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Question Number 72988 by mathmax by abdo last updated on 05/Nov/19
calculate f(x)=∫_0 ^∞ (e^(−xt^2 ) /(4+t^2 ))dt with x>0
$${calculate}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{e}^{−{xt}^{\mathrm{2}} } }{\mathrm{4}+{t}^{\mathrm{2}} }{dt}\:\:\:{with}\:{x}>\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 05/Nov/19
we have f(x)=∫_0 ^∞ (e^(−x(t^2 +4−4)) /(t^2 +4))dt =e^(4x) ∫_0 ^∞ (e^(−x(t^2 +4)) /(t^2 +4))dt =e^(4x) w(x) with w(x)=∫_0 ^∞ (e^(−x(t^2 +4)) /(t^2 +4))dt ⇒ w^′ (x)=−∫_0 ^(+∞) e^(−x(t^2 +4)) dt =−e^(−4x) ∫_0 ^∞ e^(−((√x)t)^2 ) dt =_((√x)t =u) − e^(−4x) ∫_0 ^∞ e^(−u^2 ) (du/( (√x))) =−(e^(−4x) /( (√x)))×((√π)/2) ⇒w(x)=−((√π)/2)∫_0 ^x (e^(−4u) /( (√u)))du +c =_((√u)=z) − ((√π)/2) ∫_0 ^(√x) (e^(−4z^2 ) /z)(2z)dz+c =−(√π)∫_0 ^(√x) e^(−4z^2 ) dz +c⇒ f(x)=c e^(4x) −(√π)e^(4x) ∫_0 ^(√x) e^(−4z^2 ) dz c=lim_(x→0) f(x) =∫_0 ^∞ (dt/(t^2 +4)) =_(t=2u) ∫_0 ^∞ ((2du)/(4(1+u^2 ))) =(1/2)(π/2)=(π/4) ⇒ f(x)=(π/4)e^(4x) −(√π)e^(4x) ∫_0 ^(√x) e^(−4z^2 ) dz
$${we}\:{have}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{x}\left({t}^{\mathrm{2}} +\mathrm{4}−\mathrm{4}\right)} }{{t}^{\mathrm{2}} \:+\mathrm{4}}{dt}\:={e}^{\mathrm{4}{x}} \int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{x}\left({t}^{\mathrm{2}} +\mathrm{4}\right)} }{{t}^{\mathrm{2}} \:+\mathrm{4}}{dt} \\ $$$$={e}^{\mathrm{4}{x}} {w}\left({x}\right)\:{with}\:{w}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{x}\left({t}^{\mathrm{2}} \:+\mathrm{4}\right)} }{{t}^{\mathrm{2}} \:+\mathrm{4}}{dt}\:\Rightarrow \\ $$$${w}^{'} \left({x}\right)=−\int_{\mathrm{0}} ^{+\infty} \:\:{e}^{−{x}\left({t}^{\mathrm{2}} +\mathrm{4}\right)} {dt}\:=−{e}^{−\mathrm{4}{x}} \:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left(\sqrt{{x}}{t}\right)^{\mathrm{2}} } {dt} \\ $$$$=_{\sqrt{{x}}{t}\:={u}} \:\:−\:\:{e}^{−\mathrm{4}{x}} \int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{u}^{\mathrm{2}} } \frac{{du}}{\:\sqrt{{x}}}\:=−\frac{{e}^{−\mathrm{4}{x}} }{\:\sqrt{{x}}}×\frac{\sqrt{\pi}}{\mathrm{2}}\:\Rightarrow{w}\left({x}\right)=−\frac{\sqrt{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{{x}} \:\:\frac{{e}^{−\mathrm{4}{u}} }{\:\sqrt{{u}}}{du}\:+{c} \\ $$$$=_{\sqrt{{u}}={z}} −\:\:\:\frac{\sqrt{\pi}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\sqrt{{x}}} \:\:\frac{{e}^{−\mathrm{4}{z}^{\mathrm{2}} } }{{z}}\left(\mathrm{2}{z}\right){dz}+{c}\:=−\sqrt{\pi}\int_{\mathrm{0}} ^{\sqrt{{x}}} \:{e}^{−\mathrm{4}{z}^{\mathrm{2}} } {dz}\:+{c}\Rightarrow \\ $$$${f}\left({x}\right)={c}\:{e}^{\mathrm{4}{x}} \:−\sqrt{\pi}{e}^{\mathrm{4}{x}} \:\int_{\mathrm{0}} ^{\sqrt{{x}}} \:{e}^{−\mathrm{4}{z}^{\mathrm{2}} } {dz} \\ $$$${c}={lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{4}}\:=_{{t}=\mathrm{2}{u}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{du}}{\mathrm{4}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\mathrm{2}}\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\pi}{\mathrm{4}}{e}^{\mathrm{4}{x}} \:−\sqrt{\pi}{e}^{\mathrm{4}{x}} \:\int_{\mathrm{0}} ^{\sqrt{{x}}} \:{e}^{−\mathrm{4}{z}^{\mathrm{2}} } {dz} \\ $$$$ \\ $$
Answered by mind is power last updated on 05/Nov/19
f′(x)=∫_0 ^(+∞) ((−t^2 e^(−xt^2 ) )/(4+t^2 ))=∫−e^(−xt^2 ) +4∫_0 ^(+∞) (e^(−xt^2 ) /(4+t^2 ))dt =⇒f′(x)=4f(x)−(1/( (√x)))∫_0 ^(+∞) e^(−(t(√x))^2 ) .(√x)dt ⇒f′(x)=4f(x)−(1/( (√x))).∫_0 ^(+∞) e^(−u^2 ) du=4f(x)−(1/(2(√x))).(√(π/2)) f′(x)−4f(x)+(1/(2(√x))).(√(π/2))=0 f(x)=ke^(4x) ⇒k′(x)=−(e^(−4x) /(2(√x)))(√(π/2)) =⇒k(x)=−((√π)/(2(√2)))∫(e^(−4x) /(2(√x)))dx erf(x)=∫_0 ^x e^(−t^2 ) dt (√x)=u⇒k(x)=−((√π)/(2(√2)))∫e^(−4u^2 ) du=−((√π)/(4(√2)))∫e^(−w^2 ) dw=−((√π)/(4(√2)))erf(w) =((√π)/(4(√2)))erf(2u)=−((√π)/(4(√2)))erf(2(√x))+c ⇒f(x)=ce^(−4x) −((√π)/(4(√2)))erf(2(√x))e^(−4x) f(0)=∫_0 ^(+∞) (1/(4+t^2 ))=[(1/2).arctan((t/2))]=(π/4) ⇒c=(π/4) f(x)=(π/4)e^(−4x) −((√π)/(4(√2)))erf(2(√x))e^(−4x)
$$\mathrm{f}'\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{+\infty} \frac{−\mathrm{t}^{\mathrm{2}} \mathrm{e}^{−\mathrm{xt}^{\mathrm{2}} } }{\mathrm{4}+\mathrm{t}^{\mathrm{2}} }=\int−\mathrm{e}^{−\mathrm{xt}^{\mathrm{2}} } +\mathrm{4}\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{e}^{−\mathrm{xt}^{\mathrm{2}} } }{\mathrm{4}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$=\Rightarrow\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{4f}\left(\mathrm{x}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\int_{\mathrm{0}} ^{+\infty} \mathrm{e}^{−\left(\mathrm{t}\sqrt{\mathrm{x}}\right)^{\mathrm{2}} } .\sqrt{\mathrm{x}}\mathrm{dt} \\ $$$$\Rightarrow\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{4f}\left(\mathrm{x}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}.\int_{\mathrm{0}} ^{+\infty} \mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \mathrm{du}=\mathrm{4f}\left(\mathrm{x}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}}}.\sqrt{\frac{\pi}{\mathrm{2}}} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)−\mathrm{4f}\left(\mathrm{x}\right)+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}}}.\sqrt{\frac{\pi}{\mathrm{2}}}=\mathrm{0} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ke}^{\mathrm{4x}} \\ $$$$\Rightarrow\mathrm{k}'\left(\mathrm{x}\right)=−\frac{\mathrm{e}^{−\mathrm{4x}} }{\mathrm{2}\sqrt{\mathrm{x}}}\sqrt{\frac{\pi}{\mathrm{2}}} \\ $$$$=\Rightarrow\mathrm{k}\left(\mathrm{x}\right)=−\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\frac{\mathrm{e}^{−\mathrm{4x}} }{\mathrm{2}\sqrt{\mathrm{x}}}\mathrm{dx} \\ $$$$\mathrm{erf}\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt} \\ $$$$\sqrt{\mathrm{x}}=\mathrm{u}\Rightarrow\mathrm{k}\left(\mathrm{x}\right)=−\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\mathrm{e}^{−\mathrm{4u}^{\mathrm{2}} } \mathrm{du}=−\frac{\sqrt{\pi}}{\mathrm{4}\sqrt{\mathrm{2}}}\int\mathrm{e}^{−\mathrm{w}^{\mathrm{2}} } \mathrm{dw}=−\frac{\sqrt{\pi}}{\mathrm{4}\sqrt{\mathrm{2}}}\mathrm{erf}\left(\mathrm{w}\right) \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{4}\sqrt{\mathrm{2}}}\mathrm{erf}\left(\mathrm{2u}\right)=−\frac{\sqrt{\pi}}{\mathrm{4}\sqrt{\mathrm{2}}}\mathrm{erf}\left(\mathrm{2}\sqrt{\mathrm{x}}\right)+\mathrm{c} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ce}^{−\mathrm{4x}} −\frac{\sqrt{\pi}}{\mathrm{4}\sqrt{\mathrm{2}}}\mathrm{erf}\left(\mathrm{2}\sqrt{\mathrm{x}}\right)\mathrm{e}^{−\mathrm{4x}} \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{1}}{\mathrm{4}+\mathrm{t}^{\mathrm{2}} }=\left[\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{arctan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)\right]=\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{c}=\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\pi}{\mathrm{4}}\mathrm{e}^{−\mathrm{4x}} −\frac{\sqrt{\pi}}{\mathrm{4}\sqrt{\mathrm{2}}}\mathrm{erf}\left(\mathrm{2}\sqrt{\mathrm{x}}\right)\mathrm{e}^{−\mathrm{4x}} \\ $$$$ \\ $$$$ \\ $$

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