Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 184027 by SEKRET last updated on 02/Jan/23

$$$$$$$$$$$$$${\int }_0^{\frac{\mathit{\pi }}{2}}{\mathbf{e}}^{-{\mathit{t}\mathit{g}}^2\left(\mathit{x}\right)}\mathit{d}\mathbf{x}=???$$$$$$$$$$$$$$$$$$

Answered by witcher3 last updated on 03/Jan/23

$$={\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}}{(1+x^2)}dx⩽{\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx=\frac{\sqrt{\pi }}{2}..cv$$$${\int }_0^{\mathrm{\infty }}\frac{e^{-{ax}^2}}{(1+x^2)}dx=F\left(a\right)$$$$f^{\prime }\left(a\right)-f\left(a\right)=-{\int }_0^{\mathrm{\infty }}{e}^{-{ax}^2}dx$$$$={\int }_0^{\mathrm{\infty }}{e}^{-{\left(x\sqrt{a}\right)}^2}.dx=-\frac{\sqrt{\pi }}{2\sqrt{a}}$$$$f^{\prime }-f=-\frac{\sqrt{\pi }}{2\sqrt{a}}$$$$f\left(a\right)={ke}^a$$$$k^{\prime }=-\frac{\sqrt{\pi }}{2}\frac{e^{-a}}{\sqrt{a}}\Rightarrow k=-\frac{\sqrt{\pi }}{2}\int \frac{e^{-a}}{\sqrt{a}}da=-\sqrt{\pi }\int e^{-t^2}dt$$$$=\frac{\pi }{2}erfc\left(t\right)+c,c\in \mathbb{R}$$$$f\left(a\right)=\frac{\pi }{2}erfc\left(\sqrt{a}\right)e^{-a}+{ce}^{-a}$$$$f\left(0\right)=\frac{\pi }{2}\iff c=0$$$$f\left(a\right)={\int }_0^{\mathrm{\infty }}{e}^{-{atg}^2\left(x\right)}dx=\frac{\pi }{2}erfc\left(\sqrt{a}\right)e^{-a}$$$$f\left(1\right)={\int }_0^{\mathrm{\infty }}{e}^{-{tg}^2\left(x\right)}dx=\frac{\pi }{2e}erfc\left(1\right)$$

Commented by SEKRET last updated on 05/Jan/23

$$\mathbf{thanks}\mathbf{sir}.\mathbf{beatiful}\mathbf{solution}$$

Commented by witcher3 last updated on 05/Jan/23

$$withepleasur$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com