A logarithmic sine integral via a keyhole contour

Integrals on \((0, \infty)\) containing \(\sin(\ln x)\) or \(\cos(\ln x)\) are the imaginary and real parts of \(\int_0^\infty x^i R(x)\,dx\), since \(x^i = e^{i\ln x} = \cos(\ln x) + i\sin(\ln x)\). By placing a keyhole branch cut along the negative real axis \((-\infty, 0]\) for an even rational function \(R(z)\), the integrals along the upper and lower banks of the cut both fold onto \((0, \infty)\) with different constant factors \(e^{-\pi}\) and \(e^{\pi}\). We will use this method to prove

\[ \int_0^\infty \frac{\sin(\ln x)}{x^2+4}\,dx = \frac{\pi \sin(\ln 2)}{4\cosh\left(\frac{\pi}{2}\right)}. \]

The keyhole contour

Consider the function

\[ f(z) = \frac{e^{i\log z}}{4+z^2}, \]

where the branch of the logarithm is chosen with \(-\pi < \arg z \leq \pi\), so the branch cut lies along the negative real axis \((-\infty, 0]\). We integrate \(f(z)\) around the positively oriented keyhole contour shown below:

Keyhole contour cut along the negative real axis with outer circle C_R, inner semicircle C_epsilon around the origin, and simple poles at plus 2i and minus 2i

Here \(C_R\) is a large circle of radius \(R > 2\) and \(C_\epsilon\) is a small semicircle of radius \(\epsilon < 2\), both centered at the origin. On the circle \(|z| = r\), we have \(\left|e^{i\log z}\right| = e^{-\arg z} \leq e^\pi\), so

\[ \left|\int_{C_R} f(z)\,dz\right| \leq \frac{2\pi R e^\pi}{R^2 - 4} \to 0 \quad (R \to \infty) \qquad\text{and}\qquad \left|\int_{C_\epsilon} f(z)\,dz\right| \leq \frac{\pi\epsilon e^\pi}{4 - \epsilon^2} \to 0 \quad (\epsilon \to 0^+). \]

Thus, in the limits \(R \to \infty\) and \(\epsilon \to 0^+\), only the two horizontal segments just above and just below the negative real axis contribute to the contour integral.

Computing the residues at \(\pm 2i\)

Inside the contour, \(f(z)\) has simple poles at \(z = 2i\) and \(z = -2i\). Using the branch \(-\pi < \arg z \leq \pi\), we have \(\log(2i) = \ln 2 + i\frac{\pi}{2}\) and \(\log(-2i) = \ln 2 - i\frac{\pi}{2}\). Hence,

\[ \begin{aligned} \mathop{\operatorname{Res}}_{z=2i} f(z) &= \lim_{z\to 2i} (z-2i) \frac{e^{i\log z}}{z^2+4} \\ &= \frac{e^{i\log(2i)}}{4i} \\ &= \frac{e^{i\ln 2 - \frac{\pi}{2}}}{4i}, \end{aligned} \]

and similarly,

\[ \mathop{\operatorname{Res}}_{z=-2i} f(z) = -\frac{e^{i\log(-2i)}}{4i} = -\frac{e^{i\ln 2 + \frac{\pi}{2}}}{4i}. \]

Evaluating the integral along the branch cut

On the upper bank of the negative real axis (traversed from \(-\infty\) to \(0\)), \(\arg z = +\pi\), so \(\log z = \ln|x| + i\pi\) and \(e^{i\log z} = e^{-\pi}e^{i\ln|x|}\). On the lower bank (traversed from \(0\) to \(-\infty\)), \(\arg z = -\pi\), so \(\log z = \ln|x| - i\pi\) and \(e^{i\log z} = e^{\pi}e^{i\ln|x|}\). Substituting \(t = -x > 0\) (and relabeling \(t\) as \(x\)) on both banks, the residue theorem gives

\[ \begin{aligned} \int_{-\infty}^0 \frac{e^{i(\ln|x|+i\pi)}}{4+x^2}\,dx + \int_0^{-\infty} \frac{e^{i(\ln|x|-i\pi)}}{4+x^2}\,dx &= 2\pi i \left(\mathop{\operatorname{Res}}_{z=2i} f(z) + \mathop{\operatorname{Res}}_{z=-2i} f(z)\right) \\ \implies e^{-\pi}\int_0^\infty \frac{e^{i\ln x}}{4+x^2}\,dx - e^{\pi}\int_0^\infty \frac{e^{i\ln x}}{4+x^2}\,dx &= 2\pi i \left( \frac{e^{i\ln 2}}{4i}e^{-\frac{\pi}{2}}- \frac{e^{i\ln 2}}{4i}e^{\frac{\pi}{2}}\right) \\ \implies -2\sinh(\pi) \int_0^\infty \frac{e^{i\ln x}}{4+x^2}\,dx &= -\sinh\left(\frac{\pi}{2}\right) \pi e^{i\ln 2}. \end{aligned} \]

Using the double-angle identity \(\sinh(\pi) = 2\sinh\left(\frac{\pi}{2}\right)\cosh\left(\frac{\pi}{2}\right)\) to simplify the ratio of hyperbolic sines, we obtain

\[ \int_0^\infty \frac{e^{i\ln x}}{4+x^2}\,dx = \frac{\pi e^{i\ln 2}}{4\cosh\left(\frac{\pi}{2}\right)}. \]

Taking the imaginary part of both sides yields the desired evaluation:

\[ \int_0^\infty \frac{\sin(\ln x)}{x^2+4}\,dx = \frac{\pi \sin(\ln 2)}{4\cosh\left(\frac{\pi}{2}\right)}. \]

Equating the real parts simultaneously gives the companion cosine integral:

\[ \int_0^\infty \frac{\cos(\ln x)}{x^2+4}\,dx = \frac{\pi \cos(\ln 2)}{4\cosh\left(\frac{\pi}{2}\right)}. \]