4
$\begingroup$

I am attempting to calculate \begin{equation} \int_{-\infty}^{\infty}\frac{e^{iax}}{1 + e^x} dx \end{equation} using contour integration around a rectangular region in the upper half plane containing one singularity at $i\pi$. The rectangular region has `height' $2\pi i$. Please see pictured (forgive the crude drawing - if I were feeling less lazy I would have used Tikz)

enter image description here

I believe the result should be \begin{equation}\frac{-\pi i}{\sinh{\pi a}}\end{equation} based on plotting results using this versus the case where the integral is calculated numerically, provided $a$ obeys certain criteria.

I indeed get this result by summing the contributions from $\int_{\Gamma_1}, \int_{\Gamma_3}$. All that remains is demonstrating that $\int_{\Gamma_2}, \int_{\Gamma_4} = 0$. I have been able to demonstrate $\int_{\Gamma_2} \rightarrow 0$ as $X_2 \rightarrow \infty$ provided $a > 0$.

I have been unable to demonstrate that $\int_{\Gamma_4}\rightarrow0$ as $X_1 \rightarrow \infty$ and was wondering if someone could help?

I imagine there may be a further condition on $a$. Thanks

$\endgroup$
9
  • $\begingroup$ When $x\to-\infty$ the integrand does not go to $0,$ so that direction seems like a problem. $\endgroup$ Commented Oct 26, 2022 at 20:56
  • $\begingroup$ The integral doesn't converge $\endgroup$
    – FShrike
    Commented Oct 26, 2022 at 21:41
  • $\begingroup$ @FShrike. Yes it does. $x\mapsto 1+e^x$ has no real zero and near $\infty$, $e^x$ grows faster than any polynomial. $\endgroup$
    – Medo
    Commented Oct 26, 2022 at 21:44
  • $\begingroup$ Mathematica states that it only converges $-1 < Im[a] < 0$ which makes sense. $\endgroup$ Commented Oct 26, 2022 at 21:50
  • 2
    $\begingroup$ You can find $\int_0^\infty,$ but not $\int_{-\infty}^\infty.$ $\endgroup$ Commented Oct 26, 2022 at 21:59

1 Answer 1

1
$\begingroup$

This is kinda long for a comment... For respectful friends.

...

$\Gamma_4: z=-R+iy$, $y:2\pi\rightarrow 0$.

$$\int_{\Gamma_4}\frac{e^{iaz}}{e^z+1}dz=\int_{2\pi}^0\frac{e^{-iaR}e^{-ay}}{e^{-R}e^{iy}+1}idy\approx -ie^{-iaR}\int_0^{2\pi}e^{-ay}dy=i\frac{e^{-iaR}}{a}(1-e^{-2\pi a})$$ It is not going to zero. Sorry. But if we choose $R=\frac{2\pi N}{a}$ where $N$ is integer, it has the value $\frac{i}{a}(1-e^{-2\pi a})$. Nonsense.

$\endgroup$
2
  • $\begingroup$ This should be a comment, I think $\endgroup$
    – FShrike
    Commented Oct 26, 2022 at 22:21
  • $\begingroup$ I will add WA outputs. Wait... $\endgroup$
    – Bob Dobbs
    Commented Oct 27, 2022 at 8:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .