This is a paper about Fourier cosine series approximation to option pricing problem. The coefficient $A_k$ is defined as $A_k=\frac{2}{b-a}\int_a^bf(x)\cos\left(k\pi\frac{x-a}{b-a}\right)dx$ Then the characteristic function is approximated in the following way $\phi_1(\omega):=\int_a^be^{i\omega x}f(x)dx\approx\int_Re^{i\omega x}f(x)dx=\phi(\omega)$ Then comparing those 2 equations they state $A_k\approx\frac{2}{b-a} \Re \left[ \phi_1\left(\frac{k\pi}{b-a}\right).\exp\left(-i\frac{ka\pi}{b-a}\right) \right]$
On my side , what I see is $A_k=\frac{2}{b-a}\int_a^bf(x)\cos\left(k\pi\frac{x-a}{b-a}\right)dx=\frac{2}{b-a}\int_a^bf(x)\Re \left[ e^{ik\pi \frac{x-a}{b-a}} \right]dx\\=\frac{2}{b-a}\int_a^bf(x)\Re \left[ e^{ik\pi \frac{x}{b-a}}e^{-ik\pi \frac{a}{b-a}} \right]dx$
But then it's not obvious to me that I'm allowed to take out the second exponential from the real part operator, which is the last step to recover what was announced. And I don't see how I'm allowed to take the integral inside the real part either, and here is why:
If I define g in this way $g(x)\equiv \Re \left[ e^{ ik\pi \frac { x }{ b-a } }e^{ -ik\pi \frac { a }{ b-a } } \right]=\frac { e^{ik\pi \frac { x }{ b-a } }e^{ -ik\pi \frac { a }{ b-a } } + \overline { e^{ ik\pi \frac { x }{ b-a } }e^{ -ik\pi \frac { a }{ b-a } } }}{ 2 } $ then $\frac { 2 }{ b-a } \int f(x)\Re \left[ e^{ ik\pi \frac { x }{ b-a } }e^{ -ik\pi \frac { a }{ b-a } } \right] dx=\frac { 2 }{ b-a } \int f(x)g(x)dx$ and from here I don't see why I'd be allowed to take the integral in g(x)... This might be trivial to you but not to me so far.