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Question Number 210839 by klipto last updated on 19/Aug/24

$$\mathbf{show}\mathbf{that}{\int }_0^{\mathbf{x}}{\mathbf{e}}^{\mathbf{xt}}{\mathbf{e}}^{-{\mathrm{t}}^2}\mathbf{dt}={\mathbf{e}}^{\frac{{\mathbf{x}}^2}{4}}{\underset{0}{\int }}^{\mathbf{x}}{\mathbf{e}}^{-\frac{{\mathbf{t}}^2}{4}}$$$$$$

Commented by Frix last updated on 20/Aug/24

$$\underset{0}{\stackrel{x}{\int }}{\mathrm{e}}^{xt}{\mathrm{e}}^{-t^2}dt=[u=2t-x]$$$$=\frac{{\mathrm{e}}^{\frac{x^2}{4}}}{2}\underset{-x}{\stackrel{x}{\int }}{\mathrm{e}}^{-\frac{u^2}{4}}du=$$$$={\mathrm{e}}^{\frac{x^2}{4}}\underset{0}{\stackrel{x}{\int }}{\mathrm{e}}^{-\frac{u^2}{4}}du$$

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