calculate ∫_(−∞) ^(+∞) ((cos(x^n ) +sin(x^n ))/((x^2 +9)^n )) dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 39712 by math khazana by abdo last updated on 10/Jul/18

$c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{c o s (x^{n}) + s i n (x^{n})}{{(x^{2} + 9)}^{n}} d x$
Commented by math khazana by abdo last updated on 10/Jul/18

$n i n t e g r n a t u r a l .$
Commented by abdo mathsup 649 cc last updated on 10/Jul/18
$let A_n = ∫_(−∞) ^(+∞) ((cos(x^n ) +sin(x^n ))/((x^2 +9)^n ))dx changement x=3t give A_n = 3∫_(−∞) ^(+∞) ((cos(3^n t^n ) +sin(3^n t^n ))/(9^n (t^(2 ) +1)^n )) dt =(1/3^(2n−1) ) { Re( ∫_(−∞) ^(+∞) (e^(i3^n t^n ) /((t^2 +1)))dt) +Im( ∫_(−∞) ^(+∞) (e^(i 3^n t^n ) /((t^2 +1)^n )))}dt let calculate I_n = ∫_(−∞) ^(+∞) (e^(i 3^n t^n ) /((t^2 +1)^n ))dt let ϕ(z) = (e^(i 3^n z^n ) /((z^2 +1)^n )) we have ϕ(z) = (e^(i 3^n z^n ) /((z−i)^n (z+i)^n )) the poles of ϕ are i and −i with multiplicity n ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i)$
$l e t A_{n} = \int_{- \infty}^{+ \infty} \frac{c o s (x^{n}) + s i n (x^{n})}{{(x^{2} + 9)}^{n}} d x c h a n g e m e n t$ $x = 3 t g i v e$ $A_{n} = 3 \int_{- \infty}^{+ \infty} \frac{c o s (3^{n} t^{n}) + s i n (3^{n} t^{n})}{9^{n} {(t^{2} + 1)}^{n}} d t$ $= \frac{1}{3^{2 n - 1}} {R e (\int_{- \infty}^{+ \infty} \frac{e^{i 3^{n} t^{n}}}{(t^{2} + 1)} d t) + I m (\int_{- \infty}^{+ \infty} \frac{e^{i 3^{n} t^{n}}}{{(t^{2} + 1)}^{n}})} d t$ $l e t c a l c u l a t e I_{n} = \int_{- \infty}^{+ \infty} \frac{e^{i 3^{n} t^{n}}}{{(t^{2} + 1)}^{n}} d t l e t$ $φ (z) = \frac{e^{i 3^{n} z^{n}}}{{(z^{2} + 1)}^{n}} w e h a v e$ $φ (z) = \frac{e^{i 3^{n} z^{n}}}{{(z - i)}^{n} {(z + i)}^{n}} t h e p o l e s o f φ a r e i a n d - i$ $w i t h m u l t i p l i c i t y n$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)$
Commented by abdo mathsup 649 cc last updated on 10/Jul/18
$Res(ϕ,i) =lim_(n→+∞) (1/((n−1)!)){(z−i)^n ϕ(z)}^((n−1)) =lim_(n→+∞) (1/((n−1)!)) { e^(i 3^n z^n ) (z+i)^(−n) }^((n−1)) leibniz formula give { (z+i)^(−n) e^(i 3^n z^n ) }^((n−1)) =Σ_(k=0) ^(n−1) C_(n−1) ^k ((z+i)^(−n) )^((k)) ( e^(i 3^n z^n ) )^((n−1−k)) we have (z+i)^(−n) }^((1)) =−n(z+i)^(−n−1) {(z+i)^(−n) }^((2)) =(−1)^2 n(n+1) (z+i)^(−n−2) by recurence {(z+i)^(−n) }^((k)) =(−1)^k n(n+1)...(n+k−1)(z+i)^(−n−k) also we have { e^(i 3^n z^n ) }^((1)) =ni 3^n z^(n−1) e^(i 3^n z^n ) =p_1 (z)e^(i 3^n z^n ) { e^(i 3^n z^n ) }^((2)) =p_2 (z) e^(i 3^n z^n ) ... { e^(i 3^n z^n ) }^((n−1−k)) =p_(n−1−k) (z) e^(i 3^n z^n ) and we can find a recurrence relation between the p_n Res(ϕ,i)=(1/((n−1)!))Σ_(k=0) ^(n−1) C_(n−1) ^k (−1)^k n(n+1)...(n+k−1)(2i)^(−n−k) p_(n−1−k) (i) e^(3^n i^(n+1) ) ∫_(−∞) ^(+∞) ϕ(z)dz =((2iπ)/((n−1)!)) Σ_(k=0) ^(n−1) (−1)^k C_n ^k n(n+1)...(n+k−1)p_(n−1−k) (i) e^(3^n i^(n+1) )$
$R e s (φ, i) = {l i m}_{n \to + \infty} \frac{1}{(n - 1)!} {{(z - i)}^{n} φ (z)}^{(n - 1)}$ $= {l i m}_{n \to + \infty} \frac{1}{(n - 1)!} {e^{i 3^{n} z^{n}} {(z + i)}^{- n}}^{(n - 1)} l e i b n i z$ $f o r m u l a g i v e$ ${{(z + i)}^{- n} e^{i 3^{n} z^{n}}}^{(n - 1)}$ $= \sum_{k = 0}^{n - 1} C_{n - 1}^{k} {({(z + i)}^{- n})}^{(k)} {(e^{i 3^{n} z^{n}})}^{(n - 1 - k)}$ $w e h a v e$ ${(z + i)}^{- n}}^{(1)} = - n {(z + i)}^{- n - 1}$ ${{(z + i)}^{- n}}^{(2)} = {(- 1)}^{2} n (n + 1) {(z + i)}^{- n - 2} b y r e c u r e n c e$ ${{(z + i)}^{- n}}^{(k)} = {(- 1)}^{k} n (n + 1) . . . (n + k - 1) {(z + i)}^{- n - k}$ $a l s o w e h a v e$ ${e^{i 3^{n} z^{n}}}^{(1)} = n i 3^{n} z^{n - 1} e^{i 3^{n} z^{n}} = p_{1} (z) e^{i 3^{n} z^{n}}$ ${e^{i 3^{n} z^{n}}}^{(2)} = p_{2} (z) e^{i 3^{n} z^{n}} . . .$ ${e^{i 3^{n} z^{n}}}^{(n - 1 - k)} = p_{n - 1 - k} (z) e^{i 3^{n} z^{n}} a n d w e c a n$ $f i n d a r e c u r r e n c e r e l a t i o n b e t w e e n t h e p_{n}$ $R e s (φ, i) = \frac{1}{(n - 1)!} \sum_{k = 0}^{n - 1} C_{n - 1}^{k} {(- 1)}^{k} n (n + 1) . . . (n + k - 1) {(2 i)}^{- n - k} p_{n - 1 - k} (i) e^{3^{n} i^{n + 1}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = \frac{2 i π}{(n - 1)!} \sum_{k = 0}^{n - 1} {(- 1)}^{k} C_{n}^{k} n (n + 1) . . . (n + k - 1) p_{n - 1 - k} (i) e^{3^{n} i^{n + 1}}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18
	$y=cos(x^n )+sin(x^n ) =(√2) {sin((Π/4)+x^n )} (√2) ≥y≥−(√2) (√2)∫_(−∞) ^(+∞) (dx/((x^2 +9)^n ))∫_(−−∞) ^(+∞) ((cos(x^n )+sin(x^n ))/((x^2 +9)^n ))≥−(√2)∫_(−∞) ^(+∞) (dx/((x^2 +9)^n )) let I_k =(√2) ∫_(−∞) ^(+∞) (dx/((x^2 +9)^n )) x=3tana =(√2) ∫_(−(Π/2) ) ^(Π/2) ((3sec^2 ada)/({9(sec^2 a)}^n )) =(((√2) )/3^(2n−1) )∫_(−(Π/2)) ^(Π/2) cos^(2n−2) ada =(((√2) ×2)/3^(2n−1) )∫_0 ^(Π/2) cos^(2n−2) ada now using gamma −beta function ((⌈p.⌈q)/(⌈(p+q)))=2∫_0 ^(Π/2) sin^(2p−1) θcos^(2q−1) θdθ 2p−1=0 p=(1/2) 2q−1=2n−2 q=((2n−1)/2) (((√2) ×2)/3^(2n−1) )∫_0 ^(Π/2) sin^(2×(1/2)−1) a cos^(2×((2n−1)/2)−1) a da =(((√2) )/3^(2n−1) )×((⌈((1/2))⌈(((2n−1)/2)))/(⌈((1/2)+((2n−1)/2)))) =((√2)/3^(2n−1) )×((⌈((1/2))⌈(((2n−1)/2)))/(⌈n)) so (((√2) )/3^(2n−1) )×((⌈((1/2))⌈(((2n−1)/2)))/(⌈n))≥∫_(−∞) ^(+∞) ((cos(x^n )+sin(x^n ))/((x^2 +9)^n ))≥ ((−(√2))/3^(2n−1) )×((⌈((1/2))⌈(((2n−1)/2)))/(⌈n))$
	$y = c o s (x^{n}) + s i n (x^{n})$ $= \sqrt{2} {s i n (\frac{Π}{4} + x^{n})}$ $\sqrt{2} ⩾ y ⩾ - \sqrt{2}$ $\sqrt{2} \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 9)}^{n}} \int_{- - \infty}^{+ \infty} \frac{c o s (x^{n}) + s i n (x^{n})}{{(x^{2} + 9)}^{n}} ⩾ - \sqrt{2} \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 9)}^{n}}$ $l e t I_{k} = \sqrt{2} \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 9)}^{n}} x = 3 t a n a$ $= \sqrt{2} \int_{- \frac{Π}{2}}^{\frac{Π}{2}} \frac{3 {s e c}^{2} a d a}{{9 ({s e c}^{2} a)}^{n}}$ $= \frac{\sqrt{2}}{3^{2 n - 1}} \int_{- \frac{Π}{2}}^{\frac{Π}{2}} {c o s}^{2 n - 2} a d a$ $= \frac{\sqrt{2} \times 2}{3^{2 n - 1}} \int_{0}^{\frac{Π}{2}} {c o s}^{2 n - 2} a d a$ $n o w u s i n g g a m m a - b e t a f u n c t i o n$ $\frac{⌈ p . ⌈ q}{⌈ (p + q)} = 2 \int_{0}^{\frac{Π}{2}} {s i n}^{2 p - 1} θ {c o s}^{2 q - 1} θ d θ$ $2 p - 1 = 0 p = \frac{1}{2}$ $2 q - 1 = 2 n - 2$ $q = \frac{2 n - 1}{2}$ $\frac{\sqrt{2} \times 2}{3^{2 n - 1}} \int_{0}^{\frac{Π}{2}} {s i n}^{2 \times \frac{1}{2} - 1} a {c o s}^{2 \times \frac{2 n - 1}{2} - 1} a d a$ $= \frac{\sqrt{2}}{3^{2 n - 1}} \times \frac{⌈ (\frac{1}{2}) ⌈ (\frac{2 n - 1}{2})}{⌈ (\frac{1}{2} + \frac{2 n - 1}{2})}$ $= \frac{\sqrt{2}}{3^{2 n - 1}} \times \frac{⌈ (\frac{1}{2}) ⌈ (\frac{2 n - 1}{2})}{⌈ n}$ $s o$ $\frac{\sqrt{2}}{3^{2 n - 1}} \times \frac{⌈ (\frac{1}{2}) ⌈ (\frac{2 n - 1}{2})}{⌈ n} ⩾ \int_{- \infty}^{+ \infty} \frac{c o s (x^{n}) + s i n (x^{n})}{{(x^{2} + 9)}^{n}} ⩾$ $\frac{- \sqrt{2}}{3^{2 n - 1}} \times \frac{⌈ (\frac{1}{2}) ⌈ (\frac{2 n - 1}{2})}{⌈ n}$
	Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18

	$p l s c h e c k$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum