Limit of a difference of integrals that both look almost identical, Let $\gamma (t) = t+i(e^t-1)$ for $-1\le t \le 1$.
find
$$\lim_{\epsilon \to 0^+} \left[\int_{\gamma} \frac{\sin(z)}{(z-i\epsilon)^2} dz - \int_{\gamma} \frac{\sin(z)}{(z+i\epsilon)^2} dz\right]$$
I have tried using directly the parametrization given and integrating from -1 to 1 against $dt$.  Also, instead of sin(z), I used $e^{iz}$, and planned to take the imaginary part of the result.  No luck...
I then integrated by parts -- also not insightful.
Now I'm wondering whether dominated convergence theorem can be applied.
Any ideas are welcome.
Thanks,
 A: (A very rough outline.)
Write $\gamma_{\epsilon}(t)=\gamma(t)-i\epsilon$. Then the above is:
$$\int_{\gamma_{\epsilon}} \frac{\sin (z+i\epsilon)}{z^2}dz - \int_{\gamma_{-\epsilon}} \frac{\sin (z-i\epsilon)}{z^2}dz$$
Replacing with $\sin(a+b)$ formula, we get:
$$\cos(i\epsilon)\left(\left(\int_{\gamma_{\epsilon}}-\int_{\gamma_{-\epsilon}}\right)\frac{\sin z}{z^2}dz\right)+\\\sin(i\epsilon)
\left(\left(\int_{\gamma_{\epsilon}}+\int_{\gamma_{-\epsilon}}\right)\frac{\cos z}{z^2}\,dz\right)$$
The first part is "almost" a loop around zero, counter-clockwise, with the parts missing approaching zero when $\epsilon\to 0^+$. So the limit is actually the integral $$\int_{\Gamma} \frac{\sin z}{z^2}\,dz$$
for $\Gamma$ a loop around zero clockwise. This is just $2\pi i$, if I remember that step.
I suspect you can use that $\frac{\cos z}{z^2}$ is even to bound the integrals in the second part. Basically, for even functions:
$$\int_{\gamma_{-\epsilon}} =-\int_{-\gamma_{-\epsilon}}$$
So:
$$\left(\int_{\gamma_{\epsilon}}+\int_{\gamma_{-\epsilon}}\right)\frac{\cos z}{z^2}\,dz =\left(\int_{\gamma_{\epsilon}}-\int_{-\gamma_{-\epsilon}}\right)\frac{\cos z}{z^2}\,dz$$
These two paths, $\gamma_{\epsilon}$ and $-\gamma_{-\epsilon}$ are now both on the same side of $0$, and we can connect their endpoints with curves bounded away from zero, to show that these integral are bounded, and thus this term approaches zero when multiplied by $\sin(i\epsilon)$.
So the final result is, I believe, $2\pi i$.
