Fourier transform of $ \log(x^{2}+a^{2}) $ I would like to evaluate the Fourier cosine transform of $\log(x^{2}+a^{2})$
or the integral
$$\int_{0}^{\infty}\cos(ux)\log(x^{2}+a^{2})\,dx$$
for any real $u,a$.
However, it seems that this integral may be divergent due to the behaviour of the logarithm so what should I do?

Let us assume $a>0$ without loss of generality.
It is possible to evaluate this Fourier transform in the sense of distributions as follows. I am going to assume that, due to the parity of the integrand, the one-sided transform from $0$ to $+\infty$ is equal to $1/2$ the Fourier transform from $-\infty$ to $+\infty$,
$$
\int_0^\infty \cos(ux)\log(x^2+a^2)dx=\frac{1}{2}\mathcal F\big[\log(x^2+a^2)\big]\equiv T(u)\,,
$$
where
$$
\mathcal F[\,f\,] \equiv \int_{-\infty}^{+\infty} f(x)\,e^{-iux}dx\,.
$$
Multiplying by $iu$ and using the properties of the Fourier transform
$$
iu T(u)=\frac{1}{2} \mathcal F\big[\partial_x \log(x^2+a^2)\big] =i \partial_u \mathcal F\big[(x^2+a^2)^{-1}\big]\,.
$$
Now, we can use the well known integral
$$
\mathcal F\big[(x^2+a^2)^{-1}\big]=\int_{-\infty}^{+\infty}\frac{e^{-iux}}{x^2+a^2}dx=\frac{\pi}{a}e^{-a|u|}\,,
$$
which can be obtained by contour integration. Substituting above,
$$
uT(u)=\frac{\pi}{a}\partial_u e^{-a|u|}\implies uT(u)=-\pi\, \mathrm{sgn}(u)e^{-a|u|}\,.
$$
The general solution to this equation is
$$\boxed{
T(u)=-\frac{\pi}{|u|}e^{-a|u|}+C\delta(u)\,
}$$
where $|u|^{-1}$ is understood in the sense that, for any test function $\varphi$,
$$
\langle |u|^{-1}, \varphi(u) \rangle =  \int_{|u|>1} \frac{\varphi(u)}{|u|} du + \int_{|u|<1} \frac{\varphi(u)-\varphi(0)}{|u|}du\,,
$$
while $C$ is, for now, an arbitrary real constant. In order to fix $C$, we can calculate the inverse Fourier transform:
$$
\frac{1}{2}\log(x^2+a^2)=(\mathcal F^{-1} T)(x)\,.
$$
Contracting with an arbitrary test function $\varphi$ and denoting the Fourier transform by $\tilde{}$,
$$
\langle \log(x^2+a^2),\varphi(x)\rangle = \langle T(u), \frac{1}{\pi}\tilde \varphi(-u)\rangle\,.
$$
Using the definition of $|u|^{-1}$,
the right-hand side can be recast as
$$
\langle 2\Re\int_0^\infty \frac{du}{u}(\theta(1-u)-e^{-(a-ix)u})+\frac{C}{\pi}, \varphi(x)\rangle\,.
$$
We introduce a regulator to calculate the integrals explicitly:
$$
\int_0^1 \frac{du}{u^{1-\epsilon}}=\frac{1}{\epsilon}
$$
$$
\int_0^\infty \frac{du}{u^{1-\epsilon}}e^{-(a-ix)u}=\frac{\Gamma(\epsilon)}{(a-ix)^\epsilon}=\frac{1}{\epsilon}-\gamma_E-\log(a-ix)+\ldots
$$
Thus,
$$
\log(x^2+a^2)=2\gamma_E+\log(a^2+x^2)+\frac{C}{\pi}\implies C=-2\pi \gamma_E\,.
$$
The final answer is thus
$$\boxed{
\int_0^\infty \cos(ux)\log(x^2+a^2)dx=-\frac{\pi}{|u|}e^{-a|u|}-2\pi \gamma_E\delta(u)\,}.
$$
In this way we have obtained the same result as this answer (or this answer). As emphasized there by Maxim, and also by Mark Viola in the comments, the assumption of trading the original integral with the two-sided transform is not as innocuous as it may look. Upon making that assumption, we indeed recover the same result, while a more rigorous and general approach is presented here by Mark Viola.
