Find integral $\int_{-\infty}^{\infty}\cos{\lambda x}\frac{e^x}{1+e^{3x}}dx$ I want to find integral
$$ \int_{-\infty}^{\infty}\cos(\lambda x)\frac{e^x}{1+e^{3x}}dx $$
where $\lambda$ is real number. First I replaced $e^x$ with $t$. Then the integral changed to
$$ \int_{0}^{\infty}\frac{\cos(\lambda \log t)}{1+t^3}dt$$
There are three poles $\frac{\sqrt{3}+1}{2},-1,\frac{-\sqrt{3}-1}{2}$ and $0$, so I tried to use the residue theorem.The integral route is  the quarter circle of the first quadrant with radius $R$, positive real axis, the qurter circle of the first quadrant with radius $\epsilon$ and positive imaginary axis. But I had a questions.
・How do I evaluate the integral on the imaginary axis? It looks like the integral depends on both $R$ and $\epsilon$. 
Please give me some advice.
 A: I will assume $\lambda$ is real throughout. Integrate the function $$f(z) = \frac{e^{i\lambda \ln z}}{1+z^3}$$
where $\ln z$ is the principal branch of the logarithm, around the following contour:

The assumption that $\lambda$ is real guarantees the integral along the arc tends to $0$. Thus we obtain:
$$\begin{aligned}
\int_0^\infty  {\frac{{{e^{i\lambda \ln x}}}}{{1 + {x^3}}}dx}  - {e^{\frac{{2\pi i}}{3}}}\int_0^\infty  {\frac{{{e^{i\lambda \ln ({e^{\frac{{2\pi i}}{3}}}x)}}}}{{1 + {x^3}}}dx}  &= (2\pi i) \text{Res}\left[f(z),e^{\tfrac{\pi i}{3}} \right] \\
\int_0^\infty  {\frac{{{e^{i\lambda \ln x}}}}{{1 + {x^3}}}dx}  - {e^{\frac{{2\pi i}}{3}}}\int_0^\infty  {\frac{{{e^{i\lambda \left[ {\frac{{2\pi i}}{3} + \ln x} \right]}}}}{{1 + {x^3}}}dx}  &=  - 2\pi i\frac{{{e^{ - \frac{\pi }{3}\lambda }}}}{3}(\frac{1}{2} + \frac{{\sqrt 3 }}{2}i) \\
\int_0^\infty  {\frac{{{e^{i\lambda \ln x}}}}{{1 + {x^3}}}dx}  &=  - \frac{{\pi i}}{3}{e^{ - \frac{\pi }{3}\lambda }}\frac{{1 + \sqrt 3 i}}{{1 - {e^{\frac{{2\pi i}}{3}}}{e^{ - \frac{{2\pi }}{3}\lambda }}}}
\end{aligned}$$
Taking real part yields the value of integral
$$\int_0^{\infty} \frac{\cos(\lambda \ln x)}{1+x^3} dx = \frac{\pi }{{\sqrt 3 }}\frac{{{e^{ - \frac{\pi }{3}\lambda }}(1 + {e^{ - \frac{{2\pi }}{3}\lambda }})}}{{1 + {e^{ - \frac{{2\pi }}{3}\lambda }} + {e^{ - \frac{{4\pi }}{3}\lambda }}}} = \frac{\pi }{{\sqrt 3 }}\frac{{2\cosh \left( {\tfrac{\pi }{3}\lambda } \right)}}{{1 + 2\cosh \left( {\tfrac{{2\pi }}{3}\lambda } \right)}}$$
