Suppose $D$ is the Dirac operator on a closed spin manifold $M$, with spinors $S$. One can take the functional calculus of $D$ with respect to the continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ given by $x\mapsto\frac{x}{\sqrt{x^2+1}}$. Let's denote this operator by

$$f(D):L^2(S)\rightarrow L^2(S),$$

which is now bounded.

**Part 1 of question:** I've seen it implicitly written that the index of $D$ is unchanged by this modification. Why is this true (perhaps from a functional calculus perspective)?

**Part 2 of question:** Is it possible to see that the index of $D$ is unchanged using the integral expression

$$f(D)\psi = \frac{2}{\pi}\int_0^\infty D(D^2+1+\lambda^2)^{-1}\psi\,\,d\lambda?$$

For instance, if $D$ is invertible as an operator $H^1\rightarrow L^2$ (where $H^1$ is the first Sobolev space), with inverse $D^{-1}:L^2\rightarrow H^1$, can we write down an inverse for $f(D):L^2\rightarrow L^2$ using an integral expression that involves $D^{-1}$?

I should say that although I'm asking this question about Dirac operators in particular, it really applies to more general unbounded operators.

Thanks.