Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 143083 by Mathspace last updated on 09/Jun/21

$$calculate\mathrm{\Psi }(a,b)={\int }_0^{\mathrm{\infty }}\frac{e^{-{ax}^2}}{{(x^2+b^2)}^2}dx$$$$witha>0andb>0$$

Answered by Olaf_Thorendsen last updated on 10/Jun/21

$$f_b\left(a\right)=\mathrm{\Psi }(a,b)={\int }_0^{\mathrm{\infty }}\frac{e^{-{ax}^2}}{{(x^2+b^2)}^2}dx$$$$f_b^{\prime }\left(a\right)=\frac{\partial \mathrm{\Psi }}{\partial a}(a,b)=-{\int }_0^{\mathrm{\infty }}\frac{x^2{e}^{-{ax}^2}}{{(x^2+b^2)}^2}dx$$$$f_b^{″}\left(a\right)=\frac{{\partial }^2\mathrm{\Psi }}{\partial a^2}(a,b)=+{\int }_0^{\mathrm{\infty }}\frac{x^4{e}^{-{ax}^2}}{{(x^2+b^2)}^2}dx$$$$$$$$f_b^{″}\left(a\right)-2b^2{f}_b^{\prime }\left(a\right)+b^4{f}_b\left(a\right)$$$$={\int }_0^{\mathrm{\infty }}\frac{(x^4+2b^2{x}^2+b^4)e^{-{ax}^2}}{{(x^2+b^2)}^2}dx$$$$={\int }_0^{\mathrm{\infty }}\frac{{(x^2+b^2)}^2{e}^{-{ax}^2}}{{(x^2+b^2)}^2}dx$$$$={\int }_0^{\mathrm{\infty }}{e}^{-{ax}^2}dx$$$$=\frac{1}{\sqrt{a}}{\int }_0^{\mathrm{\infty }}{e}^{-t^2}dt$$$$=\frac{1}{2}\sqrt{\frac{\pi }{a}}\left(\frac{2}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}{e}^{-t^2}dt\right)$$$$=\frac{1}{2}\sqrt{\frac{\pi }{a}}$$$$$$$$f_b^{″}\left(a\right)-2b^2{f}_b^{\prime }\left(a\right)+b^4{f}_b\left(a\right)=\frac{1}{2}\sqrt{\frac{\pi }{a}}$$$$$$$$f_b\left(a\right)=\mathrm{\Psi }(a,b)=$$$$(c_1+c_{2}a)e^{{ab}^2}+\frac{\sqrt{\pi }}{2}{e}^{{ab}^2}[a\mathrm{\Gamma }(\frac{1}{2},b^{2}a)+\frac{\mathrm{\Gamma }(\frac{3}{2},{ab}^2)}{b^3}]$$$$...\mathrm{to}\mathrm{be}\mathrm{continued}...$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com