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

$${\int }_0^{+\mathrm{\infty }}\frac{e^{-x^2}}{{(x^2+\frac{1}{2})}^2}dx$$

Commented by Frix last updated on 18/Apr/24

$$\sqrt{\pi }$$

Commented by Shrodinger last updated on 18/Apr/24

$$howplease?$$

Answered by Frix last updated on 18/Apr/24

$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{{\mathrm{e}}^{-x^2}}{{(x^2+\frac{1}{2})}^2}dx=2\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-x^2}(1-\frac{x^4+x^2-\frac{1}{4}}{{(x^2+\frac{1}{2})}^2})dx=$$$$=2\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-x^2}dx-2\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-x^2}\frac{x^4+x^2-\frac{1}{4}}{{(x^2+\frac{1}{2})}^2}dx$$$$2\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-x^2}dx=\sqrt{\pi }$$$$-2\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-x^2}\frac{x^4+x^2-\frac{1}{4}}{{(x^2+\frac{1}{2})}^2}dx={\left[\frac{x{\mathrm{e}}^{-x^2}}{x^2+\frac{1}{2}}\right]}_0^{\mathrm{\infty }}=0$$

Answered by Berbere last updated on 18/Apr/24

$${\int }_0^{\mathrm{\infty }}\frac{e^{-{tx}^2}}{x^2+a}dx=f\left(t\right);f\left(0\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{x^2+a}=\frac{1}{\sqrt{a}}.\frac{\pi }{2}$$$$f^{\prime }\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{-x^2}{x^2+a}{e}^{-{tx}^2}dx=-{\int }_0^{\mathrm{\infty }}{e}^{-{tx}^2}dx+af\left(t\right)$$$$f^{\prime }\left(t\right)=-\frac{1}{2}\sqrt{\frac{\pi }{t}}+af\left(t\right)$$$$f\left(t\right)={ke}^{at}\Rightarrow k^{\prime }=-\frac{1}{2}\sqrt{\frac{\pi }{t}}{e}^{-at}.\Rightarrow k=-\sqrt{\pi }\int \frac{e^{-at}}{2\sqrt{t}}dt;t=x^2$$$$=-\sqrt{\pi }\int e^{-{ax}^2}dx\stackrel{y=x\sqrt{a}}{=}-\frac{\pi }{2}.\frac{1}{\sqrt{a}}erf\left(x\sqrt{a}\right)=-\sqrt{\frac{2\pi }{a}}erf\left(\sqrt{at}\right)+c$$$$f\left(t\right)=(-\frac{\pi }{2\sqrt{a}}erf\left(\sqrt{at}\right)+\frac{\pi }{2\sqrt{a}})e^{at}=\frac{\pi }{2\sqrt{a}}(1-erf\left(\sqrt{at}\right))e^{at}$$$$=\frac{\pi }{2\sqrt{a}}erfc\left(\sqrt{at}\right)e^{at}$$$$\frac{\partial f}{\partial a}=-{\int }_0^{\mathrm{\infty }}\frac{1}{{(x^2+a)}^2}{e}^{-{tx}^2}{\mid }_{(t=1,a=\frac{1}{2})}=-A$$$$-\frac{\partial f}{\partial a}(1,\frac{1}{2})=-\frac{\partial }{\partial a}\frac{\pi }{2\sqrt{a}}erfc\left(\sqrt{a}\right)e^a{\mid }_{a=\frac{1}{2}}$$$$=-[-\frac{\pi }{4a\sqrt{a}}erfc\left(\sqrt{a}\right)e^a+\frac{\pi }{2\sqrt{a}}(-\frac{1}{\sqrt{a\pi }}{e}^{-a}{e}^a+erfc\left(\sqrt{a}\right)e^a).]$$$$\frac{\pi }{2\sqrt{\frac{1}{2}}}erfc\left(\sqrt{\frac{1}{2}}\right)-\frac{\pi }{\sqrt{2}}(-\frac{\sqrt{2}}{\sqrt{\pi }}+erc\left(\frac{1}{\sqrt{2}}\right)e^{\frac{1}{2}})=\sqrt{\pi }$$$${\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}}{{(x^2+\frac{1}{2})}^2}=\sqrt{\pi }$$$$$$$$$$$$$$$$$$

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