Mellin transforms of sine and cosine via a quarter-circle contour

Rotating the positive real axis onto the positive imaginary axis turns the oscillatory factor \(e^{ix}\) into the rapidly decaying factor \(e^{-y}\), converting oscillatory improper integrals into Euler's gamma integral. Using this idea, we will establish the formulas

\[ \begin{aligned} \int_0^\infty x^{s-1} \sin(x)\,dx &= \begin{cases} \sin\left(\frac{\pi s}{2}\right)\Gamma(s), & s\in (-1,0)\cup (0,1), \\ \frac{\pi}{2}, & s=0, \end{cases} \\ \int_0^\infty x^{s-1} \cos(x)\,dx &= \cos\left(\frac{\pi s}{2}\right)\Gamma(s), \qquad 0 < s < 1. \end{aligned} \]

Contour integration for \(0 < s < 1\)

Let \(f(z) = z^{s-1} e^{iz}\), where \(0 < s < 1\) and the principal branch of the logarithm (\(-\pi < \arg z \leq \pi\)) defines \(z^{s-1} = e^{(s-1)\log z}\). We integrate \(f(z)\) around the positively oriented indented quarter-circle contour shown below, and then let \(R \to \infty\) and \(\epsilon \to 0^+\):

Quarter-circle contour in the first quadrant bounded by arcs C_epsilon and C_R and segments along the positive real and imaginary axes

Here \(C_R\) and \(C_\epsilon\) are quarter-circles centered at \(z=0\) of radii \(R\) and \(\epsilon\), respectively, and \(C_\epsilon\) avoids the branch point at \(z=0\). Since \(f(z)\) is analytic inside and on the contour, Cauchy's integral theorem gives

\[ \int_\epsilon^R f(z)\,dz + \int_{C_R} f(z)\,dz + \int_{iR}^{i\epsilon} f(z)\,dz + \int_{C_\epsilon}f(z)\,dz = 0. \tag{1} \]

On the vertical segment from \(iR\) down to \(i\epsilon\), parameterizing by \(z = iy\) (\(y \in [\epsilon, R]\), \(dz = i\,dy\)) and using \((iy)^{s-1} i = i^s y^{s-1} = e^{\frac{i\pi s}{2}} y^{s-1}\) gives

\[ \int_{iR}^{i\epsilon}f(z)\,dz = -i \int_\epsilon^R f(iy)\,dy = -e^{\frac{i\pi s}{2}}\int_\epsilon^R y^{s-1} e^{-y}\,dy. \tag{2} \]

Bounding the circular arcs

On \(C_R\), parameterize \(z = R e^{i\theta}\) for \(0 \leq \theta \leq \frac{\pi}{2}\). Then \(\left|e^{i R e^{i\theta}}\right| = e^{-R\sin\theta}\), and by Jordan's inequality \(\sin\theta \geq \frac{2\theta}{\pi}\) on \(\left[0, \frac{\pi}{2}\right]\), we obtain

\[ \begin{aligned} \left|\int_{C_R} f(z)\,dz\right| &= R^s \left| \int_0^{\frac{\pi}{2}} e^{i R e^{i\theta}} e^{i\theta s}\,d\theta\right| \\ &\leq R^s \int_0^{\frac{\pi}{2}} \left| e^{i R e^{i\theta}} \right|\,d\theta \\ &\leq R^s \int_0^{\frac{\pi}{2}} e^{-R \frac{2\theta}{\pi}}\,d\theta \\ &= \frac{\pi}{2}R^{s-1} \left(1-e^{-R} \right). \end{aligned} \]

Since \(s < 1\), we have \(R^{s-1} \to 0\) as \(R \to \infty\), so

\[ \lim_{R\to \infty}\left|\int_{C_R} f(z)\,dz\right| = 0. \tag{3} \]

By the exact same estimate on \(C_\epsilon\) (or simply noting \(\left|e^{i\epsilon e^{i\theta}}\right| \leq 1\) so the integral is bounded by \(\frac{\pi}{2}\epsilon^s\)), the condition \(s > 0\) ensures that \(1 - e^{-\epsilon} = O(\epsilon)\) and hence

\[ \left|\int_{C_\epsilon} f(z)\,dz\right| \leq \frac{\pi}{2}\epsilon^{s-1} \left(1-e^{-\epsilon} \right) = O(\epsilon^s) \to 0 \quad \text{as} \quad \epsilon \to 0^+. \tag{4} \]

Combining equations (1), (2), (3), and (4) in the limits \(R \to \infty\) and \(\epsilon \to 0^+\) yields

\[ \begin{aligned} \int_0^\infty x^{s-1} e^{ix}\,dx &= e^{\frac{i\pi s}{2}} \int_0^\infty y^{s-1} e^{-y}\,dy \\ &= \left(\cos\left(\frac{\pi s}{2}\right) + i \sin\left(\frac{\pi s}{2}\right) \right) \Gamma(s). \end{aligned} \]

Separating the imaginary and real parts gives, for \(0 < s < 1\),

\[ \int_0^\infty x^{s-1} \sin(x)\,dx = \sin\left(\frac{\pi s}{2}\right)\Gamma(s), \tag{5} \]

\[ \int_0^\infty x^{s-1} \cos(x)\,dx = \cos\left(\frac{\pi s}{2}\right)\Gamma(s). \tag{6} \]

Extending the sine integral to \(-1 < s \leq 0\)

Because \(\sin x = O(x)\) near \(x=0\), the sine integral \(I(s) = \int_0^\infty x^{s-1} \sin(x)\,dx\) converges on the wider strip \(-1 < s < 1\). We have already proved that \(I(s) = \sin\left(\frac{\pi s}{2}\right)\Gamma(s)\) for \(0 < s < 1\).

Now suppose \(-1 < s < 0\). Integrating by parts with \(u = \sin x\) and \(dv = x^{s-1}\,dx\) gives

\[ I(s) = \left[ \frac{x^s \sin(x)}{s} \right]_0^\infty - \frac{1}{s}\int_0^\infty x^s \cos(x)\,dx. \]

At the upper limit \(x \to \infty\), \(|x^s \sin x| \leq x^s \to 0\) because \(s < 0\); at the lower limit \(x \to 0^+\), \(x^s \sin x \sim x^{s+1} \to 0\) because \(s > -1\). Moreover, since \(0 < s+1 < 1\), we can apply equation (6) with \(s+1\) in place of \(s\), together with the functional equation \(\Gamma(s+1) = s\Gamma(s)\):

\[ \begin{aligned} I(s) &= -\frac{1}{s} \int_0^\infty x^{(s+1)-1} \cos(x)\,dx \\ &= -\frac{1}{s} \cos\left(\frac{\pi}{2}(s+1) \right)\Gamma(s+1) \\ &= \sin\left(\frac{\pi s}{2}\right)\Gamma(s). \end{aligned} \]

Finally, the case \(s=0\) is the classic Dirichlet integral

\[ I(0) = \int_0^\infty \frac{\sin(x)}{x}\,dx = \frac{\pi}{2}, \]

which matches the removable singularity of \(\sin\left(\frac{\pi s}{2}\right)\Gamma(s)\) at \(s=0\):

\[ \lim_{s\to 0} \left[ \sin\left(\frac{\pi s}{2}\right)\Gamma(s) \right] = \lim_{s\to 0} \left[ \frac{\sin\left(\frac{\pi s}{2}\right)}{s}\Gamma(s+1) \right] = \frac{\pi}{2}. \]

In a companion post, we differentiate this identity at \(s=0\) to evaluate a logarithmic Fresnel integral.