Menu Close

let-f-t-0-sin-x-2-e-tx-2-x-2-dx-with-t-gt-0-find-a-simple-form-of-f-t-

Tinku Tara 0 Comments
Pin




Question Number 33978 by abdo imad last updated on 28/Apr/18
let f(t) = ∫_0 ^∞ ((sin(x^2 )e^(−tx^2 ) )/x^2 ) dx with t>0 find a simple form of f^′ (t) .
$${let}\:{f}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{sin}\left({x}^{\mathrm{2}} \right){e}^{−{tx}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }\:{dx}\:\:\:\:\:\:\:{with}\:{t}>\mathrm{0} \\ $$$${find}\:{a}\:{simple}\:{form}\:{of}\:{f}^{'} \left({t}\right)\:. \\ $$
Commented by math khazana by abdo last updated on 01/May/18
after verifying condition of derivabity for f we have f^′ (t) =∫_0 ^∞ −x^2 ((sin(x^2 )e^(−tx^2 ) )/x^2 )dx =−∫_0 ^∞ sin(x^2 ) e^(−tx^2 ) dx ⇒ 2 f^′ (x)= −∫_(−∞) ^(+∞) sin(x^2 )e^(−tx^2 ) dx =Im(∫_(−∞) ^(+∞) e^(−ix^2 ) e^(−tx^2 ) dx) but ∫_(−∞) ^(+∞) e^(−ix^2 ) e^(−tx^2 ) dx = ∫_(−∞) ^(+∞) e^(−(t +i)x^2 ) dx =_((√(t+i))x=u) ∫_(−∞) ^(+∞) e^(−u^2 ) (du/( (√(t+i)))) = ((√π)/( (√(t+i)))) but ∣t+i∣=(√(1+t^2 )) ⇒t+i =(√(1+t^2 ))( (t/( (√(1+t^2 )))) +(i/( (√(1+t^2 ))))) =r e^(iθ) ⇒r=(√(1+t^2 )) and cosθ= (t/( (√(1+t^2 )))) and sinθ =(1/( (√(1+t^2 )))) ⇒tanθ= (1/t) ⇒θ=arctan((1/t)) t+i =(1+t^2 )^(1/2) e^(i((π/2)−arctant)) ⇒ (√(t+i))= (1+t^2 )^(1/4) e^(i( (π/4) −(1/2)arctant)) (1/( (√(t+i)))) = (1+t^2 )^(−(1/4)) e^(i((1/2)arctant−(π/4))) ⇒ 2f^′ (x)= (1+t^2 )^(−(1/4)) sin((1/2)arctant −(π/4)) ⇒ f^′ (x) = (1/2) (1+t^2 )^(−(1/4)) sin((1/2) arctant −(π/4)) .
$${after}\:{verifying}\:{condition}\:{of}\:{derivabity}\:{for}\:{f}\:{we}\:{have} \\ $$$${f}^{'} \left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} −{x}^{\mathrm{2}} \:\:\:\frac{{sin}\left({x}^{\mathrm{2}} \right){e}^{−{tx}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}^{\mathrm{2}} \right)\:{e}^{−{tx}^{\mathrm{2}} } {dx}\:\Rightarrow\:\mathrm{2}\:{f}^{'} \left({x}\right)=\:−\int_{−\infty} ^{+\infty} \:{sin}\left({x}^{\mathrm{2}} \right){e}^{−{tx}^{\mathrm{2}} } {dx} \\ $$$$={Im}\left(\int_{−\infty} ^{+\infty} \:\:{e}^{−{ix}^{\mathrm{2}} } \:{e}^{−{tx}^{\mathrm{2}} } {dx}\right)\:{but} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:{e}^{−{ix}^{\mathrm{2}} } {e}^{−{tx}^{\mathrm{2}} } {dx}\:=\:\int_{−\infty} ^{+\infty} \:\:\:{e}^{−\left({t}\:+{i}\right){x}^{\mathrm{2}} } {dx} \\ $$$$=_{\sqrt{{t}+{i}}{x}={u}} \:\int_{−\infty} ^{+\infty} \:\:\:{e}^{−{u}^{\mathrm{2}} } \:\frac{{du}}{\:\sqrt{{t}+{i}}}\:=\:\frac{\sqrt{\pi}}{\:\sqrt{{t}+{i}}}\:\:{but} \\ $$$$\mid{t}+{i}\mid=\sqrt{\mathrm{1}+{t}^{\mathrm{2}} \:}\:\:\:\Rightarrow{t}+{i}\:=\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\left(\:\frac{{t}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:+\frac{{i}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\right) \\ $$$$={r}\:{e}^{{i}\theta} \:\Rightarrow{r}=\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:\:\:{and}\:{cos}\theta=\:\frac{{t}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:\:{and}\: \\ $$$${sin}\theta\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:\Rightarrow{tan}\theta=\:\frac{\mathrm{1}}{{t}}\:\Rightarrow\theta={arctan}\left(\frac{\mathrm{1}}{{t}}\right) \\ $$$${t}+{i}\:=\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:{e}^{{i}\left(\frac{\pi}{\mathrm{2}}−{arctant}\right)} \:\Rightarrow \\ $$$$\sqrt{{t}+{i}}=\:\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:\:{e}^{{i}\left(\:\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{2}}{arctant}\right)} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{t}+{i}}}\:=\:\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} \:\:{e}^{{i}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctant}−\frac{\pi}{\mathrm{4}}\right)} \:\Rightarrow \\ $$$$\mathrm{2}{f}^{'} \left({x}\right)=\:\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} \:\:{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}{arctant}\:−\frac{\pi}{\mathrm{4}}\right)\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} \:{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{arctant}\:−\frac{\pi}{\mathrm{4}}\right)\:. \\ $$
Commented by math khazana by abdo last updated on 01/May/18
f^′ (t)= (1/2)(1+t^2 )^(−(1/4)) sin((1/2) arctant −(π/4)) .
$${f}^{'} \left({t}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} \:{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{arctant}\:−\frac{\pi}{\mathrm{4}}\right)\:. \\ $$
Answered by candre last updated on 30/Apr/18
f(t)=∫_0 ^∞ ((sin x^2 e^(−tx^2 ) )/x^2 )dx lets use the fact that ∫_(−∞) ^(+∞) e^(−ax^2 ) dx=(√(π/a)) sin x=((e^(ix) −e^(−ix) )/(2i)) we get (df/dt)=∫_0 ^∞ (∂/∂t)((sin x^2 e^(−tx^2 ) )/x^2 )dx =−∫_0 ^∞ sin x^2 e^(−tx^2 ) dx =−∫_0 ^∞ ((e^(ix^2 ) −e^(−ix^2 ) )/(2i))e^(−tx^2 ) dx =−(1/(2i))[∫_0 ^∞ e^((i−t)x^2 ) dx−∫_0 ^∞ e^(−(i+t)x^2 ) dx] =((√π)/(4i))((√(1/(t+i)))−(√(1/(t−i))))
$${f}\left({t}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{sin}\:{x}^{\mathrm{2}} {e}^{−{tx}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }{dx} \\ $$$${lets}\:{use}\:{the}\:{fact}\:{that} \\ $$$$\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−{ax}^{\mathrm{2}} } {dx}=\sqrt{\frac{\pi}{{a}}} \\ $$$$\mathrm{sin}\:{x}=\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}} \\ $$$${we}\:{get} \\ $$$$\frac{{df}}{{dt}}=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\partial}{\partial{t}}\frac{\mathrm{sin}\:{x}^{\mathrm{2}} {e}^{−{tx}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }{dx} \\ $$$$=−\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{sin}\:{x}^{\mathrm{2}} {e}^{−{tx}^{\mathrm{2}} } {dx} \\ $$$$=−\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{e}^{{ix}^{\mathrm{2}} } −{e}^{−{ix}^{\mathrm{2}} } }{\mathrm{2}{i}}{e}^{−{tx}^{\mathrm{2}} } {dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{i}}\left[\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{\left({i}−{t}\right){x}^{\mathrm{2}} } {dx}−\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−\left({i}+{t}\right){x}^{\mathrm{2}} } {dx}\right] \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{4}{i}}\left(\sqrt{\frac{\mathrm{1}}{{t}+{i}}}−\sqrt{\frac{\mathrm{1}}{{t}−{i}}}\right) \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *