Is the Euler number of a vector bundle always finite? Let $E$ be an oriented vector bundle with an even rank $n$ over a smooth oriented $n$-dimensional manifold $M$, $e(E)$ denotes the corresponding Euler class, then the Euler number is defined by $\int_Me(E)$. If $E=TM$ with $M$ closed and oriented, then the Euler number coincides with the Euler characteristic $\chi(M)$. When $M$ is compact, the Euler number and the Euler characteristic are trivially finite. When $M$ has a finite good cover, the finiteness of $\chi(M)$ follows from the isomorphsim between the de Rham cohomology and the Cech cohomology. 
QUESTION:  Is $\int_Me(E)$ finite for every smooth oriented manifold $M$? 
 A: This is not the usual definition of the Euler number (usually called the Euler characteristic). The standard definition is the following: if $X$ is a topological space such that $b_i = \operatorname{rank}H_i(X;\mathbb{Z})$ is finite for all $i$ and non-zero for only finitely many $i$, we define the Euler characteristic of $X$ to be the integer $\chi(X) = \sum_{i=0}^{\infty}(-1)^ib_i$.
If $E$ is an oriented rank $k$ vector bundle over $M$, then $e(E) \in H^k(M; \mathbb{Z})$. If $M$ is a compact oriented $n$-dimensional smooth manifold, then $e(TM) \in H^n(M; \mathbb{Z})$ and 
$$\int_Me(TM) = \langle e(TM), [M]\rangle = \chi(M).$$
The tangent bundle of a non-compact oriented $n$-dimensional smooth manifold still has an Euler class, but note that $e(TM) \in H^n(M; \mathbb{Z}) = 0$ so $e(TM) = 0$. 
In general, non-compact manifolds do not even have a well-defined Euler characteristic as they need not have finite rank homology groups. Even when they do have an Euler characteristic, it need not be zero (e.g. $\chi(\mathbb{R}^n) = 1$), so there is no relation to the Euler class which is zero.
