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I found the following integrals in "Tables of Integrals, Series and Products Vol. 8" p. 494-495

$$\int_0^{\infty}e^{-\beta \sqrt{\gamma^2+x^2}} \, \mathrm{cos}(b \, x) \, \mathrm{d}x= \frac{\beta \, \gamma}{A} \, \mathrm{K}_1(\gamma A)$$

$$\int_0^{\infty} \sqrt{\gamma^2+x^2} e^{-\beta \sqrt{\gamma^2+x^2}} \, \mathrm{cos}(b \, x) \, \mathrm{d}x= \frac{\beta^2 \, \gamma^2}{A^2} \, \mathrm{K}_0(\gamma A)+\left(\frac{2\beta^2 \, \gamma}{A^3}-\frac{\gamma}{A}\right) \mathrm{K}_1(\gamma A)$$

$$\int_0^{\infty} (\gamma^2+x^2) e^{-\beta \sqrt{\gamma^2+x^2}} \, \mathrm{cos}(b \, x) \, \mathrm{d}x= \left(\frac{-3\beta \, \gamma^2}{A^2}+\frac{4\beta^3 \gamma^2}{A^4}\right) \, \mathrm{K}_0(\gamma A)+ \\ +\left(\frac{-6 \beta \, \gamma}{A^3}+\frac{8\beta^3\gamma}{A^5}+\frac{\beta^3\gamma^3}{A^3}\right)\mathrm{K}_1(\gamma A)$$

$$\int_0^{\infty}\frac{1}{\sqrt{\gamma^2+x^2}}e^{-\beta \sqrt{\gamma^2+x^2}} \, \mathrm{cos}(b \, x) \, \mathrm{d}x= \mathrm{K}_0(\gamma A)$$

with $A=\sqrt{\beta^2+b^2}$, Re$[\beta]>0$, Re$[\gamma]>0$, $b>0$.

It seems to me that there is some kind of system behind it. Can anybody give me a hint to proof this formulars? There are similar ones for $\mathrm{sin}(b \, x)$.

I would like to solve the integral

$$\int_0^{\infty} \frac{1}{\gamma^2+x^2} e^{-\beta \sqrt{\gamma^2+x^2}} \, \mathrm{cos}(b \, x)\, \mathrm{d}x$$

There is another one in the tables

$$\int_0^{\infty} \left(\frac{1}{\beta(\gamma^2+x^2)^{3/2}}+\frac{1}{\gamma^2+x^2}\right) e^{-\beta \sqrt{\gamma^2+x^2}} \, \mathrm{cos}(b \, x) \, \mathrm{d}x= \frac{1}{\beta \gamma} A \, \mathrm{K}_1(\gamma A)$$

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Let $F(b,\beta,\gamma)$ be given by the integral

$$\begin{align} F(b,\beta,\gamma)&=\int_0^\infty \frac{e^{-\beta \sqrt{x^2+\gamma^2}}}{\sqrt{x^2+\gamma^2}}\,\cos(bx)\,dx\tag 1 \end{align}$$

In a similar approach to the one that I took in developing THIS ANSWER, we enforce the substitution $x=\gamma \sinh(t)$ in $(1)$ to obtain

$$\begin{align} F(b,\beta,\gamma)&=\int_0^\infty e^{-\beta \gamma \cosh(t)}\,\cos(b\gamma \sinh(t))\,dt \\\\ &=\frac12\int_0^\infty e^{-\beta \gamma \cosh(t)+ib\gamma \sinh(t)}\,dt+\frac12\int_0^\infty e^{-\beta \gamma \cosh(t)-ib\gamma \sinh(t)}\,dt\\\\ &=\frac12\int_0^\infty e^{-\beta \gamma \cosh(t)+ib\gamma \sinh(t)}\,dx+\frac12\int_{-\infty}^0 e^{-\beta \gamma \cosh(t)+ib\gamma \sinh(t)}\,dx\\\\ &= \frac12 \int_{-\infty}^\infty e^{-\gamma \sqrt{\beta^2+b^2} \cosh(t)}\,dt\\\\ &=\int_0^\infty e^{-\gamma \sqrt{\beta^2+b^2} \cosh(t)}\,dt \tag 2\\\\ &=K_0(\gamma \sqrt{\beta^2+b^2}) \tag 3 \end{align}$$

where in going from $(2)$ to $(3)$ we used the well-known integral representation for the modified Bessel function of the second kind as found, for example, HERE. Therefore, we find the alternative integral representation for $K_0(\gamma \sqrt{\beta^2+b^2})$ as

$$\bbox[5px,border:2px solid #C0A000]{K_0(\gamma \sqrt{\beta^2+b^2})=\int_0^\infty \frac{e^{-\beta \sqrt{x^2+\gamma^2}}}{\sqrt{x^2+\gamma^2}}\,\cos(bx)\,dx} \tag 4$$


We can use the integral representation given in $(4)$ to derive several relationships by differentiating under the integral sign. Proceeding, we have

$$\begin{align} -\frac{\partial}{\partial \beta}\left(K_0(\gamma \sqrt{\beta^2+b^2})\right)&=\int_0^\infty e^{-\beta \sqrt{x^2+\gamma^2}}\,\cos(bx)\,dx\\\\ \end{align}$$

Then, using $K_0'(x)=-K_1(x)$, we can write

$$\bbox[5px,border:2px solid #C0A000]{\frac{\gamma \beta}{\sqrt{\beta^2+b^2}}K_1(\gamma \sqrt{\beta^2+b^2})=\int_0^\infty e^{-\beta \sqrt{x^2+\gamma^2}}\,\cos(bx)\,dx}\tag 5$$

Note that $(5)$ is the first relationship presented in the OP. And continuing to differentiate under the integral and exploiting the identities for $K_0'(x)=-K_1(x)$ and $K_1'(x)=-K_0(x)-\frac1x K_1(x)$ we can arrive at the second and third identities rather straightforwardly.

Finally, the integral of interest $I(b,\beta,\gamma)=\int_0^\infty \frac{e^{-\beta \sqrt{x^2+\gamma^2}}}{x^2+\gamma^2}\,\cos(bx)\,dx$ satisfies the relationship

$$-\frac{\partial I(b,\beta,\gamma)}{\partial \beta}=K_0(\gamma \sqrt{\beta^2+b^2})$$

Further development is left to the reader

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  • $\begingroup$ Thank you. Sadly it does not resolve my problem, since I can not get the form of my integral with that approach. $\endgroup$ – Michael_K Aug 2 '16 at 13:04
  • $\begingroup$ I am wondering if $\int_0^{\infty} \frac{\mathrm{cos}(xt)}{t^2+z^2} \mathrm{d}t = \mathrm{K}_{\frac{1}{2}}(x \, z)\, \frac{\sqrt{\pi} \, \sqrt{x}}{\sqrt{2z}}$ can help. $\endgroup$ – Michael_K Aug 2 '16 at 13:40
  • $\begingroup$ Thank you very much for your help. I guess you used $\beta \, \cosh(t) - i \, b \sinh(t)=\sqrt{\beta^2+b^2} \, \cosh(t+ \, \mathrm{arctanh}(\frac{-i \, b}{\beta}))$ $\endgroup$ – Michael_K Aug 3 '16 at 6:49
  • $\begingroup$ with $\beta>0$ and $b>0$. And then used another substitution $\tilde{t}=t+ \, \mathrm{arctanh}(\frac{-i \, b}{\beta})$ $\endgroup$ – Michael_K Aug 3 '16 at 7:04
  • $\begingroup$ Sorry for all the comments, I was not able to edit them. I tried to solve $$-\frac{\partial I(b,\beta,\gamma)}{\partial \beta}=K_0(\gamma \sqrt{\beta^2+b^2})$$ before, but I can not find any information how to do this. That is why I wanted to solve it directly. Can it be that the solution has nothing to do with Bessel Functions? Would it be possible to use the residue theorem to solve the original integral? Or maybe even $\int_{x=-\infty}^\infty \, \frac{1}{x^2-k^2} \, e^{-\sqrt{x^2-k^2} \, a} \, \mathrm{sinh}(\sqrt{x^2-k^2} \, b) \, e^{\mathrm{i}\, x \, y} \,\mathrm{d}x$ $\endgroup$ – Michael_K Aug 3 '16 at 8:26

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