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Question Number 133490 by mnjuly1970 last updated on 22/Feb/21
.... advnced calculus.... prove: 𝛗=∫_0 ^( ∞) ((sin(tan(x)))/x)dx=(π/2)(1−(1/e)) prove that: 𝚽=∫_0 ^( ∞) ((1−e^(−x^2 ) )/x^2 )dx=(√π)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\:{advnced}\:\:{calculus}…. \\ $$$$\:\:\:\:{prove}:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({tan}\left({x}\right)\right)}{{x}}{dx}=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{{e}}\right) \\ $$$$\:{prove}\:{that}:\:\boldsymbol{\Phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }{dx}=\sqrt{\pi} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\: \\ $$
Answered by Dwaipayan Shikari last updated on 22/Feb/21
I(α)=∫_0 ^∞ ((e^(−αx^2 ) −e^(−βx^2 ) )/x^2 )dx I′(α)=−∫_0 ^∞ e^(−αx^2 ) dx=−(√(π/(4α))) ⇒I(α)=−(√(πα)) +C I(β)=−(√(πβ))+C=0⇒C=(√(πβ)) I(α)=(√π) ((√β)−(√α)) α=0 β=1 I(0)=∫_0 ^∞ ((1−e^(−x^2 ) )/x^2 )dx=(√π)
$${I}\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−\alpha{x}^{\mathrm{2}} } −{e}^{−\beta{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }{dx} \\ $$$${I}'\left(\alpha\right)=−\int_{\mathrm{0}} ^{\infty} {e}^{−\alpha{x}^{\mathrm{2}} } {dx}=−\sqrt{\frac{\pi}{\mathrm{4}\alpha}}\:\Rightarrow{I}\left(\alpha\right)=−\sqrt{\pi\alpha}\:+{C} \\ $$$${I}\left(\beta\right)=−\sqrt{\pi\beta}+{C}=\mathrm{0}\Rightarrow{C}=\sqrt{\pi\beta} \\ $$$${I}\left(\alpha\right)=\sqrt{\pi}\:\left(\sqrt{\beta}−\sqrt{\alpha}\right) \\ $$$$\alpha=\mathrm{0}\:\:\beta=\mathrm{1} \\ $$$${I}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }{dx}=\sqrt{\pi} \\ $$
Commented by mnjuly1970 last updated on 22/Feb/21
grateful..mr payan very nice..
$${grateful}..{mr}\:{payan} \\ $$$${very}\:\:{nice}.. \\ $$
Answered by mathmax by abdo last updated on 22/Feb/21
Φ=∫_0 ^∞ ((1−e^(−x^2 ) )/x^2 )dx let f(a) =∫_0 ^∞ ((1−e^(−ax^2 ) )/x^2 ) with a>0 f^′ (a)=∫_0 ^∞ e^(−ax^2 ) dx =∫_0 ^∞ e^(−((√a)x)^2 ) dx =_((√a)x=y) ∫_0 ^∞ e^(−y^2 ) (dy/( (√a))) =(1/( (√a)))∫_0 ^∞ e^(−y^2 ) dy =(1/( (√a)))((√π)/2) ⇒f(a) =k+(√(πa)) lim_(a→o) f(a)=0 ⇒f(a)=(√(πa)) and Φ=f(1)=(√π)
$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } }{\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−\mathrm{e}^{−\mathrm{ax}^{\mathrm{2}} } }{\mathrm{x}^{\mathrm{2}} }\:\:\mathrm{with}\:\mathrm{a}>\mathrm{0}\: \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{ax}^{\mathrm{2}} } \mathrm{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\left(\sqrt{\mathrm{a}}\mathrm{x}\right)^{\mathrm{2}} } \mathrm{dx}\:=_{\sqrt{\mathrm{a}}\mathrm{x}=\mathrm{y}} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{y}^{\mathrm{2}} } \frac{\mathrm{dy}}{\:\sqrt{\mathrm{a}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}}\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{y}^{\mathrm{2}} } \mathrm{dy}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}}\frac{\sqrt{\pi}}{\mathrm{2}}\:\Rightarrow\mathrm{f}\left(\mathrm{a}\right)\:=\mathrm{k}+\sqrt{\pi\mathrm{a}} \\ $$$$\mathrm{lim}_{\mathrm{a}\rightarrow\mathrm{o}} \mathrm{f}\left(\mathrm{a}\right)=\mathrm{0}\:\Rightarrow\mathrm{f}\left(\mathrm{a}\right)=\sqrt{\pi\mathrm{a}}\:\mathrm{and}\:\Phi=\mathrm{f}\left(\mathrm{1}\right)=\sqrt{\pi} \\ $$

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