# Computing $\int_{-\infty}^{\infty}\frac{e^{iax}}{1 + e^x} dx$

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) 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

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

$$\Gamma_4: z=-R+iy$$, $$y:2\pi\rightarrow 0$$.
$$\int_{\Gamma_4}\frac{e^{iaz}}{e^z+1}dz=\int_{2\pi}^0\frac{e^{-iaR}e^{-ay}}{e^{-R}e^{iy}+1}idy\approx -ie^{-iaR}\int_0^{2\pi}e^{-ay}dy=i\frac{e^{-iaR}}{a}(1-e^{-2\pi a})$$ 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.