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Question Number 143083 by Mathspace last updated on 09/Jun/21

calculate Ψ(a,b)=∫_0 ^∞  (e^(−ax^2 ) /((x^2  +b^2 )^2 ))dx  with a>0 and b>0

$${calculate}\:\Psi\left({a},{b}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{ax}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$ $${with}\:{a}>\mathrm{0}\:{and}\:{b}>\mathrm{0} \\ $$

Answered by Olaf_Thorendsen last updated on 10/Jun/21

f_b (a) = Ψ(a,b) = ∫_0 ^∞ (e^(−ax^2 ) /((x^2 +b^2 )^2 )) dx  f_b ′(a) = (∂Ψ/∂a)(a,b) = −∫_0 ^∞ ((x^2 e^(−ax^2 ) )/((x^2 +b^2 )^2 )) dx  f_b ′′(a) = (∂^2 Ψ/∂a^2 )(a,b) = +∫_0 ^∞ ((x^4 e^(−ax^2 ) )/((x^2 +b^2 )^2 )) dx    f_b ′′(a)−2b^2 f_b ′(a)+b^4 f_b (a)  = ∫_0 ^∞ (((x^4 +2b^2 x^2 +b^4 )e^(−ax^2 ) )/((x^2 +b^2 )^2 )) dx  = ∫_0 ^∞ (((x^2 +b^2 )^2 e^(−ax^2 ) )/((x^2 +b^2 )^2 )) dx  = ∫_0 ^∞ e^(−ax^2 ) dx  = (1/( (√a)))∫_0 ^∞ e^(−t^2 ) dt  =(1/2) (√(π/a))((2/( (√π)))∫_0 ^∞ e^(−t^2 ) dt)  =(1/2) (√(π/a))    f_b ′′(a)−2b^2 f_b ′(a)+b^4 f_b (a) = (1/2)(√(π/a))    f_b (a) = Ψ(a,b) =  (c_1 +c_2 a)e^(ab^2 ) +((√π)/2)e^(ab^2 ) [aΓ((1/2),b^2 a)+((Γ((3/2),ab^2 ))/b^3 )]  ...to be continued...

$${f}_{{b}} \left({a}\right)\:=\:\Psi\left({a},{b}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{ax}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx} \\ $$ $${f}_{{b}} '\left({a}\right)\:=\:\frac{\partial\Psi}{\partial{a}}\left({a},{b}\right)\:=\:−\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} {e}^{−{ax}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx} \\ $$ $${f}_{{b}} ''\left({a}\right)\:=\:\frac{\partial^{\mathrm{2}} \Psi}{\partial{a}^{\mathrm{2}} }\left({a},{b}\right)\:=\:+\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{4}} {e}^{−{ax}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx} \\ $$ $$ \\ $$ $${f}_{{b}} ''\left({a}\right)−\mathrm{2}{b}^{\mathrm{2}} {f}_{{b}} '\left({a}\right)+{b}^{\mathrm{4}} {f}_{{b}} \left({a}\right) \\ $$ $$=\:\int_{\mathrm{0}} ^{\infty} \frac{\left({x}^{\mathrm{4}} +\mathrm{2}{b}^{\mathrm{2}} {x}^{\mathrm{2}} +{b}^{\mathrm{4}} \right){e}^{−{ax}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx} \\ $$ $$=\:\int_{\mathrm{0}} ^{\infty} \frac{\left({x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} {e}^{−{ax}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx} \\ $$ $$=\:\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}^{\mathrm{2}} } {dx} \\ $$ $$=\:\frac{\mathrm{1}}{\:\sqrt{{a}}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}^{\mathrm{2}} } {dt} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\:\sqrt{\frac{\pi}{{a}}}\left(\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}^{\mathrm{2}} } {dt}\right) \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\:\sqrt{\frac{\pi}{{a}}} \\ $$ $$ \\ $$ $${f}_{{b}} ''\left({a}\right)−\mathrm{2}{b}^{\mathrm{2}} {f}_{{b}} '\left({a}\right)+{b}^{\mathrm{4}} {f}_{{b}} \left({a}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\pi}{{a}}} \\ $$ $$ \\ $$ $${f}_{{b}} \left({a}\right)\:=\:\Psi\left({a},{b}\right)\:= \\ $$ $$\left({c}_{\mathrm{1}} +{c}_{\mathrm{2}} {a}\right){e}^{{ab}^{\mathrm{2}} } +\frac{\sqrt{\pi}}{\mathrm{2}}{e}^{{ab}^{\mathrm{2}} } \left[{a}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},{b}^{\mathrm{2}} {a}\right)+\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}},{ab}^{\mathrm{2}} \right)}{{b}^{\mathrm{3}} }\right] \\ $$ $$...\mathrm{to}\:\mathrm{be}\:\mathrm{continued}... \\ $$

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