How to present the function by the Fourier integral? I have a function $f(x) = \log(4+x^2)-2\log(x)$ defined on the interval $[0; \infty)$
I should to present it by the Fourier integral continuing it evenly on the interval $(-\infty;0]$ in this form:
$$\int_0^\infty[a(z)\cos zx+b(z)\sin zx]dx$$
I continued it in an even way and got the following schedule
The function is even so $b(z) = 0$
But I can`t find a(z). My teacher said that I should find it using inverse Fourier transformation. Can you help me,please?
 A: We have $$f(x)=\log \left(1+\frac{4}{x^2} \right)$$
I will use the exponential form of Fourier transform:
$$\hat{f}(\omega)=\int_{-\infty}^\infty f(x) e^{-2 \pi i \omega x} dx$$
Because the function is even, the sign of the imaginary part doesn't really matter.
As the OP says, it is correct to write:
$$\hat{f}(\omega)=\int_{-\infty}^\infty f(x) \cos (2 \pi \omega x) dx$$
So we need to find:
$$\int_{-\infty}^\infty \log \left(1+\frac{4}{x^2} \right) \cos (2 \pi \omega x) dx$$
Mathematica takes this integral and gives:
$$\hat{f}(\omega)=\frac{1-e^{-4 \pi | \omega|}}{| \omega|}$$
It's almost an (even) hyperbolic function, slightly modified by the vanishing exponential term, so it is finite at $0$. $\hat{f}(0)=4 \pi$.

If we don't want to use the angular frequency, we can change it to:
$$2 \pi \omega=z$$
$$\hat{f}(z)=2 \pi \frac{1-e^{-2 | z|}}{| z|}$$

Update:
The inverse transform is easy to prove by https://en.wikipedia.org/wiki/Frullani_integral, using the exponential inverse Fourier transform.
Update 2:
The original integral can also be found, if we use the Feinman's trick:
$$I(a)=\int_{-\infty}^\infty \log \left(1+\frac{a^2}{x^2} \right) \cos (2 \pi \omega x) dx$$
$$\frac{d}{da}I(a)=2a\int_{-\infty}^\infty \frac{\cos (2 \pi \omega x)}{a^2+x^2 }  dx=2\int_{-\infty}^\infty \frac{\cos (2 \pi \omega a x)}{1+x^2 }  dx=2 \pi e^{-2 \pi |\omega a|}$$
Now we integrate w.r.t. $a$:
$$I(a)=-\frac{1}{|\omega|} e^{-2 \pi |\omega a|} +\frac{1}{|\omega|}$$
Where the constant of integration was determined by the original integral (it's not that obvious, but I will leave the discussion out).
Finally we set $a=2$ and obtain:
$$I(2)=\frac{1-e^{-4 \pi |\omega |}}{|\omega|}  $$
