- An integral of the form
\[ F (z) = \int_ {0} ^ {\infty} f (t) t ^ {z - 1} d t \]
is called a Mellin transform, and we shall write \(\mathcal { M } ( f ) ( z ) = F ( z )\) . For example, the gamma function is the Mellin transform of the function \(e ^ { - t }\)
- Prove that
\[ \mathcal {M} (\cos) (z) = \int_ {0} ^ {\infty} \cos (t) t ^ {z - 1} d t = \Gamma (z) \cos \left(\pi \frac {z}{2}\right) \quad \mathrm{for} 0 < \mathrm{Re} (z) < 1, \]
and
\[ \mathcal {M} (\sin) (z) = \int_ {0} ^ {\infty} \sin (t) t ^ {z - 1} d t = \Gamma (z) \sin \left(\pi \frac {z}{2}\right) \quad \mathrm{for} 0 < \mathrm{Re} (z) < 1. \]
- Show that the second of the above identities is valid in the larger strip \(- 1 < \operatorname { R e } ( z ) < 1\) , and that as a consequence, one has
\[ \int_ {0} ^ {\infty} \frac {\sin x}{x} d x = \frac {\pi}{2} \quad \mathrm{and} \quad \int_ {0} ^ {\infty} \frac {\sin x}{x ^ {3 / 2}} d x = \sqrt {2 \pi}. \]
This generalizes the calculation in Exercise 2 of Chapter 2.
[Hint: For the first part, consider the integral of the function \(f ( w ) = e ^ { - w } w ^ { z - 1 }\) around the contour illustrated in Figure 1. Use analytic continuation to prove the second part.]
Figure 1. The contour in Exercise 10
Solution
Let \(z\) satisfy \(0<\Re z<1\). On the principal branch, rotate the ray in the gamma integral through angle \(-\pi/2\). The integrals over the connecting circular arcs vanish at \(0\) and at infinity because \(0<\Re z<1\). Thus \[ \Gamma(z)=\int_0^\infty e^{-t}t^{z-1}\,dt =e^{-i\pi z/2}\int_0^\infty e^{it}t^{z-1}\,dt, \] so \[ \int_0^\infty e^{it}t^{z-1}\,dt=e^{i\pi z/2}\Gamma(z). \tag{1} \] Taking real and imaginary parts of (1) gives \[ \int_0^\infty \cos t\,t^{z-1}\,dt =\Gamma(z)\cos\frac{\pi z}{2}, \] \[ \int_0^\infty \sin t\,t^{z-1}\,dt =\Gamma(z)\sin\frac{\pi z}{2}. \]
For the sine integral, near \(0\) we have \(\sin t=O(t)\), so the integral converges there when \(\Re z>-1\). At infinity, integration by parts (or Dirichlet’s test, uniformly on compact vertical substrips) gives convergence when \(\Re z<1\). Hence \[ S(z)=\int_0^\infty \sin t\,t^{z-1}\,dt \] is holomorphic on \(-1<\Re z<1\). The function \(\Gamma(z)\sin(\pi z/2)\) is also holomorphic there: the pole of \(\Gamma\) at \(0\) is cancelled by the zero of the sine factor. Since the two functions agree on \(0<\Re z<1\), the identity theorem gives \[ S(z)=\Gamma(z)\sin\frac{\pi z}{2},\qquad -1<\Re z<1. \] At \(z=0\), using \(\Gamma(z)\sim1/z\) and \(\sin(\pi z/2)\sim \pi z/2\), \[ \int_0^\infty\frac{\sin x}{x}\,dx=\frac\pi2. \] At \(z=-1/2\), since \(\Gamma(-1/2)=-2\sqrt\pi\), \[ \int_0^\infty\frac{\sin x}{x^{3/2}}\,dx =\Gamma(-1/2)\sin(-\pi/4)=\sqrt{2\pi}. \]