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.

  • $\begingroup$ 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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.