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Question Number 72988 by mathmax by abdo last updated on 05/Nov/19

$$calculatef\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-{xt}^2}}{4+t^2}dtwithx>0$$

Commented by mathmax by abdo last updated on 05/Nov/19

$$wehavef\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-x(t^2+4-4)}}{t^2+4}dt=e^{4x}{\int }_0^{\mathrm{\infty }}\frac{e^{-x(t^2+4)}}{t^2+4}dt$$$$=e^{4x}w\left(x\right)withw\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-x(t^2+4)}}{t^2+4}dt\Rightarrow$$$$w^{{}^{\prime }}\left(x\right)=-{\int }_0^{+\mathrm{\infty }}{e}^{-x(t^2+4)}dt=-e^{-4x}{\int }_0^{\mathrm{\infty }}{e}^{-{\left(\sqrt{x}t\right)}^2}dt$$$${=}_{\sqrt{x}t=u}-e^{-4x}{\int }_0^{\mathrm{\infty }}{e}^{-u^2}\frac{du}{\sqrt{x}}=-\frac{e^{-4x}}{\sqrt{x}}\times \frac{\sqrt{\pi }}{2}\Rightarrow w\left(x\right)=-\frac{\sqrt{\pi }}{2}{\int }_0^x\frac{e^{-4u}}{\sqrt{u}}du+c$$$${=}_{\sqrt{u}=z}-\frac{\sqrt{\pi }}{2}{\int }_0^{\sqrt{x}}\frac{e^{-4z^2}}{z}\left(2z\right)dz+c=-\sqrt{\pi }{\int }_0^{\sqrt{x}}{e}^{-4z^2}dz+c\Rightarrow$$$$f\left(x\right)=c{e}^{4x}-\sqrt{\pi }{e}^{4x}{\int }_0^{\sqrt{x}}{e}^{-4z^2}dz$$$$c={lim}_{x\to 0}f\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+4}{=}_{t=2u}{\int }_0^{\mathrm{\infty }}\frac{2du}{4(1+u^2)}=\frac{1}{2}\frac{\pi }{2}=\frac{\pi }{4}\Rightarrow$$$$f\left(x\right)=\frac{\pi }{4}{e}^{4x}-\sqrt{\pi }{e}^{4x}{\int }_0^{\sqrt{x}}{e}^{-4z^2}dz$$$$$$

Answered by mind is power last updated on 05/Nov/19

$${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)={\int }_0^{+\mathrm{\infty }}\frac{-{\mathrm{t}}^2{\mathrm{e}}^{-{\mathrm{xt}}^2}}{4+{\mathrm{t}}^2}=\int -{\mathrm{e}}^{-{\mathrm{xt}}^2}+4{\int }_0^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{-{\mathrm{xt}}^2}}{4+{\mathrm{t}}^2}\mathrm{dt}$$$$=\Rightarrow {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=4\mathrm{f}\left(\mathrm{x}\right)-\frac{1}{\sqrt{\mathrm{x}}}{\int }_0^{+\mathrm{\infty }}{\mathrm{e}}^{-{\left(\mathrm{t}\sqrt{\mathrm{x}}\right)}^2}.\sqrt{\mathrm{x}}\mathrm{dt}$$$$\Rightarrow {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=4\mathrm{f}\left(\mathrm{x}\right)-\frac{1}{\sqrt{\mathrm{x}}}.{\int }_0^{+\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{u}}^2}\mathrm{du}=4\mathrm{f}\left(\mathrm{x}\right)-\frac{1}{2\sqrt{\mathrm{x}}}.\sqrt{\frac{\pi }{2}}$$$${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)-4\mathrm{f}\left(\mathrm{x}\right)+\frac{1}{2\sqrt{\mathrm{x}}}.\sqrt{\frac{\pi }{2}}=0$$$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ke}}^{4\mathrm{x}}$$$$\Rightarrow {\mathrm{k}}^{\prime }\left(\mathrm{x}\right)=-\frac{{\mathrm{e}}^{-4\mathrm{x}}}{2\sqrt{\mathrm{x}}}\sqrt{\frac{\pi }{2}}$$$$=\Rightarrow \mathrm{k}\left(\mathrm{x}\right)=-\frac{\sqrt{\pi }}{2\sqrt{2}}\int \frac{{\mathrm{e}}^{-4\mathrm{x}}}{2\sqrt{\mathrm{x}}}\mathrm{dx}$$$$\mathrm{erf}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}{\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dt}$$$$\sqrt{\mathrm{x}}=\mathrm{u}\Rightarrow \mathrm{k}\left(\mathrm{x}\right)=-\frac{\sqrt{\pi }}{2\sqrt{2}}\int {\mathrm{e}}^{-{4\mathrm{u}}^2}\mathrm{du}=-\frac{\sqrt{\pi }}{4\sqrt{2}}\int {\mathrm{e}}^{-{\mathrm{w}}^2}\mathrm{dw}=-\frac{\sqrt{\pi }}{4\sqrt{2}}\mathrm{erf}\left(\mathrm{w}\right)$$$$=\frac{\sqrt{\pi }}{4\sqrt{2}}\mathrm{erf}\left(2\mathrm{u}\right)=-\frac{\sqrt{\pi }}{4\sqrt{2}}\mathrm{erf}\left(2\sqrt{\mathrm{x}}\right)+\mathrm{c}$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{ce}}^{-4\mathrm{x}}-\frac{\sqrt{\pi }}{4\sqrt{2}}\mathrm{erf}\left(2\sqrt{\mathrm{x}}\right){\mathrm{e}}^{-4\mathrm{x}}$$$$\mathrm{f}\left(0\right)={\int }_0^{+\mathrm{\infty }}\frac{1}{4+{\mathrm{t}}^2}=[\frac{1}{2}.\arctan \left(\frac{\mathrm{t}}{2}\right)]=\frac{\pi }{4}$$$$\Rightarrow \mathrm{c}=\frac{\pi }{4}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\pi }{4}{\mathrm{e}}^{-4\mathrm{x}}-\frac{\sqrt{\pi }}{4\sqrt{2}}\mathrm{erf}\left(2\sqrt{\mathrm{x}}\right){\mathrm{e}}^{-4\mathrm{x}}$$$$$$$$$$

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