Fourier transform of $\frac{1-e^x}{1+e^x}$

Question

I was trying to compute $\int_{-\infty}^{\infty}e^{ikx}\frac{1-e^x}{1+e^x}\, dx$ from a Mathematical trivium. I tried first with contour integration but finding the right shape was hard. I gave it a go with a rectangle, with corners at $R, R+i\pi, -R+i\pi$ and $-R$ with a detour around $i\pi$ :

$$\oint_C=\int_{-R}^{R}{e^{ikx}\frac{1-e^x}{1+e^x}\, dx}+\int_{0}^{\pi}{e^{ik(R+iz)}\frac{1-e^{R+iz}}{1+e^{R+iz}}\, dz}+\int_{R}^{-R}{e^{ik(z+i\pi)}\frac{1-e^{z+i\pi}}{1+e^{z+i\pi}}\, dz}+\int_{\pi}^{0}{e^{ik(-R+iz)}\frac{1-e^{-R+iz}}{1+e^{-R+iz}}\, dz}-i\pi\mathrm{Res}(z=i\pi)=0$$

But I'm not sure how to make progress here. For example the 3rd integral becomes :$$\int_{-R}^{R}{e^{ikz}\frac{1+e^z}{1-e^{z}}\, dz}$$ Which is not the same as the initial integral on the real axis. Can this be continued or should a different contour/ different integration method be used?

Answer 1

Note that $f(x) = \frac{1 - e^x}{1 + e^x}$ doesn't itself have a convergent Fourier transform as $f \not\in L^1(\mathbb{R})$. Furthermore, the Fourier-Plancherel transform of $f$ does not exist as $f \not\in L^2(\mathbb{R})$.

However, since $f \in L^\infty (\mathbb{R})$, it corresponds to a tempered distribution in $\mathscr{S}'(\mathbb{R})$ (the space of tempered distributions of the Schwartz space $\mathscr{S}(\mathbb{R})$), so there exists a corresponding $\hat{f}$ function (the 'Fourier transform' of $f$) such that $\hat{f}[u] = f\left [\hat{u} \right ]$ for $u \in \mathscr{S}(\mathbb{R})$.

Now, I don't see how to directly get the Fourier transform of $f$, but observe that $Df = -\frac{1}{2}\operatorname{sech}^2 \left (\frac{x}{2}\right )$, and $\widehat{Dg} = -i\xi \hat{g}$ for all $g \in \mathscr{S}'(\mathbb{R})$ and $\widehat{g_1 g_2} = \frac{1}{2\pi} \hat{g_1} * \hat{g_2}$ for all $g_1, g_2 \in \mathscr{S}(\mathbb{R})$. As $\operatorname{sech} \in \mathscr{S}(\mathbb{R})$, we can use these results. We first have \begin{align*} \widehat{\operatorname{sech}}(\xi) &= \int_{-\infty}^\infty \frac{2e^{i\xi x}}{e^x + e^{-x}} \ \mathrm{d}x\\ &= 2\int_0^\infty \frac{u^{i\xi}}{u^2 + 1} \ \mathrm{d}u && \text{using } u = e^x\\ &= \pi \sec \left (\frac{\pi}{2} i\xi \right )\\ &= \pi \operatorname{sech} \left (\frac{\pi}{2} \xi \right ) \end{align*} The integral was evaluated using a keyhole contour about the positive real-axis (and is an elementary exercise in complex analysis).

Hence, \begin{align*} \widehat{\operatorname{sech}^2}(\xi) &= \frac{\pi}{2} \int_{-\infty}^\infty \operatorname{sech}\left (\frac{\pi}{2}x \right ) \operatorname{sech}\left (\frac{\pi}{2} (\xi - x) \right ) \ \mathrm{d}x\\ &= \pi \left [\operatorname{csch} \left (\frac{\pi}{2} \xi \right ) \log \left (\frac{\operatorname{csch} \left (\frac{\pi}{2}x \right )}{\operatorname{csch} \left (\frac{\pi}{2}(\xi - x) \right )} \right ) \right ]_{-\infty}^\infty\\ &= \pi \xi \operatorname{csch} \left (\frac{\pi}{2} \xi \right ) \end{align*} Thus, we have \begin{align*} \widehat{Df} &= -2\pi\xi \operatorname{csch} (\pi \xi)\\ -i\xi \hat{f} &= -2\pi\xi \operatorname{csch} (\pi \xi) \end{align*} Note that this is not a classical equation, but rather an equation in the distributional sense, so we have for some constant $c$ \begin{align*} \hat{f}(\xi) = -2\pi i \operatorname{csch} (\pi \xi) + c \delta_0(\xi) \end{align*} However, we see that $f$ is an odd function, and thus so is $\hat{f}$. Hence, $c = 0$, so we conclude \begin{align*} \boxed{\hat{f}(\xi) = -2\pi i \operatorname{csch} (\pi \xi)} \end{align*}

Answer 2

Using algebra and the Gaussian hypergeometric function $$I_k=\int{e^{ikz}\,\,\frac{1+e^z}{1-e^{z}}\, dz}=\frac{i\, e^{i k z}}{k}\Big(1-2 \, _2F_1\left(1,i k;i k+1;e^z\right)\Big)$$ Then $$J_k(R)=\int_{-R}^{+R}{e^{ikz}\,\,\frac{1+e^z}{1-e^{z}}\, dz}$$ which is a nice function.