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Question Number 140399 by mnjuly1970 last updated on 07/May/21

𝛏 :=∫_0 ^( ∞) ((e^(βˆ’x^2 ) βˆ’e^(βˆ’x) )/x) dx = k.Ξ³ find ” k ” ... Ξ³ := Euler constant....

$$ \\ $$ $$\:\:\:\:\:\:\:\:\:\boldsymbol{\xi}\::=\int_{\mathrm{0}} ^{\:\infty} \:\frac{{e}^{βˆ’{x}^{\mathrm{2}} } βˆ’{e}^{βˆ’{x}} }{{x}}\:{dx}\:=\:{k}.\gamma\: \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:{find}\:\:''\:{k}\:\:''\:... \\ $$ $$\:\:\:\:\:\:\:\:\:\:\gamma\::=\:\mathscr{E}{uler}\:{constant}.... \\ $$

Answered by qaz last updated on 07/May/21

βˆ’Ξ³=∫_0 ^∞ (e^(βˆ’x) βˆ’(1/(1+x)))(dx/x) βˆ’Ξ³=∫_0 ^∞ (e^(βˆ’x^2 ) βˆ’(1/(1+x^2 )))((d(x^2 ))/x^2 )=2∫_0 ^∞ (e^(βˆ’x^2 ) βˆ’(1/(1+x^2 )))(dx/x) β‡’βˆ’(Ξ³/2)=∫_0 ^∞ (e^(βˆ’x^2 ) βˆ’(1/(1+x^2 )))(dx/x) ΞΎ=∫_0 ^∞ (e^(βˆ’x^2 ) βˆ’e^(βˆ’x) )(dx/x) =∫_0 ^∞ {(e^(βˆ’x^2 ) βˆ’(1/(1+x^2 )))βˆ’(e^(βˆ’x) βˆ’(1/(1+x)))+((1/(1+x^2 ))βˆ’(1/(1+x)))}(dx/x) =βˆ’(Ξ³/2)+Ξ³+∫_0 ^∞ ((1/(1+x^2 ))βˆ’(1/(1+x)))(dx/x) =(Ξ³/2)+∫_0 ^∞ ((1/(1+x))βˆ’(x/(1+x^2 )))dx =(Ξ³/2)+ln((1+x)/( (√(1+x^2 ))))∣_0 ^∞ =(Ξ³/2) β‡’k=(1/2)

$$βˆ’\gamma=\int_{\mathrm{0}} ^{\infty} \left({e}^{βˆ’{x}} βˆ’\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)\frac{{dx}}{{x}} \\ $$ $$βˆ’\gamma=\int_{\mathrm{0}} ^{\infty} \left({e}^{βˆ’{x}^{\mathrm{2}} } βˆ’\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)\frac{{d}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \left({e}^{βˆ’{x}^{\mathrm{2}} } βˆ’\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)\frac{{dx}}{{x}} \\ $$ $$\Rightarrowβˆ’\frac{\gamma}{\mathrm{2}}=\int_{\mathrm{0}} ^{\infty} \left({e}^{βˆ’{x}^{\mathrm{2}} } βˆ’\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)\frac{{dx}}{{x}} \\ $$ $$\xi=\int_{\mathrm{0}} ^{\infty} \left({e}^{βˆ’{x}^{\mathrm{2}} } βˆ’{e}^{βˆ’{x}} \right)\frac{{dx}}{{x}} \\ $$ $$=\int_{\mathrm{0}} ^{\infty} \left\{\left({e}^{βˆ’{x}^{\mathrm{2}} } βˆ’\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)βˆ’\left({e}^{βˆ’{x}} βˆ’\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)+\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }βˆ’\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)\right\}\frac{{dx}}{{x}} \\ $$ $$=βˆ’\frac{\gamma}{\mathrm{2}}+\gamma+\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }βˆ’\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)\frac{{dx}}{{x}} \\ $$ $$=\frac{\gamma}{\mathrm{2}}+\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}}βˆ’\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx} \\ $$ $$=\frac{\gamma}{\mathrm{2}}+{ln}\frac{\mathrm{1}+{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\mid_{\mathrm{0}} ^{\infty} \\ $$ $$=\frac{\gamma}{\mathrm{2}} \\ $$ $$\Rightarrow{k}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented bymnjuly1970 last updated on 07/May/21

bravo ...mr payan...

$$\:\:\:{bravo}\:\:...{mr}\:{payan}... \\ $$

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