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Question Number 9541 by FilupSmith last updated on 14/Dec/16
∫(−1)^x xdx=??
$$\int\left(−\mathrm{1}\right)^{{x}} {xdx}=?? \\ $$
Answered by FilupSmith last updated on 15/Dec/16
∫(−1)^x xdx=∫e^(xln(−1)) xdx =∫e^(xiπ) xdx =uv−∫u′vdx u=x v′=e^(iπx) u′=1 v=−(i/π)e^(iπx) ∫e^(iπx) xdx=−(i/π)e^(iπx) x+(i/π)∫e^(iπx) dx ∫e^(iπx) xdx=−(i/π)e^(iπx) x+(i/π)(−(i/π))e^(iπx) +c ∫e^(iπx) xdx=−(i/π)e^(iπx) x+(1/π^2 )e^(iπx) +c ∫e^(iπx) xdx=e^(iπx) ((1/π^2 )−((ix)/π))+c ∴∫(−1)^x xdx=(−1)^x ((1/π^2 )−((ix)/π))+c
$$\int\left(−\mathrm{1}\right)^{{x}} {xdx}=\int{e}^{{x}\mathrm{ln}\left(−\mathrm{1}\right)} {xdx} \\ $$$$=\int{e}^{{xi}\pi} {xdx} \\ $$$$={uv}−\int{u}'{vdx} \\ $$$${u}={x}\:\:\:\:\:\:\:\:\:\:\:{v}'={e}^{{i}\pi{x}} \\ $$$${u}'=\mathrm{1}\:\:\:\:\:\:\:\:\:\:{v}=−\frac{{i}}{\pi}{e}^{{i}\pi{x}} \\ $$$$\int{e}^{{i}\pi{x}} {xdx}=−\frac{{i}}{\pi}{e}^{{i}\pi{x}} {x}+\frac{{i}}{\pi}\int{e}^{{i}\pi{x}} {dx} \\ $$$$\int{e}^{{i}\pi{x}} {xdx}=−\frac{{i}}{\pi}{e}^{{i}\pi{x}} {x}+\frac{{i}}{\pi}\left(−\frac{{i}}{\pi}\right){e}^{{i}\pi{x}} +{c} \\ $$$$\int{e}^{{i}\pi{x}} {xdx}=−\frac{{i}}{\pi}{e}^{{i}\pi{x}} {x}+\frac{\mathrm{1}}{\pi^{\mathrm{2}} }{e}^{{i}\pi{x}} +{c} \\ $$$$\int{e}^{{i}\pi{x}} {xdx}={e}^{{i}\pi{x}} \left(\frac{\mathrm{1}}{\pi^{\mathrm{2}} }−\frac{{ix}}{\pi}\right)+{c} \\ $$$$\therefore\int\left(−\mathrm{1}\right)^{{x}} {xdx}=\left(−\mathrm{1}\right)^{{x}} \left(\frac{\mathrm{1}}{\pi^{\mathrm{2}} }−\frac{{ix}}{\pi}\right)+{c} \\ $$

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