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Question-65395




Question Number 65395 by imron876 last updated on 29/Jul/19
Commented by mathmax by abdo last updated on 29/Jul/19
∫ (−1)^x dx =∫ e^(iπx) dx =(1/(iπ))e^(iπx)  +c =(((−1)^x )/(iπ)) +c
$$\int\:\left(−\mathrm{1}\right)^{{x}} {dx}\:=\int\:{e}^{{i}\pi{x}} {dx}\:=\frac{\mathrm{1}}{{i}\pi}{e}^{{i}\pi{x}} \:+{c}\:=\frac{\left(−\mathrm{1}\right)^{{x}} }{{i}\pi}\:+{c} \\ $$
Answered by ajfour last updated on 29/Jul/19
∫e^(iπx) dx=(e^(iπx) /(iπ))+c         = (((−1)^x )/(iπ))+c .
$$\int{e}^{{i}\pi{x}} {dx}=\frac{{e}^{{i}\pi{x}} }{{i}\pi}+{c} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\left(−\mathrm{1}\right)^{{x}} }{{i}\pi}+{c}\:. \\ $$

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