Why can't I use complex contour integration to evaluate the real integral $\int_0^\infty f(x)\cos(ax)\, dx$?

Question

If I want to evaluate the integral of real functions like $$\int_0^\infty f(x)\cos(ax) dx$$ where $a$ is a real constant, why can't I simply do contour integration by using the complex function $f(z)cos(az)dz$?

My Attempt

I know using Euler's formula that $\cos ax = e^{iax} - i\sin(ax)$ and based on other examples with integrals of this form, it seems like the correct corresponding complex function to solve is:

$\oint_C f(z)e^{iaz} = \pi iRe \Big\{ f(z)e^{iaz} \Big\}$, where C is the quarter circle contour. If this is true, I don't quite understand why.

Since all I know about $f(x)$ is that it's a rational function, I also don't know if $\int_0^\infty f(x)\cos(ax) dx = \frac{1}{2}\int_{-\infty}^\infty f(x)\cos(ax) dx $

I haven't covered Fourier Series yet and am working to solve this by using Residues.

Thanks!

Answer 1

Well - the real integral from $0$ to $\infty$ is a limit of a path integral along the x-axis from $0$ to $N$, taking the limit $\lim_{N -> \infty}$. The logical way to calculate this is look at a closed loop integral in the complex plane of the function that has real values $f(x)\cos(ax)$ - so we can take $g(z) = f(z)e^{iaz}$ for this. For the closed path integral then you should take something that allows you to calculate the respective path integrals. For example a quarter circle and a path back to $0$ along the imaginary axis. Then apply the residue theorem for any singular points of $f(z)$ within the closed path and take the limit of $N$ to $\infty$.

And you may need to use another contour if you are unable to calculate the respective path integrals, or when those do not vanish when $N$ goes to $\infty$.

For examples see Schaum's outline of complex variables - a truely good book for reference.

Downloadable on https://www.pdfdrive.com/schaums-complex-variables-d18720112.html