Existence of a contour from zero to complex infinity such that an integral along that contour converges Given an arbitrary analytic function that is not a polynomial, does there always exist contours $C_R$ from the origin to some point $R\exp\left[i\theta(R)\right]$, such that the integrals of the given function along $C_R$ have the property that the limit for $R\to\infty$ exists?
Edit: Trigonometric functions (multiplied by polynomials) are counterexamples. But perhaps these are exceptional cases. The question is then if generically you do have converging integrals to complex infinity.
 A: This is only a partial answer.
Your question is closely connected to asymptotic values of entire functions. Let me reinterpret your question slightly: Given an entire function $f$, does there exist a (continuous) curve $C$ parametrized by $\gamma : [0,\infty) \to \mathbb{R}$ with $\gamma(0) = 0$ and $\lim_{t\to\infty} |\gamma(t)| = \infty$ such that
$$
\lim_{R \to \infty} \int_{C_R} f(z)\,dz
$$
exists? Here $C_R$ is the curve parameterized by $\gamma(t)$ for $0 \le t \le R$.
Since $f$ is entire, $f$ has an anti-derivative $F$ (chose the one such that $F(0) = 0)$. Hence, the integral above is just
$$
F(\gamma(R)).
$$
So the question is then whether $F$ has a (finite) asymptotic value.
In general, the answer is no.
There is an old theorem by Wiman (1915) saying that if $F$ is of exponential order $\lambda < 1/2$, then $F$ has no finite asymptotic value. This was later improved to:

If $F$ is of exponential order $\lambda$, then $F$ has at most $2\lambda$ asymptotic values.

There are examples ($F(z) = e^{e^z}$ for instance) that have an infinite number of asymptotic values. I have looked a little in the literature, but haven't found much about conditions that guarantee the existence of asymptotic values. Possibly no good conditions are known.
There are other examples that fail to possess asymptotic values, see for example this paper by Gray and Shah from 1965, where they for example show that
$$
F(z) = \sum_{k=0}^\infty \frac{z^k}{2^{k(k+1)}}
$$
has no asymptotic values.
