# Do these integrals evaluate to terms involving Kronecker Deltas?

There is a famous integral representation of the Kronecker delta: $$\delta_{N,M} = \int_0^1 dx\ e^{-2 \pi i (N-M)x}$$

Noting this, I have encountered two integrals where $N,M,m \in \mathbb{Z}$. First: $$\mathcal{I}_{1} = \int_{m-\tfrac{1}{2}}^{m+\tfrac{1}{2}} dx\ x \ e^{-2 \pi i (N-M)x}$$ I have a reference which seems to be implying that $\mathcal{I}_1 = m \delta_{M,N}$.

The second integral is worse (for any $y > 0$): $$\mathcal{I}_{2} = \int_{m-\tfrac{1}{2}}^{m+\tfrac{1}{2}} dx\ \sqrt{ x^2 + y^2 } \ e^{-2 \pi i (N-M)x}$$ I have no clue how to evaluate either of these. Do they both involve a $\delta_{N,M}$?

First note that your representation of the Kronecker delta is still correct if we shift the limits of integration: $$\delta_{N,M} = \int \limits_{m-\frac{1}{2}}^{m+\frac{1}{2}} \mathrm{d} x \, \mathrm{e}^{- 2 \pi \mathrm{i} (N-M) x} \, .$$
For the first integral we get $$\mathcal{I}_1 = \int \limits_{m-\frac{1}{2}}^{m+\frac{1}{2}} \mathrm{d} x \, x = m$$ if $N=M$ . If $N \neq M$, we can integrate by parts to obtain \begin{align} \mathcal{I}_1 &= \left[- \frac{x \mathrm{e}^{- 2 \pi \mathrm{i} (N-M) x}}{2 \pi \mathrm{i} (N-M)}\right]_{x=m-\frac{1}{2}}^{x=m+\frac{1}{2}} + \frac{1}{2 \pi \mathrm{i} (N-M)}\int \limits_{m-\frac{1}{2}}^{m+\frac{1}{2}} \mathrm{d} x \, \mathrm{e}^{- 2 \pi \mathrm{i} (N-M) x} \\ &= \frac{\mathrm{i} (-1)^{N-M}}{2 \pi (N-M)} \, , \end{align} since the second integral vanishes by our initial observation. Therefore it does not lead to a Kronecker delta. We see that $\operatorname{Re} (\mathcal{I}_1) = m \delta_{N,M}$ holds, however.