# Asymptotic expansion of an integral with singularity, Hydrodynamic free-surface green function

I am frustrated by the asymptotic expansion of the free surface green function when epsilon (or H for I2)tends to zero and with a singularity K(K is a constant). Can anyone help me derive the formula posted in the picture, or give me some suggestions? Thanks very much!

$$I1 = \int_{0}^\infty \frac{e^{-\varepsilon x}} { x-K } dx$$

another integral is (similar to the proceeding one, part of finite depth green function, regular part has omitted),

$$I2 = P.V.\int_{0}^\infty k^{2n}Cosh{(kV)}e^{-kH}(\frac{1} {kSinh(kH)-Cosh(kH) }) dk$$

where, $$k^{2n}$$ comes from the series of the Bessel function of the first kind, H is the water depth, V is a coordinate refer to vertical position. The integral is divergent when H tends to zero. How to derive the asymptotic expansion when H tends to zero.

• It may help to add some background to your question: perhaps you can explain what you attempted and what your difficulties were. Furthermore, there appears to be no picture currently visible in the question. – Ricardo Andrade Jul 14 '13 at 7:44
• Thanks very much!! I have shown more details. Is it possible that I send you the paper by email? – David Jul 14 '13 at 8:02
• That form of integral is very common in hydrodynamic, and to be approximated by polynomials. In order to improve the precision, the singularity needs to be extracted. i.e $$I = f_1asymptotic + f_2polynomial$$ – David Jul 14 '13 at 8:07
• Your formulas, as written make no sense, or you omitted something. The integral is a function of epsilon and K. But in the second formula there is no epsilon and no K. Instead, the second formula contains H. That H is "water depth" explains nothing, because you do not tell what is K. – Alexandre Eremenko Jul 14 '13 at 8:28
• Thanks for your comments. They are two different integrals, but with similar property. I want to derive the asymptotic expansion of that kind. – David Jul 14 '13 at 8:34

The first integral can be written as $$I_1 = \int_{0}^\infty \frac{e^{-x}} { x-K\varepsilon}\, dx,$$ so it is a function of parameter $y=K\varepsilon$. Actually it is almost the exponential integral function $E_1(z)$: $$I_1=e^{K\varepsilon}E_1(K\varepsilon).$$ The series expansion for $E_1$ can be found here DLMF.
Also calculations with Mma indicate that $$\int_{0}^\infty \frac{x^n e^{-\varepsilon x}} { x-K }\, dx= \sum_{m=1}^{n}(m-1)!\varepsilon^{-m}K^{n-m}+ K^ne^{K\varepsilon}E_1(K\varepsilon)=$$ $$K^n\left( \sum_{m=1}^{n}(m-1)!y^{-m}+e^{y}E_1(y)\right).$$
• @David As for the second integral I don't understand the asymptotic to which parameter? Also does $SinH(kH)$ mean $\sin(kH^2)$ or $\sinh(kH)$? – Andrew Jul 14 '13 at 18:34
• @David If $V>0$ for $V>2H$ the integral is already divergent. – Andrew Jul 15 '13 at 17:43