Integration Questions

Question Number 206549 by Shrodinger last updated on 18/Apr/24

$$\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{−{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx} \\$$

Commented by Frix last updated on 18/Apr/24

$$\sqrt{\pi} \\$$

Commented by Shrodinger last updated on 18/Apr/24

$${how}\:{please}\:? \\$$

Answered by Frix last updated on 18/Apr/24

$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{e}^{−{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx}=\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−{x}^{\mathrm{2}} } \left(\mathrm{1}−\frac{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\right){dx}= \\$$$$=\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−{x}^{\mathrm{2}} } {dx}−\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−{x}^{\mathrm{2}} } \frac{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx} \\$$$$\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−{x}^{\mathrm{2}} } {dx}=\sqrt{\pi} \\$$$$−\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−{x}^{\mathrm{2}} } \frac{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx}=\left[\frac{{x}\mathrm{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}}\right]_{\mathrm{0}} ^{\infty} =\mathrm{0} \\$$

Answered by Berbere last updated on 18/Apr/24

$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{tx}^{\mathrm{2}} } }{{x}^{\mathrm{2}} +{a}}{dx}={f}\left({t}\right);{f}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{x}^{\mathrm{2}} +{a}}=\frac{\mathrm{1}}{\:\sqrt{{a}}}.\frac{\pi}{\mathrm{2}} \\$$$${f}'\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \frac{−{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{a}}{e}^{−{tx}^{\mathrm{2}} } {dx}=−\int_{\mathrm{0}} ^{\infty} {e}^{−{tx}^{\mathrm{2}} } {dx}+{af}\left({t}\right) \\$$$${f}'\left({t}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\pi}{{t}}}+{af}\left({t}\right) \\$$$${f}\left({t}\right)={ke}^{{at}} \Rightarrow{k}'=−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\pi}{{t}}}{e}^{−{at}} .\Rightarrow{k}=−\sqrt{\pi}\int\frac{{e}^{−{at}} }{\mathrm{2}\:\sqrt{{t}}}{dt};{t}={x}^{\mathrm{2}} \\$$$$=−\sqrt{\pi}\int{e}^{−{ax}^{\mathrm{2}} } {dx}\overset{{y}={x}\sqrt{{a}}} {=}−\frac{\pi}{\:\mathrm{2}}.\frac{\mathrm{1}}{\:\sqrt{{a}}}{erf}\left({x}\sqrt{{a}}\right)=−\sqrt{\frac{\mathrm{2}\pi}{{a}}}{erf}\left(\sqrt{{at}}\right)+{c} \\$$$${f}\left({t}\right)=\left(−\frac{\pi}{\:\mathrm{2}\sqrt{{a}}}{erf}\left(\sqrt{{at}}\right)+\frac{\pi}{\mathrm{2}\sqrt{{a}}}\right){e}^{{at}} =\frac{\pi}{\mathrm{2}\sqrt{{a}}}\left(\mathrm{1}−{erf}\left(\sqrt{{at}}\right)\right){e}^{{at}} \\$$$$=\frac{\pi}{\mathrm{2}\sqrt{{a}}}{erfc}\left(\sqrt{{at}}\right){e}^{{at}} \\$$$$\frac{\partial{f}}{\partial{a}}=−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +{a}\right)^{\mathrm{2}} }{e}^{−{tx}^{\mathrm{2}} } \mid_{\left({t}=\mathrm{1},{a}=\frac{\mathrm{1}}{\mathrm{2}}\right)} =−{A} \\$$$$−\frac{\partial{f}}{\partial{a}}\left(\mathrm{1},\frac{\mathrm{1}}{\mathrm{2}}\right)=−\frac{\partial}{\partial{a}}\frac{\pi}{\mathrm{2}\sqrt{{a}}}{erfc}\left(\sqrt{{a}}\right){e}^{{a}} \mid_{{a}=\frac{\mathrm{1}}{\mathrm{2}}} \\$$$$=−\left[−\frac{\pi}{\mathrm{4}{a}\sqrt{{a}}}{erfc}\left(\sqrt{{a}}\right){e}^{{a}} +\frac{\pi}{\mathrm{2}\sqrt{{a}}}\left(−\frac{\mathrm{1}}{\:\sqrt{{a}\pi}}{e}^{−{a}} {e}^{{a}} +{erfc}\left(\sqrt{{a}}\right){e}^{{a}} \right).\right] \\$$$$\frac{\pi}{\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}{erfc}\left(\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}\right)−\frac{\pi}{\:\sqrt{\mathrm{2}}}\left(−\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\pi}}+{erc}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right){e}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)=\sqrt{\pi} \\$$$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{\pi} \\$$$$\\$$$$\\$$$$\\$$$$\\$$