An integral of the form
$$\int_{-\infty}^{\infty}\frac{f(x)}{1+x^2}dx\tag{1}$$
can sometimes be easily computed if $f(x)$ can be analytically continued to the complex plane using contour integration. E.g. if $f(z)$ is an entire function that tends to zero fast enough at infinity in the upper half plane, then we have:
$$\int_{-\infty}^{\infty}\frac{f(x)}{1+x^2}dx = \pi f(i) \tag{2}$$
We may assume that $f(x)$ is an even function, as the integral in (1) over the odd part of $f(x)$ would vanish anyway. If we then assume that $f(x)$ has a series expansion around $x=0$ of the form:
$$f(x) = \sum_{k=0}^{\infty}(-1)^k c_k x^{2k}$$
then (2) can be written as:
$$\int_{-\infty}^{\infty}\frac{f(x)dx}{1+x^2} = \pi \sum_{k=0}^{\infty} c_k $$
but this formula is obviously not going to be generally valid, as the analytic continuation of $f(x)$ may not tend to zero at infinity in the upper half plane, the function may have poles or branch point singularities, making the formula not applicable.
Now, I've found a way to get around these problems using heuristic methods. The expression I've found reads:
$$\int_{-\infty}^{\infty}\frac{f(x)}{1+x^2}dx = \pi \sum_{k=0}^{\infty} \left(c_k - c_{k+\frac{1}{2}}\right) \tag{3}$$
This is then valid if $\lim_{k\to\infty}c_k=0$ (so, we don't require the summation of $c_k$ separately to be convergent, $f(z)$ is allowed to have a singularity at $z=i$). But the problem is that it requires one to analytically continue the coefficients $c_k$ to make them well defined for fractional values for $k$. This is not unique and makes (3) ambiguous. However, writing $c_k$ in terms of factorials and replacing these by gamma functions always seems to yield the correct value for the integral.
The question is then to derive (3) using rigorous methods, which will then automatically contain a rigorous definition of $c_k$ for fractional $k$.
Appendix: Heuristic derivation
Eq. (3) can be derived using Ramanujan's master theorem, specializing to the case of $\int_0^{\infty}f(x) dx$ with $f(x) = \sum_{k=0}^{\infty}(-1)^k c_k x^{2k}$, this reduces to Glaisher's theorem obtained by Glaisher in the late 19th century:
$$\int_0^{\infty}f(x) dx = \frac{\pi}{2} c_{-\frac{1}{2}}\tag{A1}$$
One can use (A1) to derive (3) directly, but since the coefficient of $x^{2n}$ is then given by a summation over the $c_n$, one has to analytically continue an upper limit of a summation to do this. This is how I originally obtained (3), but I've now found a simpler way that involves using the same sort of formal heuristic arguments that Glaisher used to obtain (A1).
As a warm-up, let's start with deriving (A1). We introduce an operator $E$ that acts on the coefficients $c_n$ as follows:
$$E c_n = c_{n+1}$$
We can then write:
$$f(x) = \sum_{k=0}^{\infty}(-1)^k c_k x^{2k} = \sum_{k=0}^{\infty}(-1)^k E^k x^{2k} c_0 = \frac{1}{1+E x^2} c_0\tag{A2}$$
If we now integrate this from $0$ to infinity pretending that $E$ is a real number, we get:
$$\int_0^{\infty}f(x)dx = \frac{\pi}{2}E^{-\frac{1}{2}}c_0$$
If we then write $E^{-\frac{1}{2}}c_0$ as $c_{-\frac{1}{2}}$ we get the result (A1).
We can then apply this heuristic logic directly to the integral:
$$\int_0^{\infty}\frac{f(x)}{1+x^2}dx$$
Substituting (A2) in here and integrating, pretending that $E$ is a real number, yields:
$$\int_0^{\infty}\frac{f(x)}{1+x^2}dx = \frac{\pi}{2}\frac{1-\sqrt{E}}{1-E}c_0$$
Expanding this in powers of $E$ yields (3).
