Proof $\int\Re(f(x))\,\mathrm{d}x=\Re(\int f(x)\,\mathrm{d}x)$ I have a function $f: \mathbb{R}\to\mathbb{C}$. How can I proof/argue that $$\int\Re(f(x))\,\mathrm{d}x=\Re\left(\int f(x)\,\mathrm{d}x\right)$$ (and the same for the imaginary part)? I'm afraid I don't have any idea how to start…
The reason I ask is that I need to proof $\widehat{\overline{f}}(n)=\overline{\widehat{f}(-n)}$ and I'm coming to a point where I need the step from $\frac{1}{2\pi}\int_{-\pi}^\pi\overline{f(x)e^{inx}}\,\mathrm{d}x$ to $\frac{1}{2\pi}\overline{\int_{-\pi}^\pi f(x)e^{inx}\,\mathrm{d}x}$.
 A: Write $f(x)=u(x)+iv(x)\,,$ $u(x)$ and  $v(x)$ are real, and $e^{inx}=\cos(nx)+i\sin(nx)$, then express it in terms of real and imaginary parts. Evaluate each integral and then compare the results.
A: Eventually I realized, it is not necessary to pull $\Re$ and $\Im$ out of the integral: I just want to proof that $\int \overline{f(x)}\,\mathrm{d}x=\overline{\int f(x)\,\mathrm{d}x}$ for some complex function $f$, so I only need to pull out the complex conjugation. I think this could be done like this:
$$
\int \overline{f(x)}\,\mathrm{d}x=
\int \overline{\Re(f(x))+i\Im(f(x))}\,\mathrm{d}x=\\
\int \Re(f(x))-i\Im(f(x))\,\mathrm{d}x=
\int \Re(f(x))\,\mathrm{d}x-\int i\Im(f(x))\,\mathrm{d}x=\\
\overline{\int \Re(f(x))\,\mathrm{d}x+i\int \Im(f(x))\,\mathrm{d}x}=
\overline{\int \Re(f(x))+i\Im(f(x))\,\mathrm{d}x}=\\
\overline{\int f(x)\,\mathrm{d}x}
$$
As this is valid for all complex functions (please comment if I'm wrong), it is also valid for $f(x)e^{inx}$.
Thanks to all who helped me getting in the right direction! :-)

Now that the question is somewhat misleading, should I delete it or edit the question/title?
