Question Number 143083 by Mathspace last updated on 09/Jun/21
$${calculate}\:\Psi\left({a},{b}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{ax}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$${with}\:{a}>\mathrm{0}\:{and}\:{b}>\mathrm{0} \\ $$
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...](https://www.tinkutara.com/question/Q143099.png)
$${f}_{{b}} \left({a}\right)\:=\:\Psi\left({a},{b}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{ax}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx} \\ $$$${f}_{{b}} '\left({a}\right)\:=\:\frac{\partial\Psi}{\partial{a}}\left({a},{b}\right)\:=\:−\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} {e}^{−{ax}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx} \\ $$$${f}_{{b}} ''\left({a}\right)\:=\:\frac{\partial^{\mathrm{2}} \Psi}{\partial{a}^{\mathrm{2}} }\left({a},{b}\right)\:=\:+\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{4}} {e}^{−{ax}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx} \\ $$$$ \\ $$$${f}_{{b}} ''\left({a}\right)−\mathrm{2}{b}^{\mathrm{2}} {f}_{{b}} '\left({a}\right)+{b}^{\mathrm{4}} {f}_{{b}} \left({a}\right) \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \frac{\left({x}^{\mathrm{4}} +\mathrm{2}{b}^{\mathrm{2}} {x}^{\mathrm{2}} +{b}^{\mathrm{4}} \right){e}^{−{ax}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \frac{\left({x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} {e}^{−{ax}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}^{\mathrm{2}} } {dx} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{{a}}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}^{\mathrm{2}} } {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\sqrt{\frac{\pi}{{a}}}\left(\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}^{\mathrm{2}} } {dt}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\sqrt{\frac{\pi}{{a}}} \\ $$$$ \\ $$$${f}_{{b}} ''\left({a}\right)−\mathrm{2}{b}^{\mathrm{2}} {f}_{{b}} '\left({a}\right)+{b}^{\mathrm{4}} {f}_{{b}} \left({a}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\pi}{{a}}} \\ $$$$ \\ $$$${f}_{{b}} \left({a}\right)\:=\:\Psi\left({a},{b}\right)\:= \\ $$$$\left({c}_{\mathrm{1}} +{c}_{\mathrm{2}} {a}\right){e}^{{ab}^{\mathrm{2}} } +\frac{\sqrt{\pi}}{\mathrm{2}}{e}^{{ab}^{\mathrm{2}} } \left[{a}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}},{b}^{\mathrm{2}} {a}\right)+\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}},{ab}^{\mathrm{2}} \right)}{{b}^{\mathrm{3}} }\right] \\ $$$$…\mathrm{to}\:\mathrm{be}\:\mathrm{continued}… \\ $$