π/4)−((arctan(n))/2))⇒ A_n =^4 (√(n^2 +1))cos(−(π/4) +((arctan(n))/2)) and B_n =^4 (√(n^2 +1))sin(−(π/4) +((arctan(n))/2)) .$
	$w e h a v e A_{n} + {i B}_{n} = \int_{0}^{\infty} e^{- {n x}^{2} + {i x}^{2}} d x$ $= \int_{0}^{\infty} e^{- (n - i) x^{2}} d x \Rightarrow 2 A_{n} + 2 i B_{n} = \int_{- \infty}^{+ \infty} e^{- (n - i) x^{2}} d x$ $=_{\sqrt{n - i} x = t} \int_{- \infty}^{+ \infty} e^{- t^{2}} \frac{d t}{\sqrt{n - i}} = \frac{\sqrt{π}}{\sqrt{n - i}}$ $b u t n - i = \sqrt{n^{2} + 1} {\frac{n}{\sqrt{n^{2} + 1}} - \frac{i}{\sqrt{n^{2} + 1}}} = r e^{i θ} \Rightarrow$ $r = \sqrt{n^{2} + 1} a n d t a n θ = - \frac{1}{n} \Rightarrow θ = - a r c t a n (\frac{1}{n}) \Rightarrow$ $\sqrt{n - i} = {(n^{2} + 1)}^{\frac{1}{4}} e^{\frac{i θ}{2}} = {(n^{2} + 1)}^{\frac{1}{4}} e^{- i \frac{a r c t a n (\frac{1}{n})}{2}}$ $=^{4} \sqrt{n^{2} + 1} {c o s (\frac{1}{2} a r c t a n (\frac{1}{n}) - i s i n (\frac{1}{2} a r c t a n (\frac{1}{n}))$ $=^{4} \sqrt{n^{2} + 1} {c o s (\frac{π}{4} - \frac{a r c t a n (n)}{2}) - i s i n (\frac{π}{4} - \frac{a r c t a n (n)}{2}) \Rightarrow$ $A_{n} =^{4} \sqrt{n^{2} + 1} c o s (- \frac{π}{4} + \frac{a r c t a n (n)}{2}) a n d$ $B_{n} =^{4} \sqrt{n^{2} + 1} s i n (- \frac{π}{4} + \frac{a r c t a n (n)}{2}) .$
	Commented by math khazana by abdo last updated on 10/Aug/18

	$e r r o r a t t h e f i n a l l i n e s$ $A_{n} = \frac{\sqrt{π}}{2}^{4} \sqrt{n^{2} + 1} c o s (\frac{a r c t a n (n)}{2} - \frac{π}{4}) a n d$ $B_{n} = \frac{\sqrt{π}}{2}^{4} \sqrt{n^{2} + 1} s i n (\frac{a r c t a n (n)}{2} - \frac{π}{4}) .$
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