UnkleRhaukus
  • UnkleRhaukus
\[\frac2{\sqrt\pi}\int\limits_0^\infty ze^{-z^2}\int\limits_0^ze^{-u^2}\,\mathrm du\,\mathrm dz \]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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UnkleRhaukus
  • UnkleRhaukus
i think i have to change the order of integration
UnkleRhaukus
  • UnkleRhaukus
is this right?\[=\frac2{\sqrt\pi}\int\limits_0^\infty e^{-u^2}\int\limits_u^\infty ze^{-z^2}\,\mathrm dz\,\mathrm du \]
UnkleRhaukus
  • UnkleRhaukus
|dw:1378391683418:dw|

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UnkleRhaukus
  • UnkleRhaukus
|dw:1378391783721:dw|
UnkleRhaukus
  • UnkleRhaukus
|dw:1378392058072:dw|
anonymous
  • anonymous
looks fine
UnkleRhaukus
  • UnkleRhaukus
\[\newcommand \p \newcommand \p \dd [1] { \,\mathrm d#1 } \p \erf [1] { \operatorname{\text{erf} }\left({#1}\right)} \begin{align} \int\limits_0^\infty ze^{-z^2}\erf z\dd z =\int\limits_0^\infty ze^{-z^2}\left[\frac2{\sqrt\pi}\int\limits_0^ze^{-u^2}\dd u\right]\dd z \\ =\frac2{\sqrt\pi}\int\limits_0^\infty ze^{-z^2}\int\limits_0^ze^{-u^2}\dd u\dd z =\frac2{\sqrt\pi}\int\limits_0^\infty e^{-u^2}\int\limits_u^\infty ze^{-z^2}\dd z\dd u \\ \qquad\qquad\text{let } t =z^2 \\ \qquad\qquad \dd t=2z\dd z \\ \\ \qquad\qquad z=\infty \to t=\infty \\ \qquad\qquad z=u \to t=u^2 \\ =\frac1{\sqrt\pi}\int\limits_0^\infty e^{-u^2}\int\limits_{u^2}^\infty e^{-t}\dd t\dd u \\ =\frac1{\sqrt\pi}\int\limits_0^\infty e^{-u^2}\left.\frac{e^{-t}}{-1}\right|_{u^2}^\infty\dd u \\ =\frac1{\sqrt\pi}\int\limits_0^\infty e^{-u^2}e^{-u^2}\dd u \\ =\frac1{\sqrt\pi}\int\limits_0^\infty e^{-2u^2}\dd u \\ =\frac1{\sqrt\pi}\int\limits_0^\infty e^{-(\sqrt2u)^2}\dd u \\ \qquad\qquad\text{let }\sqrt 2u =y \\ \qquad\qquad \sqrt2\dd u =\dd y \\ \\ \qquad\qquad u=\infty \to y=\infty \\ \qquad\qquad u=0 \to y=0 \\ =\frac1{\sqrt\pi}\int\limits_0^\infty e^{-y^2}\frac{\dd y}{\sqrt2} \\ =\frac1{\sqrt{\pi}}\frac{\sqrt\pi}{2}\frac{\erf\infty}{\sqrt2} \\ =\frac1{2\sqrt2} \\ =\frac{\sqrt2}4 \end{align}\]
UnkleRhaukus
  • UnkleRhaukus
cool it worked

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