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

Question

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

Answer 1

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

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

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