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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.

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  • $\begingroup$ i think you integral doesn't converge on the given interval $\endgroup$ – Dr. Sonnhard Graubner Aug 12 '17 at 6:34
  • $\begingroup$ You are right. So the range of $\lambda$ should be restricted. I want to find the integral when it converge. $\endgroup$ – masutarou Aug 12 '17 at 6:40
  • $\begingroup$ The integral you written in the title do not match with what you really want to find. $\endgroup$ – pisco Aug 12 '17 at 7:05
  • $\begingroup$ The integral given in title seems to converge for any $\lambda$ $\endgroup$ – Claude Leibovici Aug 12 '17 at 8:05
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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:

enter image description here

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 $$\int_0^{\infty} \frac{\cos(\lambda \ln x)}{1+x^3} dx = \frac{\pi }{{\sqrt 3 }}\frac{{{e^{ - \frac{\pi }{3}\lambda }}(1 + {e^{ - \frac{{2\pi }}{3}\lambda }})}}{{1 + {e^{ - \frac{{2\pi }}{3}\lambda }} + {e^{ - \frac{{4\pi }}{3}\lambda }}}} = \frac{\pi }{{\sqrt 3 }}\frac{{2\cosh \left( {\tfrac{\pi }{3}\lambda } \right)}}{{1 + 2\cosh \left( {\tfrac{{2\pi }}{3}\lambda } \right)}}$$

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  • $\begingroup$ Thank you for your answer. But I have one question. Why do you choice the integral route? Why not the quarter circle? $\endgroup$ – masutarou Aug 12 '17 at 8:47
  • $\begingroup$ I choose this contour because you can recover the denumerator $1+z^3$. If you integrate around quarter circle, the integrand along the vertical line will be something like $1+iz^3$, you cannot recover the denumerator in this case. $\endgroup$ – pisco Aug 12 '17 at 8:54
  • $\begingroup$ I see. Thank you. Your way is very skillful:) $\endgroup$ – masutarou Aug 12 '17 at 9:11

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