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Compute $I:=\int e^{it}\cos t\,dt$, where $i\in\mathbb{C}$.

Attempt: Let $u=e^{it}$, $dv=\cos t\,dt$. Then $du=ie^{it}\,dt$ and $v=\sin t$

Using integration by parts, we have $I=u=e^{it}\sin t-i\int e^{it}\sin t\,dt$. Say $I_1:=\int e^{it}\sin t\,dt$. By the same method, we have $I_1=-e^{it}\cos t+i\cdot I$. So we get $ I=e^{it}\sin t-i(-e^{it}\cos t+i\cdot I)=e^{it}\sin t+ie^{it}\cos t+I$.

Can anyone find my mistake?

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Your work is correct and means that, for a imaginary exponent, the classical double integration by part (*), that I suppose you well know for real exponent, does not work. Simply you have found that your primitive function $I$ is defined up to constant $i= e^{it \sin t}+i e^{it}\cos t$.


(*) For a real exponent $a$ we have: $$ \int e^{at}\cos t dt=e^{at}\sin t-a\int e^{at}\sin t dt $$ and $$ \int e^{at}\sin t dt=-e^{at}\cos t+a\int e^{at}\cos t dt $$ so that: $$ \int e^{at}\cos t dt=e^{at}\sin t+a e^{at}\cos t -a^2\int e^{at}\cos t dt $$ and we find: $$ \int e^{at}\cos t dt=\dfrac{e^{at}(\sin t+a \cos t)}{(1+a^2)} $$ up to a constant.

Note that, if $a$ is not a real number, for this derivation we need $a^2\ne -1$.

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  • $\begingroup$ I think it has nothing to do with the exponent being complex or real. It has to do with the integral being indefinite, as opposed to definite. If he had had a definite integral in his exercise, the method he uses would have worked perfectly, even with complex exponents. (See my answer.) $\endgroup$ – Alex M. Jun 10 '15 at 11:52
  • $\begingroup$ I've added to my answer. $\endgroup$ – Emilio Novati Jun 10 '15 at 12:19
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$e^{it}\sin t+ie^{it}\cos t=ie^{it}(\cos t-i\sin t)=i$
so the two integrals differ by a constant.
I think you have no mistake - any integral has an undetermined constant; your two $I$ have different constants.
If you do it as a definite integral, then they differ by $i|_a^b=0$

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