find the value of U_n =∫_(−∞) ^(+∞) e^(−nx^2 ) sin(x^2 −2x)dx find nature of the serie Σ U_n and Σe^(−n^2 ) U


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Question Number 66065 by mathmax by abdo last updated on 08/Aug/19

$f i n d t h e v a l u e o f U_{n} = \int_{- \infty}^{+ \infty} e^{- {n x}^{2}} s i n (x^{2} - 2 x) d x$ $f i n d n a t u r e o f t h e s e r i e Σ U_{n} a n d Σ e^{- n^{2}} U_{n}$
Commented by mathmax by abdo last updated on 09/Aug/19
$we have x^2 −2x =x^2 −2x +1−1 =(x−1)^2 −1 changement x−1 =t give U_n =∫_(−∞) ^(+∞) e^(−n(t+1)^2 ) sin(t^2 −1)dt =Im (∫_(−∞) ^(+∞) e^(−n(t+1)^2 +i(t^2 −1)) dt) =Im(W_n ) ∫_(−∞) ^(+∞) e^(−n(t+1)^2 +i(t^2 −1)) dt =∫_(−∞) ^(+∞) e^(−n(t^2 +2t+1)+it^2 −i) dt =e^(−i) ∫_(−∞) ^(+∞) e^(−(n−i)t^2 −2nt −n) dt =e^(−(n+i)) ∫_(−∞) ^(+∞) e^(−{((√(n−i))t)^2 +2nt}) dt =e^(−(n+i)) ∫_(−∞) ^(+∞) e^(−{ ((√(n−i))t)^2 +2(n/(√(n−i)))(√(n−i))t +((n/(√(n−i))))^2 −(n^2 /(n−i))}) dt =e^(−(n+i)) ∫_(−∞) ^(+∞) e^(−{(√(n−i))t +(n/(√(n−i)))}^2 +(n^2 /(n−i))) dt (ch. (√(n−i))t +(n/(√(n−i))) =v =e^((n^2 /(n−i))−(n+i)) ∫_(−∞) ^(+∞) e^(−v^2 ) (dv/(√(n−i))) =(√π)(1/(√(n−i))) e^((n^2 −(n^2 −i^2 ))/(n−i)) =((√π)/(√(n−i))) e^(−(1/(n−i))) =((√π)/(√(n−i))) e^(−((n+i)/(n^2 +1))) n−i =(√(1+n^2 )) e^(i arctan(((−1)/n))) =(√(1+n^2 ))e^(−iarctan((1/n))) ⇒ (√(n−i)) =(1+n^2 )^(1/4) e^(−(i/2)arctan((1/n))) ⇒ W_n =(√π)(1+n^2 )^(−(1/4)) e^((i/2)arctan((1/n))) e^(−(n/(n^2 +1))) e^(−(i/(n^2 +1))) =(√π)(n^2 +1)^(−(1/4)) e^(−(n/(n^2 +1))) e^(i((1/2)arctan((1/n))−(1/(n^2 +1)))) ⇒ U_n =(√π)(n^2 +1)^(−(1/4)) e^(−(n/(n^2 +1))) sin((1/2)arctan((1/n))−(1/(n^2 +1)))with U_0 =(√π)sin((π/4) −1)$
$w e h a v e x^{2} - 2 x = x^{2} - 2 x + 1 - 1 = {(x - 1)}^{2} - 1 c h a n g e m e n t$ $x - 1 = t g i v e U_{n} = \int_{- \infty}^{+ \infty} e^{- n {(t + 1)}^{2}} s i n (t^{2} - 1) d t$ $= I m (\int_{- \infty}^{+ \infty} e^{- n {(t + 1)}^{2} + i (t^{2} - 1)} d t) = I m (W_{n})$ $\int_{- \infty}^{+ \infty} e^{- n {(t + 1)}^{2} + i (t^{2} - 1)} d t = \int_{- \infty}^{+ \infty} e^{- n (t^{2} + 2 t + 1) + {i t}^{2} - i} d t$ $= e^{- i} \int_{- \infty}^{+ \infty} e^{- (n - i) t^{2} - 2 n t - n} d t$ $= e^{- (n + i)} \int_{- \infty}^{+ \infty} e^{- {{(\sqrt{n - i} t)}^{2} + 2 n t}} d t$ $= e^{- (n + i)} \int_{- \infty}^{+ \infty} e^{- {{(\sqrt{n - i} t)}^{2} + 2 \frac{n}{\sqrt{n - i}} \sqrt{n - i} t + {(\frac{n}{\sqrt{n - i}})}^{2} - \frac{n^{2}}{n - i}}} d t$ $= e^{- (n + i)} \int_{- \infty}^{+ \infty} e^{- {\sqrt{n - i} t + \frac{n}{\sqrt{n - i}}}^{2} + \frac{n^{2}}{n - i}} d t (c h . \sqrt{n - i} t + \frac{n}{\sqrt{n - i}} = v$ $= e^{\frac{n^{2}}{n - i} - (n + i)} \int_{- \infty}^{+ \infty} e^{- v^{2}} \frac{d v}{\sqrt{n - i}} = \sqrt{π} \frac{1}{\sqrt{n - i}} e^{\frac{n^{2} - (n^{2} - i^{2})}{n - i}}$ $= \frac{\sqrt{π}}{\sqrt{n - i}} e^{- \frac{1}{n - i}} = \frac{\sqrt{π}}{\sqrt{n - i}} e^{- \frac{n + i}{n^{2} + 1}}$ $n - i = \sqrt{1 + n^{2}} e^{i a r c t a n (\frac{- 1}{n})} = \sqrt{1 + n^{2}} e^{- i a r c t a n (\frac{1}{n})} \Rightarrow$ $\sqrt{n - i} = {(1 + n^{2})}^{\frac{1}{4}} e^{- \frac{i}{2} a r c t a n (\frac{1}{n})} \Rightarrow$ $W_{n} = \sqrt{π} {(1 + n^{2})}^{- \frac{1}{4}} e^{\frac{i}{2} a r c t a n (\frac{1}{n})} e^{- \frac{n}{n^{2} + 1}} e^{- \frac{i}{n^{2} + 1}}$ $= \sqrt{π} {(n^{2} + 1)}^{- \frac{1}{4}} e^{- \frac{n}{n^{2} + 1}} e^{i (\frac{1}{2} a r c t a n (\frac{1}{n}) - \frac{1}{n^{2} + 1})} \Rightarrow$ $U_{n} = \sqrt{π} {(n^{2} + 1)}^{- \frac{1}{4}} e^{- \frac{n}{n^{2} + 1}} s i n (\frac{1}{2} a r c t a n (\frac{1}{n}) - \frac{1}{n^{2} + 1}) w i t h$ $U_{0} = \sqrt{π} s i n (\frac{π}{4} - 1)$
Commented by mathmax by abdo last updated on 09/Aug/19

$2) w e h a v e ∣ U_{n} ∣⩽ \sqrt{π} {(n^{2} + 1)}^{- \frac{1}{4}} e^{- \frac{n}{n^{2} + 1}} = v_{n}$ $t h e s e r i e Σ v_{n} c o n v e r g e s \Rightarrow Σ U_{n} c o n v e r g e s .$
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