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Question Number 7021 by FilupSmith last updated on 06/Aug/16
z=∫_0 ^( t) i(−1)^x dx =∫_0 ^( t) ie^(iπx) dx (1) =i∫_0 ^( t) e^(iπx) dx =i((1/(iπ))(e^(iπt) −1)) z=(1/π)((−1)^t −1) (2) =∫_0 ^( t) ie^(iπx) dx u=e^(iπx) ⇒ du=(1/(iπ))e^(iπx) dx =∫_0 ^( t) ((iπ)/(iπ))ie^(iπx) dx =∫_0 ^( t) iπidu =−π∫_0 ^( t) du =−π(e^(iπt) −1) which is incorrect?
$${z}=\int_{\mathrm{0}} ^{\:{t}} {i}\left(−\mathrm{1}\right)^{{x}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\:{t}} {ie}^{{i}\pi{x}} {dx} \\ $$$$ \\ $$$$\left(\mathrm{1}\right) \\ $$$$={i}\int_{\mathrm{0}} ^{\:{t}} {e}^{{i}\pi{x}} {dx} \\ $$$$={i}\left(\frac{\mathrm{1}}{{i}\pi}\left({e}^{{i}\pi{t}} −\mathrm{1}\right)\right) \\ $$$${z}=\frac{\mathrm{1}}{\pi}\left(\left(−\mathrm{1}\right)^{{t}} −\mathrm{1}\right) \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$$=\int_{\mathrm{0}} ^{\:{t}} {ie}^{{i}\pi{x}} {dx} \\ $$$${u}={e}^{{i}\pi{x}} \:\Rightarrow\:{du}=\frac{\mathrm{1}}{{i}\pi}{e}^{{i}\pi{x}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\:{t}} \frac{{i}\pi}{{i}\pi}{ie}^{{i}\pi{x}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\:{t}} {i}\pi{idu} \\ $$$$=−\pi\int_{\mathrm{0}} ^{\:{t}} {du} \\ $$$$=−\pi\left({e}^{{i}\pi{t}} −\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{incorrect}? \\ $$
Commented by FilupSmith last updated on 07/Aug/16
Ah, i see my mistake, thank you!
$$\mathrm{Ah},\:\mathrm{i}\:\mathrm{see}\:\mathrm{my}\:\mathrm{mistake},\:\mathrm{thank}\:\mathrm{you}! \\ $$
Commented by Yozzii last updated on 06/Aug/16
In (2), for u=e^(iπx) ⇒du=iπe^(iπx) dx ⇒u=e^(πit) at x=t and u=1 at x=0. ⇒∫_0 ^t ie^(πix) dx=i×(1/(iπ))∫_1 ^e^(iπt) du=(1/π)(e^(πit) −1)=(1/π)((−1)^t −1)
$${In}\:\left(\mathrm{2}\right),\:{for}\:{u}={e}^{{i}\pi{x}} \Rightarrow{du}={i}\pi{e}^{{i}\pi{x}} {dx} \\ $$$$\Rightarrow{u}={e}^{\pi{it}} \:{at}\:{x}={t}\:{and}\:{u}=\mathrm{1}\:{at}\:{x}=\mathrm{0}. \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{{t}} {ie}^{\pi{ix}} {dx}={i}×\frac{\mathrm{1}}{{i}\pi}\int_{\mathrm{1}} ^{{e}^{{i}\pi{t}} } {du}=\frac{\mathrm{1}}{\pi}\left({e}^{\pi{it}} −\mathrm{1}\right)=\frac{\mathrm{1}}{\pi}\left(\left(−\mathrm{1}\right)^{{t}} −\mathrm{1}\right) \\ $$$$ \\ $$
Answered by Yozzii last updated on 08/Aug/16
Check for a response in the comments.
$${Check}\:{for}\:{a}\:{response}\:{in}\:{the}\:{comments}. \\ $$

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