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Question Number 141774 by mathmax by abdo last updated on 23/May/21
calculate ∫_0 ^∞ (e^(−x^2 ) /(1+x^2 ))dx
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$
Answered by Dwaipayan Shikari last updated on 24/May/21
∫_0 ^∞ e^(−x^2 ) ∫_0 ^∞ e^(−u(1+x^2 )) dudx =∫_0 ^∞ ∫_0 ^∞ e^(−x^2 (u+1)) e^(−u) dxdu =(1/2)(√π)∫_0 ^∞ (e^(−u) /( (√(u+1))))du u+1=t^2 =(1/2)e(√π) ∫_1 ^∞ e^(−t^2 ) dt=2e(√π) (((√π)/2)−∫_0 ^1 e^(−t^2 ) dt) =(1/2)(πe−πe erf(1))=(1/2)πe(1−erf(1))=((πe)/2)erfc(1)
$$\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } \int_{\mathrm{0}} ^{\infty} {e}^{−{u}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} {dudx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} \left({u}+\mathrm{1}\right)} {e}^{−{u}} {dxdu} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\pi}\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{u}} }{\:\sqrt{{u}+\mathrm{1}}}{du}\:\:\:\:{u}+\mathrm{1}={t}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{e}\sqrt{\pi}\:\int_{\mathrm{1}} ^{\infty} {e}^{−{t}^{\mathrm{2}} } {dt}=\mathrm{2}{e}\sqrt{\pi}\:\left(\frac{\sqrt{\pi}}{\mathrm{2}}−\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{t}^{\mathrm{2}} } {dt}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\pi{e}−\pi{e}\:{erf}\left(\mathrm{1}\right)\right)=\frac{\mathrm{1}}{\mathrm{2}}\pi{e}\left(\mathrm{1}−{erf}\left(\mathrm{1}\right)\right)=\frac{\pi{e}}{\mathrm{2}}{erfc}\left(\mathrm{1}\right) \\ $$
Answered by mathmax by abdo last updated on 24/May/21
Φ=∫_0 ^∞ (∫_0 ^∞ e^(−t(1+x^2 )) dt)e^(−x^2 ) dx =∫_0 ^∞ (∫_0 ^∞ e^(−tx^2 )−x^2 ) dx)e^(−t) dt we have ∫_0 ^∞ e^(−(t+1)x^2 ) dx =∫_0 ^∞ e^(−((√(t+1))x)^2 ) dx =_((√(t+1))x=z) ∫_0 ^∞ e^(−z^2 ) (dz/( (√(t+1)))) =((√π)/(2(√(t+1)))) ⇒ Φ=((√π)/2)∫_0 ^∞ (e^(−t) /( (√(t+1))))dt =_((√(t+1))=y) ((√π)/2) ∫_1 ^∞ (e^(−(y^2 −1)) /y)(2y)dy =(√π).e ∫_1 ^∞ e^(−y^2 ) dy ⇒Φ=e(√π)∫_1 ^∞ e^(−x^2 ) dx
$$\Phi=\int_{\mathrm{0}} ^{\infty} \left(\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{t}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)} \mathrm{dt}\right)\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\left(\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{\left.−\mathrm{tx}^{\mathrm{2}} \right)−\mathrm{x}^{\mathrm{2}} } \mathrm{dx}\right)\mathrm{e}^{−\mathrm{t}} \mathrm{dt} \\ $$$$\mathrm{we}\:\:\mathrm{have}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\left(\mathrm{t}+\mathrm{1}\right)\mathrm{x}^{\mathrm{2}} } \mathrm{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\left(\sqrt{\mathrm{t}+\mathrm{1}}\mathrm{x}\right)^{\mathrm{2}} } \mathrm{dx}\: \\ $$$$=_{\sqrt{\mathrm{t}+\mathrm{1}}\mathrm{x}=\mathrm{z}} \:\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{z}^{\mathrm{2}} } \frac{\mathrm{dz}}{\:\sqrt{\mathrm{t}+\mathrm{1}}}\:=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{t}+\mathrm{1}}}\:\Rightarrow \\ $$$$\Phi=\frac{\sqrt{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{e}^{−\mathrm{t}} }{\:\sqrt{\mathrm{t}+\mathrm{1}}}\mathrm{dt}\:\:=_{\sqrt{\mathrm{t}+\mathrm{1}}=\mathrm{y}} \:\:\frac{\sqrt{\pi}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{e}^{−\left(\mathrm{y}^{\mathrm{2}} −\mathrm{1}\right)} }{\mathrm{y}}\left(\mathrm{2y}\right)\mathrm{dy} \\ $$$$=\sqrt{\pi}.\mathrm{e}\:\int_{\mathrm{1}} ^{\infty} \:\mathrm{e}^{−\mathrm{y}^{\mathrm{2}} } \mathrm{dy}\:\Rightarrow\Phi=\mathrm{e}\sqrt{\pi}\int_{\mathrm{1}} ^{\infty} \:\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{dx} \\ $$

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