Menu Close

calculate-f-a-b-0-e-ax-2-x-2-b-2-dx-with-a-gt-0-and-b-gt-0-

Tinku Tara 0 Comments
Pin




Question Number 143082 by Mathspace last updated on 09/Jun/21
calculate f(a,b)=∫_0 ^∞ (e^(−ax^2 ) /(x^2 +b^2 ))dx with a>0 and b>0
$${calculate}\:{f}\left({a},{b}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{ax}^{\mathrm{2}} } }{{x}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }{dx} \\ $$$${with}\:{a}>\mathrm{0}\:{and}\:{b}>\mathrm{0} \\ $$
Answered by Dwaipayan Shikari last updated on 09/Jun/21
=∫_0 ^∞ e^(−ax^2 ) ∫_0 ^∞ e^(−t(x^2 +b)) dxdt =((√π)/2)∫_0 ^∞ (e^(−tb) /( (√(t+a))))dt t+a=u^2 =(√π)e^(ab) ∫_(√a) ^∞ e^(−u^2 ) du =(√π)e^(ab) (((√π)/2)−((√π)/2)erf((√a)))=(π/2)e^(ab) (erfc((√a)))
$$=\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}^{\mathrm{2}} } \int_{\mathrm{0}} ^{\infty} {e}^{−{t}\left({x}^{\mathrm{2}} +{b}\right)} {dxdt} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{tb}} }{\:\sqrt{{t}+{a}}}{dt}\:\:\:\:\:\:\:\:\:\:{t}+{a}={u}^{\mathrm{2}} \\ $$$$=\sqrt{\pi}{e}^{{ab}} \int_{\sqrt{{a}}} ^{\infty} {e}^{−{u}^{\mathrm{2}} } {du} \\ $$$$=\sqrt{\pi}{e}^{{ab}} \left(\frac{\sqrt{\pi}}{\mathrm{2}}−\frac{\sqrt{\pi}}{\mathrm{2}}{erf}\left(\sqrt{{a}}\right)\right)=\frac{\pi}{\mathrm{2}}{e}^{{ab}} \left({erfc}\left(\sqrt{{a}}\right)\right) \\ $$
Answered by mathmax by abdo last updated on 10/Jun/21
f(a,b)=∫_0 ^∞ (e^(−ax^2 ) /(x^2 +b^2 ))dx =∫_0 ^∞ (∫_0 ^∞ e^(−(x^2 +b^2 )t) dt)e^(−ax^2 ) dx =∫_0 ^∞ (∫_0 ^∞ e^(−(t+a)x^2 ) dx)e^(−b^2 t) dt [but ∫_0 ^∞ e^(−(t+a)x^2 ) dx =_((√(t+a))x=y) ∫_(√a) ^∞ e^(−y^2 ) (dy/( (√(t+a)))) ⇒ f(a,b)=∫_(√a) ^∞ e^(−y^2 ) dy.∫_0 ^∞ (e^(−b^2 t) /( (√(t+a))))dt (let λ_0 =∫_(√a) ^∞ e^(−y^2 ) dy) =_(t+a=z^2 ) λ_0 ∫_(√a) ^∞ (e^(−b^2 (z^2 −a)) /z)(2z)dz =2λ_0 e^(ab^2 ) ∫_(√a) ^∞ e^(−b^2 z^2 ) dz =_(bz=u) 2λ_0 e^(ab^2 ) ∫_(b(√a)) ^∞ e^(−u^2 ) (du/b) =((2λ_0 )/b)e^(ab^2 ) ∫_(b(√a)) ^∞ e^(−u^2 ) du
$$\mathrm{f}\left(\mathrm{a},\mathrm{b}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{ax}^{\mathrm{2}} } }{\mathrm{x}^{\mathrm{2}} \:+\mathrm{b}^{\mathrm{2}} }\mathrm{dx}\:=\int_{\mathrm{0}} ^{\infty} \left(\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\left(\mathrm{x}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)\mathrm{t}} \mathrm{dt}\right)\mathrm{e}^{−\mathrm{ax}^{\mathrm{2}} } \mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left(\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\left(\mathrm{t}+\mathrm{a}\right)\mathrm{x}^{\mathrm{2}} } \mathrm{dx}\right)\mathrm{e}^{−\mathrm{b}^{\mathrm{2}} \mathrm{t}} \mathrm{dt}\:\left[\mathrm{but}\right. \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\left(\mathrm{t}+\mathrm{a}\right)\mathrm{x}^{\mathrm{2}} } \mathrm{dx}\:=_{\sqrt{\mathrm{t}+\mathrm{a}}\mathrm{x}=\mathrm{y}} \:\:\int_{\sqrt{\mathrm{a}}} ^{\infty} \:\mathrm{e}^{−\mathrm{y}^{\mathrm{2}} } \frac{\mathrm{dy}}{\:\sqrt{\mathrm{t}+\mathrm{a}}}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a},\mathrm{b}\right)=\int_{\sqrt{\mathrm{a}}} ^{\infty} \:\mathrm{e}^{−\mathrm{y}^{\mathrm{2}} } \mathrm{dy}.\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{b}^{\mathrm{2}} \mathrm{t}} }{\:\sqrt{\mathrm{t}+\mathrm{a}}}\mathrm{dt}\:\:\:\left(\mathrm{let}\:\lambda_{\mathrm{0}} =\int_{\sqrt{\mathrm{a}}} ^{\infty} \:\mathrm{e}^{−\mathrm{y}^{\mathrm{2}} } \mathrm{dy}\right) \\ $$$$=_{\mathrm{t}+\mathrm{a}=\mathrm{z}^{\mathrm{2}} } \:\:\lambda_{\mathrm{0}} \int_{\sqrt{\mathrm{a}}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{b}^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} −\mathrm{a}\right)} }{\mathrm{z}}\left(\mathrm{2z}\right)\mathrm{dz}\:=\mathrm{2}\lambda_{\mathrm{0}} \mathrm{e}^{\mathrm{ab}^{\mathrm{2}} } \int_{\sqrt{\mathrm{a}}} ^{\infty} \:\:\mathrm{e}^{−\mathrm{b}^{\mathrm{2}} \mathrm{z}^{\mathrm{2}} } \mathrm{dz} \\ $$$$=_{\mathrm{bz}=\mathrm{u}} \:\:\:\mathrm{2}\lambda_{\mathrm{0}} \mathrm{e}^{\mathrm{ab}^{\mathrm{2}} } \int_{\mathrm{b}\sqrt{\mathrm{a}}} ^{\infty} \:\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \frac{\mathrm{du}}{\mathrm{b}} \\ $$$$=\frac{\mathrm{2}\lambda_{\mathrm{0}} }{\mathrm{b}}\mathrm{e}^{\mathrm{ab}^{\mathrm{2}} } \int_{\mathrm{b}\sqrt{\mathrm{a}}} ^{\infty} \:\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \mathrm{du} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *