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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.

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  • $\begingroup$ I think that what you are looking for is the Watson's lemma. $\endgroup$ – rafa11111 Apr 9 at 22:13
  • $\begingroup$ Thank you for the reference. Unfortunately this lemma will generate positive powers of z, which grow uncontrollably for large z. The $x$ in the reference you give plays the role of $1/z$, so the dominant term would come from the parts of the sum with $n \rightarrow \infty$. $\endgroup$ – lux Apr 9 at 22:45
  • $\begingroup$ You can cheat a little and look at how the antiderivative is expressed in terms of the exponential integral. Then you can repeat the steps that give an asymptotic expansion of the exponential integral at zero. For constant $n$, your integral is asymptotically equivalent to $-2 i \ln(z)/n$. $\endgroup$ – Maxim Apr 12 at 23:28
  • $\begingroup$ @Maxim - could you expand you answer to explain explicitly how you got that asymptotic form? I understand you are essentially doing the integral and then expanding the result? $\endgroup$ – lux Apr 13 at 0:32
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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 $n$.

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