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Question Number 56829 by maxmathsup by imad last updated on 24/Mar/19
let f(t) =∫_0 ^∞ ((cos(t(1+x^2 )))/(1+x^2 )) dx with t≥0 find a explicit form of f(t)
$${let}\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({t}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:\:{with}\:{t}\geqslant\mathrm{0} \\ $$$${find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({t}\right)\: \\ $$
Answered by Smail last updated on 25/Mar/19
f(t)=Im(∫_0 ^∞ (e^(−it(1+x^2 )) /(1+x^2 ))dx) z(t)=∫_0 ^∞ (e^(−it(1+x^2 )) /(1+x^2 ))dx z′(t)=−i∫_0 ^∞ e^(−it(1+x^2 )) dx=−ie^(−it) ∫_0 ^∞ e^(−ix^2 t) dx let u=(√(it))x⇒dx=(du/( (√(it)))) z′(t)=−(((√i)e^(−it) )/( (√t)))∫_0 ^∞ e^(−u^2 ) du =−(((√i)e^(−it) )/( (√t)))×((√π)/2) z(t)−z(0)=−((√(iπ))/2)∫_0 ^t (e^(−iu) /( (√u)))du z(0)=∫_0 ^∞ (dx/(1+x^2 ))=(π/2) let θ=(√(iu))⇒dθ=(((√i)du)/(2(√u))) z(t)=−(√π)∫_0 ^(√(it)) e^(−θ^2 ) dθ+(π/2) =−(√π)∫_0 ^(√(it)) Σ_(n=0) ^∞ (((−θ^2 )^n )/(n!))dθ+(π/2) =−(√π)Σ_(n=0) ^∞ (((−1)^n ((√(it)))^(2n+1) )/(n!(2n+1)))+(π/2) =(π/2)−(√π)Σ_(n=0) ^∞ (((−1)^n (it)^n .(√(it)))/(n!(2n+1))) =(π/2)−(√(πt))Σ_(n=0) ^∞ (((−t)^n e^(i((π/2))n) ×e^(i(π/4)) )/(n!(2n+1))) =(π/2)−(√(πt))Σ_(n=0) ^∞ (((−t)^n e^(i(((nπ)/2)+(π/4))) )/(n!(2n+1))) Thus f(t)=(π/2)−(√(πt))Σ_(n=0) ^∞ (((−t)^n cos(((2n+1)/4)π))/(n!(2n+1)))
$${f}\left({t}\right)={Im}\left(\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{it}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\right) \\ $$$${z}\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{it}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${z}'\left({t}\right)=−{i}\int_{\mathrm{0}} ^{\infty} {e}^{−{it}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} {dx}=−{ie}^{−{it}} \int_{\mathrm{0}} ^{\infty} {e}^{−{ix}^{\mathrm{2}} {t}} {dx} \\ $$$${let}\:\:{u}=\sqrt{{it}}{x}\Rightarrow{dx}=\frac{{du}}{\:\sqrt{{it}}} \\ $$$${z}'\left({t}\right)=−\frac{\sqrt{{i}}{e}^{−{it}} }{\:\sqrt{{t}}}\int_{\mathrm{0}} ^{\infty} {e}^{−{u}^{\mathrm{2}} } {du} \\ $$$$=−\frac{\sqrt{{i}}{e}^{−{it}} }{\:\sqrt{{t}}}×\frac{\sqrt{\pi}}{\mathrm{2}}\: \\ $$$${z}\left({t}\right)−{z}\left(\mathrm{0}\right)=−\frac{\sqrt{{i}\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{{t}} \frac{{e}^{−{iu}} }{\:\sqrt{{u}}}{du}\:\: \\ $$$${z}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}} \\ $$$${let}\:\:\theta=\sqrt{{iu}}\Rightarrow{d}\theta=\frac{\sqrt{{i}}{du}}{\mathrm{2}\sqrt{{u}}} \\ $$$${z}\left({t}\right)=−\sqrt{\pi}\int_{\mathrm{0}} ^{\sqrt{{it}}} {e}^{−\theta^{\mathrm{2}} } {d}\theta+\frac{\pi}{\mathrm{2}} \\ $$$$=−\sqrt{\pi}\int_{\mathrm{0}} ^{\sqrt{{it}}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\theta^{\mathrm{2}} \right)^{{n}} }{{n}!}{d}\theta+\frac{\pi}{\mathrm{2}} \\ $$$$=−\sqrt{\pi}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \left(\sqrt{{it}}\right)^{\mathrm{2}{n}+\mathrm{1}} }{{n}!\left(\mathrm{2}{n}+\mathrm{1}\right)}+\frac{\pi}{\mathrm{2}} \\ $$$$=\frac{\pi}{\mathrm{2}}−\sqrt{\pi}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \left({it}\right)^{{n}} .\sqrt{{it}}}{{n}!\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$=\frac{\pi}{\mathrm{2}}−\sqrt{\pi{t}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−{t}\right)^{{n}} {e}^{{i}\left(\frac{\pi}{\mathrm{2}}\right){n}} ×{e}^{{i}\frac{\pi}{\mathrm{4}}} }{{n}!\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$=\frac{\pi}{\mathrm{2}}−\sqrt{\pi{t}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−{t}\right)^{{n}} {e}^{{i}\left(\frac{{n}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)} }{{n}!\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$${Thus}\: \\ $$$${f}\left({t}\right)=\frac{\pi}{\mathrm{2}}−\sqrt{\pi{t}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−{t}\right)^{{n}} {cos}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{4}}\pi\right)}{{n}!\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 26/Mar/19
thanks sir smail.
$${thanks}\:{sir}\:{smail}. \\ $$

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