I would like to solve the following integral using the Fourier Cosine Transform and its Parseval identitiy.
$$I(\gamma,b,\beta)=\int_0^{\infty} \frac{1}{\gamma^2+x^2} e^{-\beta \sqrt{\gamma^2+x^2}} \, \mathrm{cos}(b \, x)\, \mathrm{d}x$$
Therefore I first use the symmetry of the integral to write
$$I(\gamma,b,\beta)=\frac{1}{2}\int_{-\infty}^{\infty} \frac{1}{\gamma^2+x^2} e^{-\beta \sqrt{\gamma^2+x^2}+\mathrm{i}\,b\,x} \, \mathrm{d}x$$
and then use the substitution $x=\gamma \, \mathrm{sinh}(t)$ which leads to $$I(\gamma,b,\beta)=\frac{1}{2}\int_{-\infty}^{\infty} \frac{1}{\gamma \, \mathrm{cosh}(t)} e^{-\beta \, \gamma \, \mathrm{cosh}(t)+\mathrm{i}\, \gamma \,b\,\mathrm{sinh}(t)} \, \mathrm{d}t$$
Now I would use the identity $$\beta \, \gamma \, \mathrm{cosh}(t)-\mathrm{i}\,b\,\mathrm{sinh}(t) = \sqrt{\beta^2+b^2} \, \mathrm{cosh}(t-\varphi)$$ with $\varphi=\mathrm{arctanh}\left(\frac{\mathrm{i}\,b}{\beta}\right)$ to get $$I(\gamma,b,\beta)=\frac{1}{2}\int_{-\infty}^{\infty} \frac{1}{\gamma \, \mathrm{cosh}(t)} e^{- \gamma \, \sqrt{\beta^2+b^2} \, \mathrm{cosh}(t-\varphi)} \, \mathrm{d}t$$
Now I have my first question. If I use the substitution $\tilde{t}=t-\varphi$ and $\varphi$ is purely imaginary are the limits of the integral still $-\infty$ to $\infty$? Assuming this is correct, I use the identity $$\frac{1}{\gamma \, \mathrm{cosh}(\tilde{t}+\varphi)}=\frac{1}{\gamma} \, \frac{\mathrm{cosh}(\tilde{t}) \, \mathrm{cosh}(\varphi)}{\mathrm{cosh}(\tilde{t})^2+\mathrm{sinh}(\varphi)^2}-\, \frac{\mathrm{sinh}(\tilde{t}) \, \mathrm{sinh}(\varphi)}{\mathrm{cosh}(\tilde{t})^2+\mathrm{sinh}(\varphi)^2}$$
The first fraction is symmetric the second one anti-symmetric, which leads to
$$I(\gamma,b,\beta)=\int_{0}^{\infty} \frac{1}{\gamma} \, \frac{\mathrm{cosh}(\tilde{t}) \, \mathrm{cosh}(\varphi)}{\mathrm{cosh}(\tilde{t})^2+\mathrm{sinh}(\varphi)^2} e^{- \gamma \, \sqrt{\beta^2+b^2} \, \mathrm{cosh}(\tilde{t})} \, \mathrm{d}\tilde{t}$$ I would then use the substitution $x=\gamma \, \mathrm{sinh}(\tilde{t})$ which results in
$$I(\gamma,b,\beta)=\int_{0}^{\infty} \frac{\mathrm{cosh}(\varphi)}{x^2+\gamma^2\,(1+\mathrm{sinh}(\varphi)^2)} \, e^{- \sqrt{\beta^2+b^2} \, \sqrt{x^2+\gamma^2}} \, \mathrm{d}x$$ Finally using $$\mathrm{cosh}(\varphi)=\frac{\beta}{\sqrt{\beta^2+b^2}}$$ $$\mathrm{sinh}(\varphi)^2=-\frac{b^2}{\beta^2+b^2}$$ gives $$I(\gamma,b,\beta)=\frac{\beta}{\sqrt{\beta^2+b^2}}\int_{0}^{\infty} \frac{1}{x^2+\gamma^2\,\frac{\beta^2}{b^2+\beta^2}} \, e^{- \sqrt{\beta^2+b^2} \, \sqrt{x^2+\gamma^2}} \, \mathrm{d}x$$
Are this manipulations correct? I would then use the Parseval identity of the Fourier Cosine Transform to further rewrite the integral.
$$\int_{0}^{\infty} \frac{1}{x^2+\gamma^2\,\frac{\beta^2}{b^2+\beta^2}} \,\mathrm{cos}(x \omega) \, \mathrm{d}x =\pi \, \frac{\sqrt{b^2+\beta^2}}{\gamma \, \beta} \, e^{-\omega \frac{\gamma \, \beta}{\sqrt{b^2+\beta^2}}}$$ $$\int_{0}^{\infty} e^{- \sqrt{\beta^2+b^2} \, \sqrt{x^2+\gamma^2}} \,\,\mathrm{cos}(x \omega)\, \mathrm{d}x=\frac{\sqrt{\beta^2+b^2} \, \gamma}{\sqrt{\omega^2+\beta^2+b^2}} \,\mathrm{K}_1\left(\gamma \, \sqrt{\omega^2+\beta^2+b^2}\right)$$
And therefore the integral gives $$I(\gamma,b,\beta)=\pi \,\sqrt{\beta^2+b^2}\int_{0}^{\infty} e^{-\omega \frac{\gamma \, \beta}{\sqrt{b^2+\beta^2}}} \, \frac{1}{\sqrt{\omega^2+\beta^2+b^2}} \, \mathrm{K}_1\left(\gamma \, \sqrt{\omega^2+\beta^2+b^2}\right) \mathrm{d}\omega$$ I would go further from there. Are there any wrong assumpions in this evaluations? Thank you for your help.