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Question Number 39712 by math khazana by abdo last updated on 10/Jul/18
calculate ∫_(−∞) ^(+∞) ((cos(x^n ) +sin(x^n ))/((x^2 +9)^n )) dx
$${calculate}\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left({x}^{{n}} \right)\:+{sin}\left({x}^{{n}} \right)}{\left({x}^{\mathrm{2}} \:+\mathrm{9}\right)^{{n}} }\:{dx} \\ $$
Commented by math khazana by abdo last updated on 10/Jul/18
n integr natural.
$${n}\:{integr}\:{natural}. \\ $$
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)
$${let}\:{A}_{{n}} =\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({x}^{{n}} \right)\:+{sin}\left({x}^{{n}} \right)}{\left({x}^{\mathrm{2}} \:+\mathrm{9}\right)^{{n}} }{dx}\:{changement} \\ $$$${x}=\mathrm{3}{t}\:{give} \\ $$$${A}_{{n}} =\:\mathrm{3}\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\mathrm{3}^{{n}} {t}^{{n}} \right)\:+{sin}\left(\mathrm{3}^{{n}} {t}^{{n}} \right)}{\mathrm{9}^{{n}} \left({t}^{\mathrm{2}\:} \:+\mathrm{1}\right)^{{n}} }\:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}{n}−\mathrm{1}} }\:\:\left\{\:\:{Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{i}\mathrm{3}^{{n}} {t}^{{n}} } }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}{dt}\right)\:+{Im}\left(\:\int_{−\infty} ^{+\infty} \:\frac{{e}^{{i}\:\mathrm{3}^{{n}} {t}^{{n}} } }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}} }\right)\right\}{dt} \\ $$$${let}\:{calculate}\:{I}_{{n}} =\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{i}\:\mathrm{3}^{{n}} \:{t}^{{n}} } }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}} }{dt}\:{let} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{{i}\:\mathrm{3}^{{n}} {z}^{{n}} } }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}} }\:\:{we}\:{have} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{{i}\:\mathrm{3}^{{n}} \:{z}^{{n}} } }{\left({z}−{i}\right)^{{n}} \left({z}+{i}\right)^{{n}} }\:\:{the}\:{poles}\:{of}\:\varphi\:{are}\:{i}\:{and}\:−{i} \\ $$$${with}\:{multiplicity}\:{n} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$
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) )
$${Res}\left(\varphi,{i}\right)\:={lim}_{{n}\rightarrow+\infty} \frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{{n}} \varphi\left({z}\right)\right\}^{\left({n}−\mathrm{1}\right)} \\ $$$$={lim}_{{n}\rightarrow+\infty} \:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\:\left\{\:{e}^{{i}\:\mathrm{3}^{{n}} {z}^{{n}} } \left({z}+{i}\right)^{−{n}} \right\}^{\left({n}−\mathrm{1}\right)} \:{leibniz} \\ $$$${formula}\:{give}\: \\ $$$$\left\{\:\left({z}+{i}\right)^{−{n}} \:\:{e}^{{i}\:\mathrm{3}^{{n}} {z}^{{n}} } \right\}^{\left({n}−\mathrm{1}\right)} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:{C}_{{n}−\mathrm{1}} ^{{k}} \:\left(\left({z}+{i}\right)^{−{n}} \right)^{\left({k}\right)} \:\left(\:{e}^{{i}\:\mathrm{3}^{{n}} {z}^{{n}} } \right)^{\left({n}−\mathrm{1}−{k}\right)} \\ $$$${we}\:{have} \\ $$$$\left.\left({z}+{i}\right)^{−{n}} \right\}^{\left(\mathrm{1}\right)} \:=−{n}\left({z}+{i}\right)^{−{n}−\mathrm{1}} \\ $$$$\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left(\mathrm{2}\right)} =\left(−\mathrm{1}\right)^{\mathrm{2}} {n}\left({n}+\mathrm{1}\right)\:\left({z}+{i}\right)^{−{n}−\mathrm{2}} \:{by}\:{recurence} \\ $$$$\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left({k}\right)} =\left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)…\left({n}+{k}−\mathrm{1}\right)\left({z}+{i}\right)^{−{n}−{k}} \\ $$$${also}\:{we}\:{have} \\ $$$$\left\{\:{e}^{{i}\:\mathrm{3}^{{n}} \:{z}^{{n}} } \right\}^{\left(\mathrm{1}\right)} \:={ni}\:\mathrm{3}^{{n}} \:{z}^{{n}−\mathrm{1}} \:{e}^{{i}\:\mathrm{3}^{{n}} \:{z}^{{n}} } \:\:={p}_{\mathrm{1}} \left({z}\right){e}^{{i}\:\mathrm{3}^{{n}} {z}^{{n}} } \\ $$$$\left\{\:{e}^{{i}\:\mathrm{3}^{{n}} {z}^{{n}} } \right\}^{\left(\mathrm{2}\right)} ={p}_{\mathrm{2}} \left({z}\right)\:{e}^{{i}\:\mathrm{3}^{{n}} \:{z}^{{n}} } … \\ $$$$\left\{\:{e}^{{i}\:\mathrm{3}^{{n}} \:{z}^{{n}} } \right\}^{\left({n}−\mathrm{1}−{k}\right)} ={p}_{{n}−\mathrm{1}−{k}} \left({z}\right)\:{e}^{{i}\:\mathrm{3}^{{n}} \:{z}^{{n}} } \:{and}\:{we}\:{can} \\ $$$${find}\:{a}\:{recurrence}\:{relation}\:{between}\:{the}\:{p}_{{n}} \\ $$$${Res}\left(\varphi,{i}\right)=\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{C}_{{n}−\mathrm{1}} ^{{k}} \:\left(−\mathrm{1}\right)^{{k}} {n}\left({n}+\mathrm{1}\right)…\left({n}+{k}−\mathrm{1}\right)\left(\mathrm{2}{i}\right)^{−{n}−{k}} \:{p}_{{n}−\mathrm{1}−{k}} \left({i}\right)\:{e}^{\mathrm{3}^{{n}} {i}^{{n}+\mathrm{1}} } \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\frac{\mathrm{2}{i}\pi}{\left({n}−\mathrm{1}\right)!}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} \:{C}_{{n}} ^{{k}} {n}\left({n}+\mathrm{1}\right)…\left({n}+{k}−\mathrm{1}\right){p}_{{n}−\mathrm{1}−{k}} \left({i}\right)\:{e}^{\mathrm{3}^{{n}} {i}^{{n}+\mathrm{1}} } \\ $$
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))
$${y}={cos}\left({x}^{{n}} \right)+{sin}\left({x}^{{n}} \right) \\ $$$$=\sqrt{\mathrm{2}}\:\left\{{sin}\left(\frac{\Pi}{\mathrm{4}}+{x}^{{n}} \right)\right\} \\ $$$$\sqrt{\mathrm{2}}\:\:\geqslant{y}\geqslant−\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{9}\right)^{{n}} }\int_{−−\infty} ^{+\infty} \frac{{cos}\left({x}^{{n}} \right)+{sin}\left({x}^{{n}} \right)}{\left({x}^{\mathrm{2}} +\mathrm{9}\right)^{{n}} }\geqslant−\sqrt{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{9}\right)^{{n}} } \\ $$$${let}\:{I}_{{k}} =\sqrt{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{9}\right)^{{n}} }\:\:{x}=\mathrm{3}{tana} \\ $$$$=\sqrt{\mathrm{2}}\:\int_{−\frac{\Pi}{\mathrm{2}}\:\:} ^{\frac{\Pi}{\mathrm{2}}} \frac{\mathrm{3}{sec}^{\mathrm{2}} {ada}}{\left\{\mathrm{9}\left({sec}^{\mathrm{2}} {a}\right)\right\}^{{n}} } \\ $$$$=\frac{\sqrt{\mathrm{2}}\:}{\mathrm{3}^{\mathrm{2}{n}−\mathrm{1}} }\int_{−\frac{\Pi}{\mathrm{2}}} ^{\frac{\Pi}{\mathrm{2}}} {cos}^{\mathrm{2}{n}−\mathrm{2}} {ada} \\ $$$$=\frac{\sqrt{\mathrm{2}}\:×\mathrm{2}}{\mathrm{3}^{\mathrm{2}{n}−\mathrm{1}} }\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {cos}^{\mathrm{2}{n}−\mathrm{2}} {ada} \\ $$$${now}\:{using}\:{gamma}\:−{beta}\:{function} \\ $$$$\frac{\lceil{p}.\lceil{q}}{\lceil\left({p}+{q}\right)}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} \theta{cos}^{\mathrm{2}{q}−\mathrm{1}} \theta{d}\theta \\ $$$$ \\ $$$$\mathrm{2}{p}−\mathrm{1}=\mathrm{0}\:\:\:\:{p}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\: \\ $$$$\mathrm{2}{q}−\mathrm{1}=\mathrm{2}{n}−\mathrm{2}\:\: \\ $$$${q}=\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\sqrt{\mathrm{2}}\:×\mathrm{2}}{\mathrm{3}^{\mathrm{2}{n}−\mathrm{1}} }\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {sin}^{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {a}\:\:{cos}^{\mathrm{2}×\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {a}\:{da} \\ $$$$=\frac{\sqrt{\mathrm{2}}\:}{\mathrm{3}^{\mathrm{2}{n}−\mathrm{1}} }×\frac{\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\lceil\left(\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\right)}{\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{3}^{\mathrm{2}{n}−\mathrm{1}} }×\frac{\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\lceil\left(\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\right)}{\lceil{n}} \\ $$$${so}\: \\ $$$$\frac{\sqrt{\mathrm{2}}\:}{\mathrm{3}^{\mathrm{2}{n}−\mathrm{1}} }×\frac{\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\lceil\left(\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\right)}{\lceil{n}}\geqslant\int_{−\infty} ^{+\infty} \frac{{cos}\left({x}^{{n}} \right)+{sin}\left({x}^{{n}} \right)}{\left({x}^{\mathrm{2}} +\mathrm{9}\right)^{{n}} }\geqslant \\ $$$$\:\:\frac{−\sqrt{\mathrm{2}}}{\mathrm{3}^{\mathrm{2}{n}−\mathrm{1}} }×\frac{\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\lceil\left(\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}\right)}{\lceil{n}} \\ $$$$ \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18
pls check
$${pls}\:{check} \\ $$

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