1. Duistermaat-Heckman Formula In this section, we consider the case of that $\left(M^{2l},\omega\right)$ is a symplectic manifold. Let $(M,\omega)$ be a symplectic manifold with $\omega$ is a symplectic structure. It means

1. $\omega$ is a non-singular 2-form. i.e. If for any $Y\in \Gamma(TM)$ there always have $\omega(X,Y)=0$, then $X=0$.
   
2. $\rd \omega=0$.

Assume $\omega$ is $S^1$-invariant and the $S^1$-action on $(M,\omega)$ is Hamiltonian. It means there exists a smooth function $\mu\in C^{\infty}(M)$ such that $\rd \mu=i_K\omega$. the $\mu$ is called with momentum map. We still assume that the zero set of $K$ is discrete.


About the generalization of above theorem in the case where the zero set of $K$ may not be discrete you can see[DH]: “J.J.Duistermaat and G.Heckman, Onthevaroation inthe cohomology of the symplectic from of the reduce phase space”.
2. Bott’s Original Idea From Bismut’s lemma, If $\rd_K\Omega =0$, one has
$$\int_M\omega=\int_M\omega\exp\left(-T\rd_K\theta\right), \quad for~~any~~T>0.$$ when $\mathrm{zero(K)}=\emptyset$,
$$\int_M\omega=\lim_{T\rightarrow+\infty}\int_M\omega\exp\left(-T\rd_K\theta\right)=0.$$ Bott’s original proof is different from above. Now we give a description of Bott’s idea.
Let $\omega$ be a $\rd_K$-closed form on $M$ with $K$ has no zeros on $M$. Since $\rd_K(\theta)=|K|^2+\rd\theta=|K|^2\left(1+\frac{\rd\theta}{|K|^2}\right)$, by $\frac{1}{1+x}=\sum_{i=0}^{\infty}(-1)^ix^i$, we can define
$$\left(\rd_K\theta\right)^{-1}= \frac{1}{|K|^2}\left[\sum_{i=0}^{\infty}(-1)^i\left(\frac{\rd\theta}{|K|^2}\right)^i\right].$$ From $K$ is $S^1$-invariant, the $\theta$ is also $S^1$-invariant. It shows $\mathcal{L}_K\theta=0$, i.e., $\rd_K^2\theta =0$. one can verify that
$$\begin{align*} \rd_K\left[\rd_K^{-1}\wedge\theta\wedge\omega\right]=& \rd_K\left(\rd_K\theta\right)^{-1}\wedge\theta\wedge\omega+ \left(\rd_K\right)^{-1}\theta \wedge\rd_K\theta\wedge\omega-\left(\rd_k\theta\right)^{-1} \wedge\theta\wedge\rd\omega\\ =&\left(\rd_K\theta\right)^{-2}\wedge \left(\rd_K\theta\right)^2+\omega-0\\ =&\omega. \end{align*}$$
From this formula and the Stokes formula, one can directly calculate
$$\int_M\omega=0.$$