I am faced with the following integral, that looks like essentially finding a Laplace transform, and would like to know how to extract the asymptotic behaviour for large argument. The integral in question is, for $n \in \mathbb{Z}^{+}$ $$ \int_{0}^{\infty} ds \left(\frac{1}{s - i n } - \frac{1}{s + in} - \frac{2i}{n} \right) \frac{\textrm{e}^{- s / z}}{s}\,.$$ Here I have removed the small-$s$ divergence from the integrand because I know how to treat it. I would like to get a grip of the functional form of this result for large $z$.
I believe this can be evaluated in terms of the Cosine and Sine exponential integrals, and then one can expand them for large argument; could consider this as the Laplace transform (with argument $1/z$) of the remainder of the integrand and look at its asymptotic form. However, I'd like to analyse the asymptotic behaviour without, necessarily, the need to evaluate the integral exactly or write it in terms of special functions; rather just by looking at the integrand itself...
I wanted to use contour integration, but one can't close the counter in a keyhole because the exponent grows for arguments with large negative real part. The other choice it to return along the imaginary axis, but that's exactly where the poles are.
Any suggestions greatly appreciated.
Let $$g(s) = \frac 1 {s - i n} - \frac 1 {s + i n} - \frac {2 i} n, \\ f(s) = \frac {g(s)} s e^{-s/z}.$$ We have $$I = \int_0^\infty f(s) ds = \left( \int_0^z + \int_z^\infty \right) f(s) ds, \\ \left| \int_z^\infty f(s) ds \right| \leq \int_z^\infty \frac 4 n \frac {e^{-s/z}} s ds = \frac 4 n \int_1^\infty \frac {e^{-s}} s ds = O(1), \\ \int_0^z f(s) ds = \underbrace {\int_0^z \frac {g(s)} s ds}_{ = I_1} + \underbrace {\int_0^z g(s) \frac {e^{-s/z} - 1} s ds}_{= I_2}, \\ |I_2| \leq \int_0^z \frac 4 n \frac {1 - e^{-s/z}} s ds = \frac 4 n \int_0^1 \frac {1 - e^{-s}} s ds = O(1).$$ It remains to estimate $I_1$, which gives $I = -2 i \ln(z)/n + O(1)$ for $z \to \infty$ and constant $n$.