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Question Number 136750 by mnjuly1970 last updated on 25/Mar/21

.....advanced calculus..... prove that :: ... 𝛗=∫_0 ^( ∞) ((1−e^(−x^2 ) )/x^2 )dx=(√π)

$$\:\:\:\:\:\:\:\:\:\:\:.....{advanced}\:\:\:\:{calculus}..... \\ $$$$\:\:\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:...\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }{dx}=\sqrt{\pi}\:\:\:\: \\ $$

Answered by Ñï= last updated on 25/Mar/21

∫_0 ^∞ ((1−e^(−x^2 ) )/x^2 )dx=∫_0 ^∞ ((e^(−0) −e^(−x^2 ) )/x^2 )dx=∫_0 ^∞ ∫_0 ^1 e^(−ax^2 ) dadx =∫_0 ^1 ((√π)/(2(√a)))da=(√π)

$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−\mathrm{0}} −{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{ax}^{\mathrm{2}} } {dadx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\sqrt{\pi}}{\mathrm{2}\sqrt{{a}}}{da}=\sqrt{\pi} \\ $$

Commented by mnjuly1970 last updated on 25/Mar/21

thanks alot...

$$\:\:\:{thanks}\:{alot}... \\ $$

Answered by Dwaipayan Shikari last updated on 25/Mar/21

η(φ)=∫_0 ^∞ ((e^(−φx^2 ) −e^(−x^2 ) )/x^2 )dx⇒η′(φ)=−∫_0 ^∞ e^(−φx^2 ) dx=−((√π)/(2(√φ))) η(φ)=−(√(πφ))+C⇒η(1)=0⇒C=(√π) η(φ)=(√π) (1−(√φ)) ⇒η(0)=(√π)

$$\eta\left(\phi\right)=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−\phi{x}^{\mathrm{2}} } −{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }{dx}\Rightarrow\eta'\left(\phi\right)=−\int_{\mathrm{0}} ^{\infty} {e}^{−\phi{x}^{\mathrm{2}} } {dx}=−\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\phi}} \\ $$$$\eta\left(\phi\right)=−\sqrt{\pi\phi}+{C}\Rightarrow\eta\left(\mathrm{1}\right)=\mathrm{0}\Rightarrow{C}=\sqrt{\pi}\: \\ $$$$\eta\left(\phi\right)=\sqrt{\pi}\:\left(\mathrm{1}−\sqrt{\phi}\right)\:\Rightarrow\eta\left(\mathrm{0}\right)=\sqrt{\pi} \\ $$

Answered by mathmax by abdo last updated on 25/Mar/21

let f(a) =∫_0 ^∞ ((e^(−ax^2 ) −e^(−x^2 ) )/x^2 )dx with a>0 ⇒f^′ (a)=−∫_0 ^∞ e^(−ax^2 ) dx =−∫_0 ^∞ e^(−((√a)x)^2 ) dx =_((√a)x=t) − ∫_0 ^∞ e^(−t^2 ) (dt/( (√a))) =−(1/( (√a))).((√π)/2) ⇒f(a)=(√π)∫ (da/(2(√a))) +C =−(√(πa)) +C f(1)=0=−(√π) +C ⇒C=(√π) ⇒f(a)=(√π)−(√(πa)) Φ=f(0)=(√π)

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{ax}^{\mathrm{2}} } −\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } }{\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\:\mathrm{with}\:\mathrm{a}>\mathrm{0}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)=−\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{ax}^{\mathrm{2}} } \mathrm{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\left(\sqrt{\mathrm{a}}\mathrm{x}\right)^{\mathrm{2}} } \mathrm{dx}\:=_{\sqrt{\mathrm{a}}\mathrm{x}=\mathrm{t}} −\:\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \frac{\mathrm{dt}}{\:\sqrt{\mathrm{a}}}\:=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}}.\frac{\sqrt{\pi}}{\mathrm{2}}\:\Rightarrow\mathrm{f}\left(\mathrm{a}\right)=\sqrt{\pi}\int\:\frac{\mathrm{da}}{\mathrm{2}\sqrt{\mathrm{a}}}\:+\mathrm{C} \\ $$$$=−\sqrt{\pi\mathrm{a}}\:+\mathrm{C} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)=\mathrm{0}=−\sqrt{\pi}\:+\mathrm{C}\:\Rightarrow\mathrm{C}=\sqrt{\pi}\:\Rightarrow\mathrm{f}\left(\mathrm{a}\right)=\sqrt{\pi}−\sqrt{\pi\mathrm{a}} \\ $$$$\Phi=\mathrm{f}\left(\mathrm{0}\right)=\sqrt{\pi} \\ $$

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