7.7 The expansion of a class of periodic functions in a series of cotangents.

Let \(f(z)\) be a periodic function of \(z\), analytic except at a certain number of simple poles; for convenience, let \(\pi\) be the period of \(f(z)\) so that \(f(z) = f(z + \pi)\).

Let \(z = x + iy\) and let \(f(z) \rightarrow l\) uniformly with respect to \(x\) as \(y \rightarrow +\infty\), when \(0 \leq x \leq \pi\); similarly let \(f(z) \rightarrow l'\) uniformly as \(y \rightarrow -\infty\).

Let the poles of \(f(z)\) in the strip \(0 < x \leq \pi\) be at \(a_{1}\), \(a_{2}\), \(\ldots, a_{n}\); and let the residues at them be \(c_{1}\), \(c_{2}\), \(\ldots, c_{n}\).

Further, let \(ABCD\) be a rectangle whose corners are \(-i\rho\), \(\pi - i\rho\), \(\pi + i\rho'\), and \(i\rho'\) in order.[1]

[1]If any of the poles are on \(x = \pi\), shift the rectangle slightly to the right; \(\rho, \rho'\) are to be taken so large that \(a_{1}\), \(a_{2}, \ldots, a_{n}\) are inside the rectangle. ↩

Consider \[ \frac{1}{2 \pi i} \!\int f(t) \cot (t-z) \, d t \] taken round this rectangle; the residue of the integrand at \(a_{r}\) is \(c_{r} \cot (a_{r}-z)\), and the residue at \(z\) is \(f(z)\).

Also the integrals along \(DA\) and \(CB\) cancel on account of the periodicity of the integrand; and as \(\rho \rightarrow \infty\), the integrand on \(AB\) tends uniformly to \(l'\! i\), while as \(\rho' \rightarrow \infty\) the integrand on \(CD\) tends uniformly to \(-li\); therefore \[ \frac{1}{2} (l' + l) = f(z) + \sum_{r=1}^{n} c_{r} \cot \:\!(a_{r} - z). \]

That is to say, we have the expansion \[ f(z) = \frac{1}{2} (l' + l) + \sum_{r=1}^{n} c_{r} \cot (z - a_{r}). \]

Example 1. Shew \[\begin{align*} \cot (x - a_{1}) \cot (x - a_{2})& \cdots \cot (x - a_{n}) \\ =& \sum_{r=1}^{n} \cot (a_{r} - a_{1}) \cdots * \cdots \cot (a_{r} - a_{n}) \cot (x - a_{r}) + (-1)^{\frac{1}{2} n}, \\ \textrm{or}\qquad =& \sum_{r=1}^{n} \cot (a_{r} - a_{1}) \cdots * \cdots \cot (a_{r} - a_{n}) \cot (x - a_{r}), \end{align*}\] according as \(n\) is even or odd, where the ‘\(*\)’ means that the factor \(\cot (a_{r} - a_{r})\) is omitted.

Example 2. Prove that \[\begin{align*} \frac{ \sin (x - b_{1}) \sin (x - b_{2}) \cdots \sin (x - b_{n}) }{ \sin (x - a_{1}) \sin (x - a_{2}) \cdots \sin (x - a_{n}) } \\ &\mkern-90mu = \frac{ \sin (a_{1} - b_{1}) \cdots \sin (a_{1} - b_{n}) }{ \sin (a_{1} - a_{2}) \cdots \sin (a_{1} - a_{n}) } \cot (x - a_{1}) \\ &\mkern-36mu + \frac{ \sin (a_{2} - b_{1}) \cdots \sin (a_{2} - b_{n}) }{ \sin (a_{2} - a_{1}) \cdots \sin (a_{2} - a_{n}) } \cot (x - a_{2}) + \cdots \\ &\mkern-54mu \cdots \, + \frac{ \sin (a_{n} - b_{1}) \cdots \sin (a_{n} - b_{n}) }{ \sin (a_{n} - a_{1}) \cdots \sin (a_{n} - a_{n-1}) } \cot (x - a_{n}) \\ &\mkern-36mu + \cos (a_{1} + a_{2} + \cdots + a_{n} - b_{1} - b_{2} - \cdots - b_{n}). \end{align*}\]

7.8 Borel’s theorem.[2]

[2]Leçons sur les séries divergentes (1901), p. 94. See also the memoirs there cited. ↩

Let \(f(z) = \sum\limits_{n=0}^{\infty} a_{n} z^{n}\) be analytic when \(\left|\, z \,\right| \leq r\), so that, by §5.23, \(\left|\, a_{n} r^{n} \,\right| < M\) where \(M\) is independent of \(n\).

Hence, if \(\phi(z) = \sum\limits_{n=0}^{\infty} \dfrac{a_{n} z^{n}}{n!}\), \(\phi(z)\) is an integral function, and \[ \left|\, \phi(z) \,\right| < \sum_{n=0}^{\infty} \frac{M \left|\, z^{n} \,\right| }{ r^{n} \cdot n!} = M e^{\left.\left|\:\! z \:\!\right|\middle/r\right.}, \] and similarly \(\left|\, \phi^{(n)}(z) \,\right| < \dfrac{M e^{\left.\left|\:\! z \:\!\right|\middle/r\right.}}{r^{\:\!n}}\).

Now consider \(f_{1}(z) = \int_{0}^{\infty}\! e^{-t} \phi(zt) \, d t\); this integral is an analytic function of \(z\) when \(\left|\, z \,\right| < r\), by §5.32.

Also, if we integrate by parts, \[\begin{align*} f_{1}(z) =& \left[\vphantom{\sum}- e^{-t} \phi(zt) \right]_{0}^{\infty} + z \!\int_{0}^{\infty}\! e^{-t} \phi'(zt) \, d t \\ =& \sum_{m=0}^{n} z^{m} \left[\vphantom{\sum} - e^{-t} \phi^{(m)}(zt) \right]_{0}^{\infty} + z^{n+1} \!\int_{0}^{\infty}\! e^{-t} \phi^{(n+1)}(zt) \, d t. \end{align*}\]

But \(\lim\limits_{t \rightarrow 0} e^{-t} \phi^{(m)}(zt) = a_{m}\); and, when \(\left|\, z \,\right| < r\), \(\lim\limits_{t \rightarrow \infty} e^{-t} \phi^{(m)}(zt) = 0\).

Therefore \[ f_{1}(z) = \sum_{m=0}^{n} a_{m} z^{m} + R_{n}, \]

where \[\begin{align*} \left|\,R_{n}\,\right| &< \left|\,z^{n+1}\right|\int_0^\infty \! e^{-t} \dfrac{M e^{\left.\left|\:\! zt \:\!\right|\middle/r\right.}}{r^{\:\!n+1}} \, d t \\ &= \left|\,\frac{z}{r}\,\right|^{n+1}\!\! \frac{M}{1-\left.\left|\,z\,\right|\middle/ r\right.} \rightarrow 0 \,\text{ as } \, n \rightarrow \infty . \end{align*}\]

Consequently, when \(\left|\, z \,\right| < r\), \[ f_{1}(z) = \sum_{m=0}^{\infty} a_{m} z^{m} = f(z); \] and so \[ f(z) = \!\int_{0}^{\infty}\! e^{-t} \phi(zt) \, d t, \] where \( \phi(z) = \sum\limits_{n=0}^{\infty} \dfrac{a_{n} z^{n}}{n!}; \) is called Borel’s function associated with \(\sum\limits_{n=0}^{\infty} a_{n} z^{n}\).

If \(S = \sum\limits_{n=0}^{\infty} a_{n}\) and \(\phi(z) = \sum\limits_{n=0}^{\infty} \dfrac{a_{n} z^{n}}{n!}\) and if we can establish the relation \(S = \int_{0}^{\infty}\! e^{-t} \phi(t) \, d t\), the series \(S\) is said (§8.41) to be ‘summable \((B)\)’; so that the theorem just proved shews that a Taylor’s series representing an analytic function is summable \((B)\).

7.81 Borel’s integral and analytic continuation.

We next obtain Borel’s result that his integral represents an analytic function in a more extended region than the interior of the circle \(\left|\, z \,\right| = r\).

Figure 2: Extended region of convergence for Borels integral.
Figure 2: Extended region of convergence for Borel’s integral.

This extended region is obtained as follows: take the singularities \(a\), \(b\), \(c, \ldots\) of \(f(z)\) and through each of them draw a line perpendicular to the line joining that singularity to the origin. The lines so drawn will divide the plane into regions of which one is a polygon with the origin inside it.

Then Borel’s integral represents an analytic function (which, by §5.5 and §7.8, is obviously that defined by \(f(z)\) and its continuations) throughout the interior of this polygon. The reader will observe that this is the first actual formula obtained for the analytic continuation of a function, except the trivial one of §5.5, example.

Figure 2a: Showing Borels integral is analytic at any point \(P\:(=\!\zeta)\\) in the polygon.
Figure 2a: Showing Borel’s integral is analytic at any point \(P\:(=\!\zeta)\) in the polygon.
[3]The reader will see this from the figure; for if there were such a singularity the corresponding side of the polygon would pass between \(O\) and \(P\); \ie, \(P\) would be outside the polygon. ↩

For, take any point \(P\) with affix \(\zeta\) inside the polygon; then the circle with \(OP\) as diameter has no singularity on or inside it;[3] and consequently we can draw a slightly larger concentric circle \(C\) with no singularity on or inside it.[4] Then, by §5.4, \[ a_{n} = \frac{1}{2 \pi i} \!\int_{C}\! \frac{f(z)}{z^{n+1}} \, d z, \] and so \[ \phi(\zeta t) = \frac{1}{2 \pi i} \sum_{n=0}^{\infty} \frac{\zeta^{n} t^{n}}{n!} \!\int_{C}\! \frac{f(z)}{z^{n+1}} \, d z; \] but \(\sum\limits_{n=0}^{\infty} \dfrac{\zeta^{n} t^{n}}{n!} \dfrac{f(z)}{z^{n+1}}\) converges uniformly (§3.34) on \(C\) since \(f(z)\) is bounded and \(\left|\, z \,\right| \geq \delta > 0\), where \(\delta\) is independent of \(z\); therefore, by §4.7, \[ \phi(\zeta t) = \frac{1}{2 \pi i} \!\int_{C}\! z^{-1} f(z) \exp(\zeta \:\! t \:\! z^{-1}) \, d z, \] and so, when \(t\) is real, \(\left|\, \phi(\zeta t) \,\right| < F(\zeta) e^{\lambda t}\), where \(F(\zeta)\) is bounded in any closed region lying wholly inside the polygon and is independent of \(t\); and \(\lambda\) is the greatest value of the real part of \(\left.\zeta \middle/ z\right.\) on \(C\).

[4]The differeuce of the radii of the circles being, say, \(\delta\). ↩

If we draw the circle traced out by the point \(\left.\zeta \middle/ z\right.\), we see that the real part of \(\left.\zeta \middle/ z\right.\) is greatest when \(z\) is at the extremity of the diameter through \(\zeta\), and so the value of \(\lambda\) is \( \left|\, \zeta \,\right| \cdot \left\{\left|\, \zeta \,\right| + \delta\right\}^{-1} < 1\).

We can get a similar inequality for \(\phi'(\zeta t)\) and hence, by §5.32, \(\int_{0}^{\infty}\! e^{-t} \phi(\zeta t) \, d t\) is analytic at \(\zeta\) and is obviously a one-valued function of \(\zeta\).

This is the result stated above.

[5]Mém de l’Acad. des Sci. (Paris, 1717); see Tweedie, Proc. Edin. Math. Soc. xxxvi. (1917). ↩

7.82 Expansions in series of inverse factorials.

[6]Methodus Differentialis (London, 1730) (or see James Stirling’s Methodus Differentialis: An Annotated Translation of Stirling’s Text (2003) –Editor). ↩

A mode of development of functions, which, after being used by Nicole[5] and Stirling[6] in the eighteenth century, was systematically investigated by Schlömilch[7] in 1863, is that of expansion in a series of inverse factorials.

[7]Compendium der höheren Analysis. ii, p. 89. More recent investigations are due to Kluyver, Nielsen and Pincherle. See Comptes Rendus, cxxxiii. (1901), p. 1273, cxxxiv. (1902), pp. 157, 587, Annales de l’École norm. sup. (3), xix. p. 409, xxii. p. 449, xxiii. p. 9, Rendiconti del Lincei, (5), xi. (1902), pp. 139, 417, and Palermo Rendiconti, xxxiv. (1912), p. 41. Properties of functions defined by series of inverse factorials have been studied in an important memoir by Nörlund, Acta Math. xxxvii. (1914), pp. 327–387. ↩

To obtain such an expansion of a function analytic when \(\left|\, z \,\right| > r\), we let the function be \(f(z) = \sum\limits_{n=0}^{\infty} a_{n} z^{-n}\),[8] and use the formula \(f(z) = \int\limits_{0}^{\infty}\! z e^{-t\:\!z} \phi(t) \, d t\), where \(\phi(t) = \sum\limits_{n=0}^{\infty} \left.a_{n} t^{n} \middle/ n!\right.\); this result may be obtained in the same way as that of §7.8.

Modify this by writing \(e^{-t} = 1 - \xi\), \(\phi(t) = F(\xi)\); then \[ f(z) = \!\int_{0}^{1}\! z (1 - \xi)^{z-1} F(\xi) \, d \xi. \]

[8]Editor’s Note: The convergence of this series, in this case implies \(\left|\, a_{n}r^{-n} \,\right| < M\), for some \(M\). ↩

Now if \(t = u + iv\) and if \(t\) be confined to the strip \(-\pi < v < \pi\), \(t\) is a one-valued function of \(\xi\) and \(F(\xi)\) is an analytic function of \(\xi\); and \(\xi\) is restricted so that \(-\pi < \arg (1-\xi) < \pi\). Also the interior of the circle \(\left|\, \xi \,\right| = 1\) corresponds to the interior of the curve traced out by the point \(t = - \log \left(2 \cos \frac{1}{2} \theta\right) + \frac{1}{2} i \theta\), ( writing \(\xi= -\exp (-i \theta \:\!)\) ); and inside this curve \[ \left|\, t \,\right| - \mathfrak{Re}(t) \leq \sqrt{ \left\{\mathfrak{Re}(t)\right\}^{2} + \pi^{2} } - \mathfrak{Re}(t) \rightarrow 0, \,\text{ as }\, \mathfrak{Re}(t) \rightarrow \infty .\] It follows that, when \(\left|\, \xi \,\right| \leq 1\), \(\left|\, F(\xi) \,\right| < M e^{r\:\!\left|\:\! t \:\!\right|} < M_{1} \left|\, e^{r\:\! t} \,\right|\), where \(M_{1}\) is independent of \(t\:\!\); and so \(F(\xi) < M_{1} \left|\, (1-\xi)^{-r} \;\!\right|\).

Now suppose that \(0 \leq \xi < 1\); then, by §5.23, \(\left|\, F^{(n)}(\xi) \,\right| < M_{2}\:\! n! \:\!\rho^{-n}\) where \(M_{2}\) is the upper bound of \(\left|\, F(z) \,\right|\) on a circle with centre \(\xi\) and radius \(\rho < 1 - \xi\).

[9]\((1 + x^{-1})^{x}\) increases with \(x\); for \(\frac{1}{1-y} > e^{y}\), when \(y < 1\), and so \(\log \left(\frac{1}{1-y}\right) > y\). That is to say, putting \(y^{-1} = 1+x\), \( \frac{d}{d x} x \log (1 + x^{-1}) = \log(1 + x^{-1}) - \frac{1}{1+x} > 0 \). ↩

Taking \(\rho = \frac{n}{n+1} (1-\xi)\) and observing that \((1 + n^{-1})^{n} < e\),[9] we find that \[\begin{align*} \left|\, F^{(n)}(\xi) \,\right| &< M_{1} \left[ 1 - \left\{ \xi + \frac{n}{n+1} \xi \right\} \right]^{-r} \!\!\!\cdot n! \cdot \left\{ \frac{n (1-\xi)}{n+1} \right\}^{-n} \\ &< M_{1} e (n+1)^{r} n! (1-\xi)^{-r-n}. \end{align*}\]

Remembering that, by §4.5, \(\int_{0}^{1}\) means \(\lim\limits_{\epsilon \rightarrow +0} \int_{0}^{1-\epsilon}\), we have, by repeated integrations by parts, \[\begin{align*} f(z) &= \lim_{\epsilon \rightarrow +0} \left[ \vphantom{\sum} -(1 - \xi)^{z} F(\xi) \right]_{0}^{1-\epsilon} + \!\int_{0}^{1 - \epsilon}\! (1 - \xi)^{z} F'(\xi) \, d \xi \\ &= \lim_{\epsilon \rightarrow +0} \left[ \vphantom{\sum} -(1 - \xi)^{z} F(\xi) \right]_{0}^{1 - \epsilon} + \frac{1}{z + 1} \left[ \vphantom{\sum} -(1 - \xi)^{z+1} F'(\xi) \right]_{0}^{1 - \epsilon} \\ & \phantom{ \,=\left[ \vphantom{\sum} -(1 - \xi)^{z} F(\xi) \right]_{0}^{1 - \epsilon}\,} + \frac{1}{z + 1} \!\int_{0}^{1 - \epsilon}\! (1 - \xi)^{z+1} F''(\xi) \, d \xi \\ &= \vphantom{\int}\quad \cdots \\ &= b_{0} + \frac{b_{1}}{z+1} + \frac{b_{2}}{(z+1)(z+2)} + \cdots + \frac{b_{n}}{(z+1)(z+2)\cdots(z+n)} + R_{n}, \end{align*}\] where \[\begin{align*} b_{n} &= \lim_{\epsilon \rightarrow 0} \left[ \vphantom{\sum} -(1-\xi)^{z+n} F^{(n)}(\xi) \right]_{0}^{1 - \epsilon} \\ &= F^{(n)}(0), \end{align*}\] if the real part of \(z+n-r-n>0\), i.e. if \(\mathfrak{Re}(z) > r\); further \[\begin{align*} \left|\, R_{n} \,\right| &\leq \frac{1}{ \left|\, (z+1)(z+2)\cdots(z+n) \,\right|} \lim_{\epsilon \rightarrow 0} \!\int_{0}^{1 - \epsilon}\! \left|\, (1-\xi)^{z+n} F^{(n+1)}(\xi) \,\right| \, d \xi \\ &< \frac{ M_{1} e (n+2)^{r} n!}{ \left|\, (z+1)(z+2)\cdots(z+n) \,\right| \mathfrak{Re}(z-r)} \\ &< \frac{ M_{1} e (n+2)^{r} n!}{ (r+1+\delta)(r+2+\delta) \cdots (r+n+\delta) \:\!\delta }, \end{align*}\] where \(\delta = \mathfrak{Re}(z - r)\).

Now \[ \prod_{m=1}^{n} \left\{ \left( 1 + \frac{r + \delta}{m} \right) e^{-\frac{r+\delta}{m}} \right\} \] tends to a limit (§2.71) as \(n \rightarrow \infty\), and so \(\left|\, R_{n} \,\right| \rightarrow 0 \) if \( (n+2)^{r} e^{-(r+\delta) \sum_{1}^{n} \left.1\middle/m\right.} \) tends to zero; but \[ \sum_{m=1}^{n} \left.1\middle/m\right. > \!\int_{1}^{n+1}\! \frac{\, d x}{x} = \log(n+1), \] by §4.43 (II), and \((n + 2)^{r} (n+1)^{-r-\delta} \rightarrow 0\) when \(\delta > 0\); therefore \(R_{n} \rightarrow 0\) as \(n \rightarrow \infty\), and so, when \(\mathfrak{Re}(z) > r\), we have the convergent expansion \[ f(z) = b_{0} + \frac{b_{1}}{z+1} + \frac{b_{2}}{(z+1)(z+2)} + \cdots + \frac{b_{n}}{(z+1)(z+2)\cdots(z+n)} + \cdots. \]

Example 1. Obtain the same expansion by using the results \[ \frac{1}{(z+1)(z+2)\cdots(z+n+1)} = \frac{1}{n!} \!\int_{0}^{1}\! u^{n} (1-u)^{z} \, d u, \] \[ \int_{C}\! \frac{ f(t) \, d t }{z - t} = \!\int_{C} \!\, d t \!\int_{0}^{1}\! f(t) (1-u)^{z-t-1} \, d u. \]

Example 2. Obtain the expansion \[ \log\left(1 + \frac{1}{z}\right) = \frac{1}{z} - \frac{a_{1}}{z(z+1)} - \frac{a_{2}}{z(z+1)(z+2)} - \cdots, \] where \[ a_{n} = \!\int_{0}^{1}\! t \:\! (1-t) (2-t) \cdots (n-1-t) \, d t, \] and discuss the region in which it converges. \(\vphantom{\\ 3\\}\)
(Schlömilch.)