# Interchanging integral with real and imaginary operators?

Mathematics Asked on December 24, 2020

$$int sin(3x) cos(nx) dx to Re int sin(3x) e^{inx} dx$$

$$sin(3x) to Im e^{i3x}$$

Hence,

$$Re left( Im int e^{i(3+n)x} dx right)$$

Or,

$$Re left(Im frac{e^{i(3+n)}}{i(3+n)} right) to Re left(Im frac{-ie^{i(3+n)}}{(3+n)} right)$$

Considering,

$$Im frac{-ie^{i(3+n)}}{(3+n)} to -frac{cos(3+n)}{3+n}$$

Hence,

$$int sin(3x) cos(nx) dx = – frac{ cos(3+n)}{3+n}$$

Now this is wrong.. why?
Btw I am using result from here

You have indeed

$$sin 3x = Im e^{i3x}$$ And therefore:

$$(sin 3x) e^{inx} = (Im e^{i3x}) e^{inx}$$ and $$Re[(sin 3x) e^{inx}] = Re[(Im e^{i3x}) e^{inx}]$$ which is not equal to $$Re[Im (e^{i(3+n)x})]$$

In general for two complex numbers

$$Im(zz^prime) neq Im(z) z^prime$$

Example

$$Im( i cdot i) = 0 neq i = Im(i) cdot i$$

Correct answer by mathcounterexamples.net on December 24, 2020

As @MarkViola noted, the problem is nothing to do with calculus, only with how you manipulate complex numbers. To wit, with $$w:=exp3ix,,z:=exp inx$$ you seem to argue $$Im wRe z=ReIm(wz)$$ (which would be $$Im(wz)$$, by the way). But$$w=a+ib,,z=c+id,,a,,b,,c,,dinBbb RimpliesIm wRe z=bc,,Im(wz)=ad+bc.$$The original problem can be solved without complex numbers using$$2sin3xcos nx=sin[(n+3)x]-sin[(n-3)x].$$

Answered by J.G. on December 24, 2020

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