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Question Number 140977 by Mathspace last updated on 14/May/21
find ∫_0 ^∞ (e^(−t(1+x^2 )) /(1+x^2 ))dx with t≥0
$${find}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{t}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:{with}\:{t}\geqslant\mathrm{0} \\ $$
Answered by mathmax by abdo last updated on 14/May/21
let f(t)=∫_0 ^∞ (e^(−(1+x^2 )t) /(1+x^2 ))dx ⇒f^′ (t)=−∫_0 ^∞ e^(−(1+x^2 )t) dx =−e^(−t) ∫_0 ^∞ e^(−tx^2 ) dx =_(u=(√t)x) −e^(−t) ∫_0 ^∞ e^(−u^2 ) (du/( (√t))) =−(e^(−t) /( (√t))).((√π)/2) =−((√π)/2) (e^(−t) /( (√t))) ⇒f(t)=−((√π)/2)∫_0 ^t (e^(−u) /( (√u)))du +k f(0)=(π/2) =k ⇒f(t)=(π/2)−((√π)/2)∫_0 ^t (e^(−u) /( (√u)))du ((√u)=x) =(π/2)−((√π)/2)∫_0 ^(√t) (e^(−x^2 ) /x)(2x)dx =(π/2)−(√π)∫_0 ^(√t) e^(−x^2 ) dx ⇒ f(t)=(π/2)−(√π)∫_0 ^(√t) e^(−x^2 ) dx
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{t}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{−\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{t}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{t}\right)=−\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\mathrm{t}} \:\mathrm{dx} \\ $$$$=−\mathrm{e}^{−\mathrm{t}} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{tx}^{\mathrm{2}} } \mathrm{dx}\:=_{\mathrm{u}=\sqrt{\mathrm{t}}\mathrm{x}} \:−\mathrm{e}^{−\mathrm{t}} \int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \frac{\mathrm{du}}{\:\sqrt{\mathrm{t}}} \\ $$$$=−\frac{\mathrm{e}^{−\mathrm{t}} }{\:\sqrt{\mathrm{t}}}.\frac{\sqrt{\pi}}{\mathrm{2}}\:=−\frac{\sqrt{\pi}}{\mathrm{2}}\:\frac{\mathrm{e}^{−\mathrm{t}} }{\:\sqrt{\mathrm{t}}}\:\Rightarrow\mathrm{f}\left(\mathrm{t}\right)=−\frac{\sqrt{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{t}} \:\frac{\mathrm{e}^{−\mathrm{u}} }{\:\sqrt{\mathrm{u}}}\mathrm{du}\:+\mathrm{k} \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{2}}\:=\mathrm{k}\:\Rightarrow\mathrm{f}\left(\mathrm{t}\right)=\frac{\pi}{\mathrm{2}}−\frac{\sqrt{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{t}} \:\frac{\mathrm{e}^{−\mathrm{u}} }{\:\sqrt{\mathrm{u}}}\mathrm{du}\:\left(\sqrt{\mathrm{u}}=\mathrm{x}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}−\frac{\sqrt{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{t}}} \:\:\:\frac{\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } }{\mathrm{x}}\left(\mathrm{2x}\right)\mathrm{dx}\:=\frac{\pi}{\mathrm{2}}−\sqrt{\pi}\int_{\mathrm{0}} ^{\sqrt{\mathrm{t}}} \:\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{dx}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{t}\right)=\frac{\pi}{\mathrm{2}}−\sqrt{\pi}\int_{\mathrm{0}} ^{\sqrt{\mathrm{t}}} \:\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{dx} \\ $$
Answered by qaz last updated on 14/May/21
∫_0 ^∞ (e^(−t(1+x^2 )) /(1+x^2 ))dx =−∫_0 ^t dt∫_0 ^∞ e^(−t(1+x^2 )) dx+(π/2) =−∫_0 ^t e^(−t) dt∫_0 ^∞ e^(−tx^2 ) dx+(π/2) =−∫_0 ^t e^(−t) ((Γ((1/2)))/(2t^(1/2) ))dt+(π/2) =−((√π)/2)∫_0 ^t t^(−(1/2)) e^(−t) dt+(π/2) =(π/2)[1−erf(t^2 )]
$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{t}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=−\int_{\mathrm{0}} ^{{t}} {dt}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} {dx}+\frac{\pi}{\mathrm{2}} \\ $$$$=−\int_{\mathrm{0}} ^{{t}} {e}^{−{t}} {dt}\int_{\mathrm{0}} ^{\infty} {e}^{−{tx}^{\mathrm{2}} } {dx}+\frac{\pi}{\mathrm{2}} \\ $$$$=−\int_{\mathrm{0}} ^{{t}} {e}^{−{t}} \frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}{t}^{\frac{\mathrm{1}}{\mathrm{2}}} }{dt}+\frac{\pi}{\mathrm{2}} \\ $$$$=−\frac{\sqrt{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{{t}} {t}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{t}} {dt}+\frac{\pi}{\mathrm{2}} \\ $$$$=\frac{\pi}{\mathrm{2}}\left[\mathrm{1}−{erf}\left({t}^{\mathrm{2}} \right)\right] \\ $$

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