I would like to understand how the following two integrals are taken: $$ I_\infty(n,m)=\int\limits_{-\infty}^{\infty}\operatorname{d}\xi \int\limits_{-\infty}^{\infty}\operatorname{d}\xi'\epsilon(\xi-\xi')\operatorname{e}^{i(n\xi+m\xi')} $$ $$ I_\pi(n,m)=\int\limits_{-\pi}^{\pi}\operatorname{d}\xi \int\limits_{-\pi}^{\pi}\operatorname{d}\xi'\epsilon(\xi-\xi')\operatorname{e}^{i(n\xi+m\xi')} $$ where $$ \epsilon(\xi)=1-2\theta(\xi) $$ is the antisymmetric step function.
I know the answer for the second case but don't know how to derive it:
I guess one should change variables to $t=\xi-\xi'$ and $s=\xi+\xi'$, which leads to something like $\int \epsilon(r) \operatorname{d}r \operatorname{e}^{ir(n-m)/2}\int \operatorname{d}s \operatorname{e}^{is(n+m)/2}$. The second integral gives the desired delta but I'm not sure about the first one.
By the definition, $\epsilon(t)=-1$ for $t> 0$ and $\epsilon(t)=1$ for $t<0$. It follows that $$\begin{align} I_\pi(n,m)&=-\int_{-\pi}^{\pi}dx \int_{-\pi}^{x}e^{i(nx+my)}dy+\int_{-\pi}^{\pi}dx \int_{x}^{\pi}e^{i(nx+my)}dy\\ &=-\int_{-\pi}^{\pi}e^{inx}dx \int_{-\pi}^{x}e^{imy}dy+\int_{-\pi}^{\pi}e^{inx}dx \int_{x}^{\pi}e^{imy}dy\\ &=-\int_{-\pi}^{\pi}e^{imy}dy \int_{y}^{\pi}e^{inx}dx+\int_{-\pi}^{\pi}e^{inx}dx \int_{x}^{\pi}e^{imy}dy\\ &=-A(m,n)+A(n,m). \end{align}$$ Hence, if $m\not=0$ and $n\not=0$ then $$\begin{align} A(n,m)&=\int_{-\pi}^{\pi}e^{inx}dx \int_{x}^{\pi}e^{imy}dy =\frac{1}{im}\int_{-\pi}^{\pi}e^{inx}((-1)^m-e^{imx})dx\\ &=\frac{i}{m}\int_{-\pi}^{\pi}e^{i(m+n)x}dx= \frac{2\pi i}{m}\delta_{m,-n} \end{align} $$ and therefore $$I_\pi(n,m)=-A(m,n)+A(n,m)= -\frac{2\pi i}{n}\delta_{-n,m}+\frac{2\pi i}{m}\delta_{m,-n} =-\frac{4\pi i}{n}\delta_{m,-n}.$$ Can you handle the other cases?
P.S. Note that $$A(m,n)+A(n,m)=\int_{-\pi}^{\pi}e^{inx}dx\cdot\int_{-\pi}^{\pi}e^{imy}dy=4\pi^2\delta_{n,0}\,\delta_{m,0}.$$