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After-looking-at-a-previous-question-I-was-wondering-if-the-following-is-correct-I-n-0-n-1-x-dx-n-R-I-n-0-n-e-ipix-dx-1-I-n-i-pi-e-ipix-0-n-I-n-i-pi-e-ipix-0-n-

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Question Number 6441 by Temp last updated on 27/Jun/16
After looking at a previous question I was wondering if the following is correct: I(n)=∫_0 ^( n) (−1)^x dx, n∈R I(n)=∫_0 ^( n) e^(iπx) dx (1) I(n)=−(i/π)(e^(iπx) )_0 ^n I(n)=−(i/π)(e^(iπx) )_0 ^n I(n)=−(i/π)(e^(iπn) −1) I(n)=∫_0 ^( n) (cos(x)+isin(x))dx (2) I(n)=(sin(x)−icos(x))_0 ^n I(n)=−i(cos(x)+isin(x))_0 ^n I(n)=−i(e^(iπn) −1) Which is correct?
$$\mathrm{After}\:\mathrm{looking}\:\mathrm{at}\:\mathrm{a}\:\mathrm{previous}\:\mathrm{question} \\ $$$$\mathrm{I}\:\mathrm{was}\:\mathrm{wondering}\:\mathrm{if}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{is}\:\mathrm{correct}: \\ $$$${I}\left({n}\right)=\int_{\mathrm{0}} ^{\:{n}} \left(−\mathrm{1}\right)^{{x}} {dx},\:\:{n}\in\mathbb{R} \\ $$$${I}\left({n}\right)=\int_{\mathrm{0}} ^{\:{n}} {e}^{{i}\pi{x}} {dx}\:\:\:\left(\mathrm{1}\right) \\ $$$${I}\left({n}\right)=−\frac{{i}}{\pi}\left({e}^{{i}\pi{x}} \right)_{\mathrm{0}} ^{{n}} \\ $$$${I}\left({n}\right)=−\frac{{i}}{\pi}\left({e}^{{i}\pi{x}} \right)_{\mathrm{0}} ^{{n}} \\ $$$${I}\left({n}\right)=−\frac{{i}}{\pi}\left({e}^{{i}\pi{n}} −\mathrm{1}\right) \\ $$$$ \\ $$$${I}\left({n}\right)=\int_{\mathrm{0}} ^{\:{n}} \left(\mathrm{cos}\left({x}\right)+{i}\mathrm{sin}\left({x}\right)\right){dx}\:\:\:\left(\mathrm{2}\right) \\ $$$${I}\left({n}\right)=\left(\mathrm{sin}\left({x}\right)−{i}\mathrm{cos}\left({x}\right)\right)_{\mathrm{0}} ^{{n}} \\ $$$${I}\left({n}\right)=−{i}\left(\mathrm{cos}\left({x}\right)+{i}\mathrm{sin}\left({x}\right)\right)_{\mathrm{0}} ^{{n}} \\ $$$${I}\left({n}\right)=−{i}\left({e}^{{i}\pi{n}} −\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{Which}\:\mathrm{is}\:\mathrm{correct}? \\ $$
Commented by Yozzii last updated on 27/Jun/16
In (2) cosx+isinx=e^(ix) ≠e^(πix) . ∫_0 ^n e^(ix) dx=(1/i)e^(ix) ∣_0 ^n =−i(e^(ni) −1) (1) is correct, (2) is incorrect.
$${In}\:\left(\mathrm{2}\right)\:{cosx}+{isinx}={e}^{{ix}} \neq{e}^{\pi{ix}} . \\ $$$$\int_{\mathrm{0}} ^{{n}} {e}^{{ix}} {dx}=\frac{\mathrm{1}}{{i}}{e}^{{ix}} \mid_{\mathrm{0}} ^{{n}} =−{i}\left({e}^{{ni}} −\mathrm{1}\right) \\ $$$$\left(\mathrm{1}\right)\:{is}\:{correct},\:\left(\mathrm{2}\right)\:{is}\:{incorrect}. \\ $$
Commented by Temp last updated on 27/Jun/16
Ahhh, thanks! Silly error
$$\mathrm{A}{hhh},\:\mathrm{thanks}!\:\mathrm{Silly}\:\mathrm{error} \\ $$
Commented by Temp last updated on 27/Jun/16
I checked and if i fix (2) it has the same result!
$$\mathrm{I}\:\mathrm{checked}\:\mathrm{and}\:\mathrm{if}\:\mathrm{i}\:\mathrm{fix}\:\left(\mathrm{2}\right)\:\mathrm{it}\:\mathrm{has}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{result}! \\ $$

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