Let $X$ be a (quasi-)projective, nonsingular, complex variety. Denote by $\mathcal{T}\_X$ its tangent sheaf and by $X^{\mathrm{an}}$ its analytification. I am looking for a proof for the equality


        $\displaystyle \int\_X c\_n(\mathcal{T}\_X) = \chi(X^{\mathrm{an}})$,


i.e. the degree of the top chern class is equal to the topological Euler characteristic of $X$. There's Example 3.2.13 in Fulton's book on intersection theory which briefly mentions this, but it does not give a reference. Can someone help me out with one? Thanks in advance.



Top Chern Class = Euler Characteristic
As an alternative to R. Budney's answer, one might also notice that the Gauss-Bonnet formula (the one you mention - mind that you must assume that $X$ is projective, otherwise the integral might not even make sense) is a consequence of the Hirzebruch-Riemann-Roch theorem. Indeed, the HRR theorem says
$$
\chi(V)=\int\_{X}{\rm Td}({\rm T}X){\rm ch}(V)
$$
where $$\chi(V):=\sum\_{l}{(-1)}^l{\rm rk}(H^l(X,V))$$ is the Euler characteristic of coherent sheaves. Now there is an universal identity of Chern classes
$$
{\rm ch}(\sum\_{r}(-1)^r\Omega\_X^r){\rm Td}(\Omega^\vee\_X)=c^{\rm top}(\Omega^\vee\_X)
$$
(called the Borel-Serre identity). Here $\Omega\_X$ is the sheaf of differential of $X$ and thus $\Omega^\vee\_X={\rm T}X$. Plugging the element $\sum\_{r}(-1)^r\Omega\_{X}^r$ into the HRR theorem, one gets
$$
\sum\_{k,l}(-1)^{l+k}{\rm rk}(H^k(X,\Omega^l))=\int\_{X}c^{\rm top}(TX)
$$
and by the Hodge decomposition theorem
$$
\sum\_{k,l}(-1)^{l+k}{\rm rk}(H^k(X,\Omega^l))=\sum\_{r}{(-1)}^r{\rm rk}(H^r(X({\bf C}),{\bf C}))
$$
where $H^r(X({\bf C}),{\bf C})$ is the $r$-th singular cohomology group. 
The quantity $\sum\_{r}{(-1)}^r{\rm rk}(H^r(X({\bf C}),{\bf C}))$ is the topological Euler characteristic, so this proves what you want. 
The HRR theorem is proved in chap. 15 of Fulton's book (or in Hirzebruch's book "Topological methods...") and the Borel-Serre identity is Ex. 3.2.5, p. 57 of the same book.


