A real integral evaluated over a rectangular contour

When an integrand combines exponential functions with a rational factor such as \(\frac{1}{x^2+\pi^2}\), a rectangular contour of height \(2\pi\) can exploit the periodicity of \(e^z\) in the imaginary direction while building the rational factor from the difference of two simple fractions. Using this technique, we will prove that for every \(a \in \mathbb{R}\),

\[ \int_{-\infty}^\infty \frac{e^x}{e^{2x}+e^{2a}}\frac{1}{x^2+\pi^2}\,dx = \frac{2\pi e^{-a}}{4a^2+\pi^2}-\frac{1}{1+e^{2a}}. \]

Setting up the rectangular contour

Consider the complex function

\[ f(z) = \frac{e^z}{(e^{2z}+e^{2a})z}. \]

Notice why \(f(z)\) is well suited to the horizontal lines \(\operatorname{Im} z = \pm \pi\): on those lines \(e^{x \pm i\pi} = -e^x\) and \(e^{2(x \pm i\pi)} = e^{2x}\), so the exponential factor \(\frac{e^z}{e^{2z}+e^{2a}}\) takes the exact same value \(-\frac{e^x}{e^{2x}+e^{2a}}\) on both horizontal edges, while the remaining factor \(\frac{1}{z}\) produces

\[ \frac{1}{x-i\pi} - \frac{1}{x+i\pi} = \frac{2\pi i}{x^2+\pi^2}. \]

Let \(\Gamma\) be the positively oriented rectangle in the complex plane with vertices \(R-i\pi\), \(R+i\pi\), \(-R+i\pi\), and \(-R-i\pi\), where \(R > |a|\):

Positively oriented rectangular contour with vertices at plus or minus R plus or minus i pi enclosing three simple poles

Computing the residues inside the contour

Inside \(\Gamma\), the denominator \((e^{2z}+e^{2a})z\) vanishes at three points, each of which is a simple pole of \(f(z)\):

At \(z = 0\), the residue is

\[ \mathop{\operatorname{Res}}_{z=0} f(z) = \lim_{z \to 0} z f(z) = \frac{1}{1+e^{2a}}. \]

The solutions of \(e^{2z} = -e^{2a} = e^{2a \pm i\pi}\) with \(\operatorname{Im} z \in (-\pi, \pi)\) are \(z_\pm = a \pm \frac{i\pi}{2}\). At a simple zero \(z_0\) of \(e^{2z}+e^{2a}\), differentiating the denominator gives \(\frac{d}{dz}(e^{2z}+e^{2a})\big|_{z=z_0} = 2e^{2z_0} = -2e^{2a}\). Since \(e^{a \pm i\pi/2} = \pm i e^a\), the residues at \(z_\pm\) are

\[ \begin{aligned} \mathop{\operatorname{Res}}_{z=a+\frac{i\pi}{2}} f(z) &= \frac{e^{a+\frac{i\pi}{2}}}{-2e^{2a}\left(a+\frac{i\pi}{2}\right)} = -\frac{i e^{-a}}{2a+i\pi}, \\ \mathop{\operatorname{Res}}_{z=a-\frac{i\pi}{2}} f(z) &= \frac{e^{a-\frac{i\pi}{2}}}{-2e^{2a}\left(a-\frac{i\pi}{2}\right)} = \frac{i e^{-a}}{2a-i\pi}. \end{aligned} \]

By the residue theorem,

\[ \begin{aligned} \int_\Gamma f(z)\,dz &= 2\pi i \left( \mathop{\operatorname{Res}}_{z=0} f(z) + \mathop{\operatorname{Res}}_{z=a+\frac{i\pi}{2}} f(z) + \mathop{\operatorname{Res}}_{z=a-\frac{i\pi}{2}} f(z)\right) \\ &= 2\pi i \left( \frac{1}{1+e^{2a}} - \frac{i e^{-a}}{2a+i\pi} + \frac{i e^{-a}}{2a-i\pi}\right) \\ &= 2\pi i \left( \frac{1}{1+e^{2a}} - \frac{2\pi e^{-a}}{4a^2 +\pi^2}\right). \end{aligned} \tag{1} \]

Relating the contour integral to the real integral

Breaking \(\Gamma\) into its four sides and parameterizing the bottom and top edges by \(z = x \mp i\pi\) for \(x \in [-R, R]\), we have

\[ \begin{aligned} \int_\Gamma f(z)\,dz &= \int_{-R-i\pi}^{R-i\pi} f(z)\,dz + \int_{R-i\pi}^{R+i\pi} f(z)\,dz + \int_{R+i\pi}^{-R+i\pi} f(z)\,dz + \int_{-R+i\pi}^{-R-i\pi} f(z)\,dz \\ &= \int_{-R}^{R} \bigl(f(x-i\pi)-f(x+i\pi)\bigr)\,dx + \int_{R-i\pi}^{R+i\pi} f(z)\,dz + \int_{-R+i\pi}^{-R-i\pi} f(z)\,dz \\ &= -2\pi i \int_{-R}^{R} \frac{e^x}{e^{2x}+e^{2a}}\frac{1}{x^2+\pi^2}\,dx + \int_{R-i\pi}^{R+i\pi} f(z)\,dz + \int_{-R+i\pi}^{-R-i\pi} f(z)\,dz. \end{aligned} \tag{2} \]

Bounding the vertical segments

On the right vertical edge \(z = R + iy\) with \(y \in [-\pi, \pi]\), we have \(|R+iy| \geq R\) and, by the reverse triangle inequality for \(R > |a|\),

\[ |e^{2R+2iy} + e^{2a}| = e^{2R}|e^{2iy} + e^{2a-2R}| \geq e^{2R}(1 - e^{2a-2R}). \]

Hence

\[ \begin{aligned} \left| \int_{R-i\pi}^{R+i\pi} f(z)\,dz\right| &= \left|\int_{-\pi}^\pi f(R+iy)\,i\,dy \right| \\ &\leq e^R \int_{-\pi}^\pi \frac{dy}{|R+iy|\cdot |e^{2R+2iy}+e^{2a}|} \\ &\leq \frac{e^{-R}}{R} \int_{-\pi}^\pi \frac{dy}{|e^{2iy}+e^{2a-2R}|} \\ &\leq \frac{e^{-R}}{R} \frac{2\pi}{1-e^{2a-2R}}. \end{aligned} \tag{3} \]

Similarly, on the left vertical edge \(z = -R + iy\), we have \(|e^z| = e^{-R}\) and \(|e^{-2R+2iy} + e^{2a}| \geq e^{2a} - e^{-2R}\) for \(R > |a|\), so

\[ \left|\int_{-R+i\pi}^{-R-i\pi} f(z)\,dz \right| \leq \frac{e^{-R}}{R}\frac{2\pi}{e^{2a}-e^{-2R}}. \tag{4} \]

From bounds (3) and (4), both vertical integrals tend to \(0\) as \(R \to \infty\). Equating (1) and (2) in the limit \(R \to \infty\) and dividing by \(-2\pi i\) yields

\[ \int_{-\infty}^\infty \frac{e^x}{e^{2x}+e^{2a}}\frac{1}{x^2+\pi^2}\,dx = \frac{2\pi e^{-a}}{4a^2+\pi^2}-\frac{1}{1+e^{2a}}. \]