Gauge Terrell

2022-07-04

How do I solve the following integral ${\int }_{0}^{\mathrm{\infty }}\mathrm{exp}\left(-x\right){\int }_{\sqrt{2kx}}^{\mathrm{\infty }}\mathrm{exp}\left(\frac{-{t}^{2}}{2}\right)\mathrm{d}t\mathrm{d}x$

Mateo Carson

Expert

Integrate by switching the order
$\begin{array}{rl}{\int }_{0}^{\mathrm{\infty }}{e}^{-x}{\int }_{\sqrt{2kx}}^{\mathrm{\infty }}{e}^{-\frac{1}{2}{t}^{2}}dt\phantom{\rule{mediummathspace}{0ex}}dx=& {\int }_{0}^{\mathrm{\infty }}{e}^{-\frac{1}{2}{t}^{2}}{\int }_{0}^{\frac{1}{2k}{t}^{2}}{e}^{-x}dx\phantom{\rule{mediummathspace}{0ex}}dt\\ =& {\int }_{0}^{\mathrm{\infty }}\left({e}^{-\frac{1}{2}{t}^{2}}-{e}^{-\frac{1}{2}\left(1+\frac{1}{k}\right){t}^{2}}\right)dt\\ =& \sqrt{\frac{\pi }{2}}\left(1-\sqrt{\frac{k}{1+k}}\right)\end{array}$

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