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Question Number 65395 by imron876 last updated on 29/Jul/19

Commented by mathmax by abdo last updated on 29/Jul/19

$$\int {(-1)}^{x}dx=\int e^{i\pi x}dx=\frac{1}{i\pi }{e}^{i\pi x}+c=\frac{{(-1)}^x}{i\pi }+c$$

Answered by ajfour last updated on 29/Jul/19

$$\int e^{i\pi x}dx=\frac{e^{i\pi x}}{i\pi }+c$$$$=\frac{{(-1)}^x}{i\pi }+c.$$

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