I have a simple integral to compute with the residue theorem, but since it has been a while since I have solved integrals with this method I am a bit confused.
The integral I have to calculate is $\int_{-\infty}^{+\infty} q\frac{(e^{iqr}-e^{-iqr})}{q^2+\xi^{-2}} \,dq$
$r, \xi$ are both greater than $0$ and real numbers.
The first thing I want to do is to split the difference into two integrals:
$\int_{-\infty}^{+\infty} q\frac{e^{iqr}}{q^2+\xi^{-2}} \,dq-\int_{-\infty}^{+\infty} q\frac{e^{-iqr}}{q^2+\xi^{-2}} \,dq$
After that, I have tried to compute the first integral using residue theorem:
the two singularities, which are the same also for the second integral, are $z_1=i\xi^{-1}$ and $z_2=-i\xi^{-1}$. Now I have to compute the first residue for $z_1$, which turns out to be $Res(f(z),z_1)=\frac{e^{-r/\xi}}{2}$, while the second residue is $Res(f(z),z_2)=\frac{e^{r/\xi}}{2}$.
With this done, hopefully without errors, I don't remember how to go on. For the first singularity, I think I have to close the path upwards while I have to do the same thing but downward, but do I have to change signs?
What I get as a final result is:
$\int_{-\infty}^{+\infty} q\frac{e^{iqr}}{q^2+\xi^{-2}} \,dq=2\pi i(\frac{e^{-r/\xi}}{2}+\frac{e^{r/\xi}}{2})$
and the same for the second integral but with a minus sign in front of it. The result is obviously wrong, I think the right result is $2\pi ie^{-r/\xi}$.
Where am I wrong? I avoided all the arguments on the circular paths etc., because I'm not interested in a rigorous mathematical threatment, I just want to be able to do the integral. I think what I'm getting wrong are the signs, I don't remember if I have to change signs when I close the path in order to include $z_2$.
Best Answer
I am going to use the following
Theorem: Let $P$ be a finite subset of $\mathbb{C}\setminus\mathbb{R}$, let $f:\mathbb{C}\setminus P\longrightarrow\mathbb{C}$ be an analytic function such that $\lim_{z\to\infty}f(z)=0$, and let $a\in\mathbb{R}\setminus\{0\}$. Then \begin{multline} \int_{-\infty}^\infty e^{iaz}f(z)\,dz=\\ =2\pi i\times\begin{cases}\text{sum of the residues of $f$ on those $\eta\in P$ with }\operatorname{Im}\eta>0&\text{ if }a>0\\-\text{sum of the residues of $f$ on those $\eta\in P$ with }\operatorname{Im}\eta<0&\text{ if }a<0.\end{cases} \end{multline}
Concerning the first integral: since $r>0$, the theorem says that the integral $$ \int_{-\infty}^\infty q\frac{e^{iqr}}{q^2+\xi^{-2}}\,dq. $$ is equal to $2\pi i$ times the sum of the residues of $q\frac {e^{irq}}{q^2+\xi^{-2}}$ on the upper half-plane. It has only one residue there, located at $\frac i\xi$, and the value of that residue is $\frac12e^{-r/\xi}$. So, the integral of this function is equal to $\pi ie^{-r/\xi}$.
Now, since $-r<0$, the theorem also says that the integral $$ \int_{-\infty}^\infty q\frac{e^{-iqr}}{q^2+\xi^{-2}}\,dq. $$ is equal to minus the sum of the residues of $q\frac {e^{-irq}}{q^2+\xi^{-2}}$ on the lower half-plane. The function has only one residue there, located it $-\frac i\xi$, and the value of that residue is, again, $\frac12e^{-r/\xi}$. Therefore, the integral of this function is equal to $-\pi ie^{-r/\xi}$. So, $$ \int_{-\infty}^\infty q\frac{e^{iqr}-e^{-iqr}}{q^2+\xi^{-2}}\,dq=2\pi ie^{-r/\xi}. $$