Integration Questions

Question Number 174727 by princeDera last updated on 09/Aug/22

$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\$$

Answered by aleks041103 last updated on 10/Aug/22

$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:\frac{{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}= \\$$ $$=\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\int\frac{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:\frac{{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}= \\$$ $$=\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}^{\mathrm{2}} } }{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }{dx}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \frac{−{x}^{\mathrm{2}} {e}^{−{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\$$ $$\int_{\mathrm{0}} ^{\infty} \frac{−{x}^{\mathrm{2}} {e}^{−{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {xe}^{−{x}^{\mathrm{2}} } \left(−\frac{\mathrm{2}{xdx}}{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\right)= \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {xe}^{−{x}^{\mathrm{2}} } {d}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\right)= \\$$ $$=\left[\frac{{xe}^{−{x}^{\mathrm{2}} } }{\mathrm{2}\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}\right]_{\mathrm{0}} ^{\infty} −\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\left({e}^{−{x}^{\mathrm{2}} } −\mathrm{2}{x}^{\mathrm{2}} {e}^{−{x}^{\mathrm{2}} } \right){dx}= \\$$ $$=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} {e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}= \\$$ $$=\int_{\mathrm{0}} ^{\infty} \left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\right){e}^{−{x}^{\mathrm{2}} } {dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}= \\$$ $$=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}−\left(\frac{\mathrm{1}}{\mathrm{2}}+{a}^{\mathrm{2}} \right)\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}^{\mathrm{2}} } {dx}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }= \\$$ $$=\frac{\sqrt{\pi}}{\mathrm{2}}−\left(\frac{\mathrm{1}}{\mathrm{2}}+{a}^{\mathrm{2}} \right)\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}^{\mathrm{2}} } {dx}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} } \\$$ $$\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=\frac{\sqrt{\pi}}{\mathrm{2}{a}^{\mathrm{2}} }+\left(\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} }−\mathrm{1}\right)\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}^{\mathrm{2}} } {dx}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} } \\$$

Commented byaleks041103 last updated on 10/Aug/22

$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}^{\mathrm{2}} } {dx}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{a}^{\mathrm{2}} \left({x}/{a}\right)^{\mathrm{2}} } {d}\left({x}/{a}\right)}{\left({x}/{a}\right)^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{{a}}{f}\left({a}\right) \\$$ $${f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{a}^{\mathrm{2}} {x}^{\mathrm{2}} } {dx}}{{x}^{\mathrm{2}} +\mathrm{1}} \\$$ $${f}\:'\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{−\mathrm{2}{ax}^{\mathrm{2}} {e}^{−{a}^{\mathrm{2}} {x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\mathrm{2}{a}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{1}\right){e}^{−\left({ax}\right)^{\mathrm{2}} } {dx}= \\$$ $$=\mathrm{2}{a}\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{a}^{\mathrm{2}} {x}^{\mathrm{2}} } {dx}}{{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{2}\int_{\mathrm{0}} ^{\infty} {e}^{−\left({ax}\right)^{\mathrm{2}} } {d}\left({ax}\right)= \\$$ $$=\mathrm{2}{af}\left({a}\right)−\sqrt{\pi} \\$$ $$\Rightarrow{f}\:'\left({a}\right)=\mathrm{2}{af}\left({a}\right)−\sqrt{\pi} \\$$ $$\Rightarrow{e}^{−{a}^{\mathrm{2}} } {f}\:'−\mathrm{2}{ae}^{−{a}^{\mathrm{2}} } {f}=−\sqrt{\pi}{e}^{−{a}^{\mathrm{2}} } \\$$ $$\left({e}^{−{a}^{\mathrm{2}} } \right){f}\:'+\left({e}^{−{a}^{\mathrm{2}} } \right)'{f}=\left({e}^{−{a}^{\mathrm{2}} } {f}\right)'=−\sqrt{\pi}{e}^{−{a}^{\mathrm{2}} } \\$$ $$\Rightarrow{e}^{−{a}^{\mathrm{2}} } {f}\left({a}\right)−\mathrm{0}=−\sqrt{\pi}\int_{\infty} ^{{a}} {e}^{−{x}^{\mathrm{2}} } {dx}=+\sqrt{\pi}\left({q}\left(\infty\right)−{q}\left({a}\right)\right) \\$$ $$\Rightarrow{q}\left({a}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}{erf}\left({a}\right) \\$$ $$\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=\frac{\sqrt{\pi}}{\mathrm{2}{a}^{\mathrm{2}} }+\left(\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} }−\mathrm{1}\right)\frac{\mathrm{1}}{{a}}{f}\left({a}\right) \\$$ $$\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=\frac{\sqrt{\pi}}{\mathrm{2}{a}^{\mathrm{2}} }−\frac{\pi}{\mathrm{2}{a}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} }\right){e}^{{a}^{\mathrm{2}} } \left(\mathrm{1}−{erf}\left({a}\right)\right) \\$$ $${this}\:{is}\:{true}\:{for}\:{a}>\mathrm{0} \\$$ $${but}\:{the}\:{initial}\:{integral}\:{is}\:{an}\:{even}\:{function}\:{in}\:{a} \\$$ $$\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}=\frac{\sqrt{\pi}}{\mathrm{2}{a}^{\mathrm{2}} }−\frac{\pi}{\mathrm{2}\mid{a}\mid}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{a}^{\mathrm{2}} }\right){e}^{{a}^{\mathrm{2}} } \left(\mathrm{1}−{erf}\left(\mid{a}\mid\right)\right) \\$$

Commented byprinceDera last updated on 10/Aug/22

$${thank}\:{you}\:{sir} \\$$

Commented bypeter frank last updated on 11/Aug/22

$$\mathrm{thank}\:\mathrm{you} \\$$

Commented byTawa11 last updated on 11/Aug/22

$$\mathrm{Great}\:\mathrm{sir} \\$$