I got the following integrals
$\int_{x=-\infty}^\infty \, \frac{1}{x^2-k^2} \, e^{-\sqrt{x^2-k^2} \, a} \, \mathrm{sinh}(\sqrt{x^2-k^2} \, b) \, e^{\mathrm{i}\, x \, y} \,\mathrm{d}x$
$\int_{x=-\infty}^\infty \, \frac{1}{x^2+k^2} \, e^{-\sqrt{x^2+k^2} \, a} \, \mathrm{sinh}(\sqrt{x^2+k^2} \, b) \, e^{\mathrm{i}\, x \, y} \, \mathrm{d}x$
with Im$[k^2]<0$ and therefore no singularities on the real axis and $a,b,y$ are positive real numbers with $a>b$, $\, y>0$.
And the integrals
$\int_{x=-\infty}^\infty \, \frac{1}{x^2-k^2} \, e^{-\sqrt{x^2-k^2} \, b} \, \mathrm{cosh}(\sqrt{x^2-k^2} \, a) \, e^{\mathrm{i}\, x \, y} \, \mathrm{d}x$
$\int_{x=-\infty}^\infty \, \frac{1}{x^2+k^2} \, e^{-\sqrt{x^2+k^2} \, b} \, \mathrm{cosh}(\sqrt{x^2+k^2} \, a) \, e^{\mathrm{i}\, x \, y} \, \mathrm{d}x$
with Im$[k^2]<0$ and therefore no singularities on the real axis and $a,b,y$ are positive real numbers with $b>a$, $\, y>0$.
I thought that the residue theory is maybe a good choice to evaluate these integrals. I guess the solution of the first and second integral as well as the third and fourth integral are more or less the same, since I only got a different sign for $k^2$.
For the third and fourth integral I would have two poles at $x_1 =k $ , $x_2 = -k $ and
$e^{-\sqrt{x^2-k^2} \, b} \, \mathrm{cosh}(\sqrt{x^2-k^2} \, a) = 1$ with $k = (x_1, x_2)$
Therefore, I have two first order poles at $(x_1, x_2)$. Can I use the residue theorem and Jordan's Lemma to evaluate these integrals?
Im am not sure if
$\lim\limits_{|z| \rightarrow +\infty}{\frac{1}{z^2-k^2} \, e^{-\sqrt{z^2-k^2} \, b} \, \mathrm{cosh}(\sqrt{z^2-k^2} \, a)}$
is fulfilled, which is requried for Jordan's Lemma. Since $b>a\;$ I guess it is fulfilled.
For the first and second integral I got one more problem. The poles from the first integral are also at $x_1 =k $ , $x_2 = -k $, but
$e^{-\sqrt{x^2-k^2} \, a} \, \mathrm{sinh}(\sqrt{x^2-k^2} \, b) = 0$ with $k = (x_1, x_2)$
If I use the Tylor series expansion for the hyperbolic sine I get
$\mathrm{sinh}(\sqrt{x^2-k^2} \, b)= (\sqrt{x^2-k^2} \, b) + \frac{(\sqrt{x^2-k^2} \, b)^3}{3!}+\frac{(\sqrt{x^2-k^2} \, b)^5}{5!}+...$
which finally leads to a pole of order $1/2$ at (x_1, x_2). Is it possible to use the residue theorem in this case? I read something about branch cuts to solve it. Would it be possible to split the integral using
$\mathrm{sinh}(\sqrt{x^2-k^2} \, b) = \frac{e^{\sqrt{x^2-k^2} \, b}-e^{-\sqrt{x^2-k^2} \, b}}{2}$,
since then I got first order poles again as far as I understand the problem.
Edit: Since the integrands are symmetric, I would change the limits to $x[0,\infty]$ and $e^{\mathrm{i}\, x \, y}$ to $\mathrm{Cos}( x \, y)$. Using
$\mathrm{sinh}(\sqrt{x^2-k^2} \, b) = \frac{e^{\sqrt{x^2-k^2} \, b}-e^{-\sqrt{x^2-k^2} \, b}}{2}$
and
$\mathrm{cosh}(\sqrt{x^2-k^2} \, b) = \frac{e^{\sqrt{x^2-k^2} \, b}+e^{-\sqrt{x^2-k^2} \, b}}{2}$
I get
$\int_{x=0}^\infty \, \frac{1}{x^2-k^2} \, (e^{-\sqrt{x^2-k^2} \, (a-b)} - e^{-\sqrt{x^2-k^2} \, (a+b)}) \, \mathrm{cos}(\, x \, y) \,\mathrm{d}x$
$\int_{x=0}^\infty \, \frac{1}{x^2+k^2} \, (e^{-\sqrt{x^2+k^2} \, (a-b)} - e^{-\sqrt{x^2+k^2} \, (a+b)}) \, \mathrm{cos}(\, x \, y) \,\mathrm{d}x$
$\int_{x=0}^\infty \, \frac{1}{x^2-k^2} \, (e^{-\sqrt{x^2-k^2} \, (b-a)} + e^{-\sqrt{x^2-k^2} \, (b+a)}) \, \mathrm{cos}(\, x \, y) \,\mathrm{d}x$
$\int_{x=0}^\infty \, \frac{1}{x^2+k^2} \, (e^{-\sqrt{x^2+k^2} \, (b-a)} + e^{-\sqrt{x^2+k^2} \, (b+a)}) \, \mathrm{cos}(\, x \, y) \,\mathrm{d}x$
I do know the integral
$\int_{x=0}^\infty \, \frac{1}{\sqrt{(x^2+k^2)}} \, e^{-\sqrt{x^2+k^2} \, (b-a)} \, \mathrm{cos}(\, x \, y) \,\mathrm{d}x = \mathrm{K}_0(k \, \sqrt{y^2+(b-a)^2})$
with $\mathrm{K}_0$ the modified Bessel function of the second kind. Sadly there is a square root in the denominator compared to my integrals. Any ideas how to solve this?
Thank you for your help.