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Question Number 148809 by jlewis last updated on 31/Jul/21

$$\underset{\mathrm{i}=0}{\sum ^{\mathrm{\infty }}}[(-0.{5\mathrm{x}}^2)\mathrm{i}/\mathrm{i}!]\mathrm{integrate}$$

Answered by qaz last updated on 31/Jul/21

$$\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{-\frac{1}{2}{\mathrm{x}}^2\mathrm{n}}{\mathrm{n}!}$$$$=-\frac{1}{2}{\mathrm{x}}^2\left(\mathrm{yD}\right){\mid }_{\mathrm{y}=1}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{y}}^{\mathrm{n}}}{\mathrm{n}!}$$$$=-\frac{1}{2}{\mathrm{x}}^2\left({\mathrm{ye}}^{\mathrm{y}}\right){\mid }_{\mathrm{y}=1}$$$$=-\frac{\mathrm{e}}{2}{\mathrm{x}}^2$$

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