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UDC 517.432.1 : 517.63
ON THE THEORY OF THE LAPLACE TRANSFORM
A. A. STEMPKOVSKAYA
In the present work some criteria are established for the existence of singularities of analytic functions represented by a Laplace integral. The interest of these questions also lies in the fact that, unlike power series, where at least one singularity always occurs on the circle of convergence, this is not always the case on the axis of convergence of a Laplace integral.
Theorem 1. Let \(a(t)=x(t)+iy(t)\), let \(x(t)\) have constant sign for sufficiently large \(t\), and
\[ \frac{|y(t)|}{x(t)}=O(t^\gamma), \qquad \gamma \ge 0, \]
then the real point of the axis of convergence \(\sigma_c\) of the Laplace integral
\[ f(s)=\int_0^\infty e^{-st}a(t)\,dt \]
will be a singular point of \(f(s)\), \(s=\sigma+i\tau\).
Without loss of generality, suppose that \(\sigma_c=0\) and \(x(t)>0\) for sufficiently large \(t\). Indeed,
\[ f(s)=\int_0^\infty e^{-st}a(t)\,dt = \int_0^\infty e^{-(s-\sigma_c)t}a_1(t)\,dt, \]
where \(a_1(t)=e^{-\sigma_c t}a(t)\). The conditions of the theorem are also fulfilled for the function \(a_1(t)\), and the abscissa of convergence of the last integral is zero.
We carry out the proof by contradiction. Suppose that \(\sigma_c=0\) is not a singular point of the function \(f(s)\); then it can be expanded in a Taylor series in the circle of radius \(1+\delta'\) with center at \(s=1\) (\(\delta'\) is sufficiently small and greater than zero)
\[ f(s)=\sum_{k=0}^{\infty}\frac{f^{(k)}(1)}{k!}(s-1)^k, \]
where
\[ f^{(k)}(1)=\int_0^\infty e^{-t}(-t)^k a(t)\,dt, \tag{1} \]
and
\[ f(s)=\sum_{k=0}^{\infty}\frac{(s-1)^k}{k!}\int_0^\infty e^{-t}(-t)^k a(t)\,dt, \tag{2} \]
\[ f(-\delta)=\sum_{k=0}^{\infty}\frac{(1+\delta)^k}{k!}\int_0^\infty e^{-t}t^k a(t)\,dt, \qquad 0<\delta<\delta'. \tag{3} \]
Since the series
\[ \sum_{k=0}^{\infty}\frac{(1+\delta)^k t^k}{k!} \]
converges uniformly for \(t \le R\), because it is majorized by the convergent series
\[ \sum_{k=0}^{\infty}\frac{(1+\delta)^k R^k}{k!}, \]
we may write
\[ \sum_{k=0}^{\infty} \frac{(1+\delta)^k}{k!}\int_{0}^{R} e^{-t}t^k a(t)\,dt = \]
\[ = \int_{0}^{R} e^{-t}a(t)\left\{\sum_{k=0}^{\infty} \frac{(1+\delta)^k t^k}{k!}\right\}\,dt = \int_{0}^{R} e^{\delta t} a(t)\,dt. \tag{4} \]
We shall show that, whatever \(\varepsilon>0\) may be, there exists such an \(R(\varepsilon)\) that, for \(R_1>R_{11}>R(\varepsilon)\),
\[ \left| \sum_{k=0}^{\infty} \frac{(1+\delta)^k}{k!} \int_{R_{11}}^{R_1} e^{-t}t^k a(t)\,dt \right|<\varepsilon . \]
But
\[ \left| \sum_{k=0}^{\infty} \frac{(1+\delta)^k}{k!} \int_{R_{11}}^{R_1} e^{-t}t^k a(t)\,dt \right| \le \sum_{k=0}^{\infty} \frac{(1+\delta)^k}{k!} \int_{R_{11}}^{R_1} e^{-t}t^k x(t)\,dt + \]
\[ + C\sum_{k=0}^{\infty} \frac{(1+\delta)^k}{k!} \int_{R_{11}}^{R_1} e^{-t}t^{k+p}x(t)\,dt = \sum_{k=0}^{N} \frac{(1+\delta)^k}{k!} \int_{R_{11}}^{R_1} e^{-t}t^k x(t)\,dt + \]
\[ + \sum_{k=N+1}^{\infty} \frac{(1+\delta)^k}{k!} \int_{R_{11}}^{R_1} e^{-t}t^k x(t)\,dt + C\sum_{k=0}^{N} \frac{(1+\delta)^k}{k!} \int_{R_{11}}^{R_1} e^{-t}t^{k+p}x(t)\,dt + \]
\[ + C\sum_{k=N+1}^{\infty} \frac{(1+\delta)^k}{k!} \int_{R_{11}}^{R_1} e^{-t}t^{k+p}x(t)\,dt \]
by virtue of the conditions of the theorem, \(p\) being an integer greater than \(\gamma\). Let us estimate each of these sums:
\[ \sum_{k=N+1}^{\infty} \frac{(1+\delta)^k}{k!} \int_{R_{11}}^{R_1} e^{-t}t^k x(t)\,dt < \sum_{k=N+1}^{\infty} \frac{(1+\delta)^k}{k!} \int_{0}^{\infty} e^{-t}t^k x(t)\,dt < \frac{\varepsilon}{4}. \]
as the remainder of the real part of the convergent series (3),
\[ \sum_{k=N+1}^{\infty} \frac{(1+\delta)^k}{k!} \int_{R_{11}}^{R_1} e^{-t}t^{k+p}x(t)\,dt = \sum_{k=N+p+1}^{\infty} \frac{(1+\delta)^{k-p}}{(k-p)!} \int_{R_{11}}^{R_1} e^{-t}t^k x(t)\,dt \le \]
\[ \le \sum_{k=N+p+1}^{\infty} \frac{(1+\delta)^{k-p}}{(k-p)!} \int_{0}^{\infty} e^{-t}t^k x(t)\,dt < \frac{\varepsilon}{4C} \]
as the remainder of the real part of the series (2), differentiated \(p\) times and convergent at the point \(s=-\delta\). Further, since the integral in (1) converges, we have
\[ \int_{R_{11}}^{R_1} e^{-t}t^k x(t)\,dt < \frac{\varepsilon k!}{4(N+1)(1+\delta)^k} \]
for \(R_1>R_{11}>R,\quad 0\le k<N,\) and
\[ \int_{R_{11}}^{R_1} e^{-t}t^{k+p}x(t)\,dt < \frac{\varepsilon k!}{4(N+1)(1+\delta)^k C}. \]
for \(R_1>R_{11}>R,\quad 0\le k\le N.\) And hence
\[ \left|\sum_{k=0}^{\infty}\frac{(1+\delta)^k}{k!}\int_{R_{11}}^{R_1} e^{-t}t^k\alpha(t)\,dt\right|<\varepsilon,\qquad R_1>R_{11}>R(\varepsilon). \]
Consequently, in (4) we have the right to pass to the limit as \(R\to\infty\). We obtain
\[ \sum_{k=0}^{\infty}\frac{(1+\delta)^k}{k!}\int_{0}^{\infty} e^{-t}t^k\alpha(t)\,dt = \int_{0}^{\infty} e^{\delta t}\alpha(t)\,dt = f(-\delta), \]
which is impossible, since the abscissa of convergence of the Laplace integral is \(\sigma_c=0\). Thus, \(\sigma_c\) is a singular point of \(f(s)=L\{\alpha(t)\}\).
A very special case of this theorem for \(\gamma=0\) was considered in [1].
The following theorem extends this result to the two-sided Laplace transform (the Mellin transform).
Theorem 2. Let \(\alpha(t)=x(t)+iy(t)\), let \(x(t)\) be of constant sign for sufficiently large values of \(|t|\),
\[ \frac{|y(t)|}{x(t)}=O(t^\gamma),\qquad \gamma\ge 0 \]
and let
\[ f(s)=\int_{-\infty}^{\infty} e^{-st}\alpha(t)\,dt \]
converge in the strip \(\sigma_1<\operatorname{Re}s<\sigma_2\). Then the function \(f(s)\) has singularities at the points \(s=\sigma_1\) and \(s=\sigma_2\).
Represent \(f(s)=f_1(s)+f_2(s)\), where
\[ f_1(s)=\int_{0}^{\infty} e^{-st}\alpha(t)\,dt \]
and
\[ f_2(s)=\int_{0}^{\infty} e^{st}\alpha(-t)\,dt, \]
and the abscissae of convergence are, respectively, \(\sigma_1\) and \(\sigma_2\). (The first integral converges in the half-plane \(\operatorname{Re}s>\sigma_1\), the second in the half-plane \(\operatorname{Re}s<\sigma_2\).) Since, by Theorem 1, the abscissa of convergence \(\sigma_1\) will be a singular point of the function \(f_1(s)\), and the function \(f_2(s)\) is analytic at this point, the point \(s=\sigma_1\) will also be singular for the function \(f(s)\). At the point \(s=\sigma_2\) the function \(f_1(s)\) is analytic; we shall show that at this point \(f_2(s)\), and consequently \(f(s)\), has a singularity. Without loss of generality one may take \(\sigma_2=0\) and \(x(-t)>0\) for sufficiently large values of \(|t|\). We carry out the proof by contradiction.
Suppose that at the point \(\sigma_2\) the function \(f_2(s)\) is analytic; then it can be expanded in a Taylor series in the disk of radius \(1+\delta'\) (\(\delta'\) sufficiently small but positive) with center at \(s=-1\):
\[ f_2(s)=\sum_{k=0}^{\infty}\frac{f_2^{(k)}(-1)}{k!}(s+1)^k, \]
where
\[ f_2^{(k)}(s)=\int_{0}^{\infty} e^{st}t^k\alpha(-t)\,dt \]
and
\[ f_2^{(k)}(-1)=\int_{0}^{\infty} e^{-t}t^k\alpha(-t)\,dt. \]
Hence
\[ f_2(s)=\sum_{k=0}^{\infty}\frac{(s+1)^k}{k!}\int_{0}^{\infty} e^{-t}t^k\alpha(-t)\,dt \]
and
\[ f_2(\delta)=\sum_{k=0}^{\infty}\frac{(1+\delta)^k}{k!}\int_{0}^{\infty} e^{-t}t^k\alpha(-t)\,dt,\qquad 0<\delta<\delta'. \]
Further, just as in Theorem 1, we conclude that \(f_2(s)\), and hence also \(f(s)\), has a singularity at the point \(s=\sigma_2\).
If the function \(a(t)\) is real, then for a singularity of a real point of the axis of convergence
\[ f(s)=\int_{0}^{\infty} e^{-st} a(t)\,dt \]
it is sufficient, as is known (see [1, 2]), that \(a(t)\) have constant sign for sufficiently large \(t\). In this connection we shall consider to what extent it is possible to replace the assumption of constant sign of the function by its sign-changing character.
Theorem 3. Let \(a(t)=a_1(t)+a_2(t)\), where
\[ \begin{aligned} a_1(t)&=a(t),\\ a_2(t)&=0 \end{aligned} \biggr\}\quad \text{if } a(t)\ge 0, \]
and
\[ \begin{aligned} a_1(t)&=0,\\ a_2(t)&=a(t) \end{aligned} \biggr\}\quad \text{if } a(t)<0,\quad \text{and } f(s)=f_1(s)+f_2(s), \]
where \(f_1(s)=L\{a_1(t)\}\), \(f_2(s)=L\{a_2(t)\}\), with abscissae of convergence \(\sigma_1\) and \(\sigma_2\), respectively. Then, if \(\sigma_1\ne\sigma_2\), the abscissa of convergence of \(f(s)=L\{a(t)\}\) is a singular point of \(f(s)\).
Let \(\sigma_1<\sigma_2\). Then
\[ f(s)=\int_{0}^{\infty} e^{-st}\{a_1(t)+a_2(t)\}\,dt \]
has abscissa of convergence \(\sigma_2\).
At the point \(s=\sigma_2\) the function \(f_1(s)\) is analytic, while for the function \(f_2(s)\) the real point of the axis of convergence is a singular point. Hence the abscissa of convergence \(\sigma_2\) is a singular point also of the function \(f(s)\).
In the case \(\sigma_1=\sigma_2=\sigma\), the function may either have or not have a singularity at this point. For example, let
\[ a_1(t)= \begin{cases} 0, & 2n\pi \le t \le (2n+1)\pi,\\ -\sin t, & (2n+1)\pi \le t \le (2n+2)\pi, \end{cases} \]
\[ a_2(t)= \begin{cases} -1, & 2n\pi < t < (2n+1)\pi,\\ 0, & (2n+1)\pi \le t \le (2n+2)\pi, \end{cases} \quad (n=0,1,2,\ldots), \]
and, respectively,
\[ f_1(s)=\frac{1}{(s^2+1)(e^{\pi s}-1)},\qquad f_2(s)=-\frac{1}{s(1+e^{-s})}. \]
It is easy to see that the abscissae of convergence of both Laplace integrals are equal to zero. The function \(f_1(s)\) has simple poles at the points \(s=2ki\) \((k=0,1,2,\ldots)\) and \(s=\pm i\). The residue of \(f_1(s)\) at zero is equal to \(1/\pi\). The function \(f_2(s)\) has simple poles at the points \(s=0\), \(s=-i\pi(2k+1)\) \((k=0,1,\ldots)\). The residue of \(f_2(s)\) at zero is equal to \(-1/2\).
Consider the function \(a(t)=A a_1(t)+B a_2(t)\); then \(f(s)=A f_1(s)+B f_2(s)\), and the abscissa of convergence of \(f(s)=L\{a(t)\}\) is equal to zero. The singular points of \(f(s)\) are the singular points of both \(f_1(s)\) and \(f_2(s)\). The residue of the function \(f(s)\) at zero is
\[ \operatorname*{res}_{0} f(s)=\frac{A}{\pi}-\frac{B}{2}. \]
Obviously, if \(A\) and \(B\) are chosen so that
\[ \frac{A}{\pi}-\frac{B}{2}\ne 0, \]
then \(f(s)\) has a singularity at zero; but if
\[ \frac{A}{\pi}-\frac{B}{2}=0, \]
then there will be no singularity.
In the following assertions the existence of singularities is established under the assumption of infinite divergence of the integrals connected with \(a(t)\).
Theorem 4. Let \(f(s)=L\{a(t)\}\) have finite abscissa of convergence \(\sigma_c\), and let
\[ \int_{0}^{\infty} e^{-\sigma_c t} a(t)\,dt \]
diverge to \(+\infty\) or \(-\infty\). Then the real point of the axis of convergence is a singular point of \(f(s)\).
Without loss of generality one may put \(\sigma_c=0\). Indeed,
\[ f(s)=\int_{0}^{\infty} e^{-st} a(t)\,dt =\int_{0}^{\infty} e^{-(s-\sigma_c)t} a_1(t)\,dt, \]
where
\[ a_1(t)=e^{-\sigma_c t}a(t)\,dt, \]
the abscissa of convergence of this integral is equal to zero and the requirement of the theorem is fulfilled.
For the proof it is enough to show that \(f(s)\to\infty\) as \(s\to +0\) along the real axis.
Consider the integral
\[ \int_0^x e^{-st}a(t)\,dt = e^{-sx}v(x)+s\int_0^x e^{-st}v(t)\,dt, \tag{5} \]
where \(v(t)=\int_0^t a(\tau)\,d\tau\). Since
\[ \lim_{x\to\infty} \frac{\log\left|\int_0^x a(t)\,dt\right|}{x}=0, \]
then for sufficiently large \(x\)
\[ \left|\int_0^x a(t)\,dt\right|<e^{\varepsilon x}. \]
We estimate the nonintegral term in (5):
\[ e^{-sx}\left|\int_0^x a(\tau)\,d\tau\right| < e^{-(s-\varepsilon)x}\to 0 \]
for real \(s\) greater than zero and \(x\to\infty\).
Let \(\lim_{t\to\infty} v(t)\to +\infty\). Then, whatever sufficiently large \(p>0\) may be, there exists such a \(T_0\) that for all \(t>T_0\), \(v(t)>p\). And, consequently,
\[ \int_0^\infty e^{-st}a(t)\,dt = s\int_0^{T_0} v(t)\,dt + s\int_{T_0}^{\infty} e^{-st}v(t)\,dt \ge s\int_0^{T_0} e^{-st}v(t)\,dt + pe^{-sT_0}. \tag{6} \]
Moreover,
\[ \left|\int_0^{T_0} e^{-st}v(t)\,dt\right| < \int_0^{T_0}|v(t)|\,dt<C, \]
since \(e^{-st}\le 1\). Thus the first term in (6), as \(s\to +0\) along the real axis, also tends to zero, while the second term tends to \(p\). Hence
\[ \int_0^\infty e^{-st}a(t)\,dt>\frac{p}{2}, \]
and since \(p\) can be arbitrarily large, it follows that \(\lim f(s)=\infty\) as \(s\to +0\) along the real axis, and \(s=0\) is a singular point of \(f(s)\).
The case when
\[ \int_0^\infty e^{-\sigma_c t}a(t)\,dt \]
diverges to \(-\infty\) is proved similarly.
In the case where
\[ \int_0^\infty e^{-\sigma_c t}a(t)\,dt \]
diverges to infinity while oscillating, the conclusion of the theorem may fail.
Indeed, let \(a(t)=t\sin t\); the Laplace integral of this function has abscissa of convergence \(\sigma_c=0\), nevertheless no real point of the axis of convergence is a singular point of the function
\[ f(s)=L\{t\sin t\}=\frac{2s}{(s^2+1)^2}. \]
Theorem 5. If
\[ f(s)=\int_0^\infty e^{-st}\{a(t)+i\beta(t)\}\,dt \]
has a finite abscissa of convergence \(\sigma_c\), and if at least one of the integrals
\[ \int_0^\infty e^{-\sigma_c t}a(t)\,dt \quad\text{or}\quad \int_0^\infty e^{-\sigma_c t}\beta(t)\,dt \]
diverges to \(+\infty\) or \(-\infty\), then the real point of the axis of convergence is a singular point of the function \(f(s)\).
Without loss of generality, in the proof we take \(\sigma_c=0\). It is enough to show that \(|f(s)|\to\infty\) as \(s\to +0\) along the real axis, since
\[ |f(s)|=\sqrt{\left(\int_{0}^{\infty} e^{-st}\alpha(t)\,dt\right)^2+ \left(\int_{0}^{\infty} e^{-st}\beta(t)\,dt\right)^2}. \]
For this it is enough to prove the divergence of one of these integrals. And this, under the conditions of the present theorem, was proved in Theorem 4.
References
- Doetsch G. Theorie und Anwendung der Laplace Transformation, D. P. New-York, 1943.
- Widder D. W. The Laplace Transform, Princeton, 1946.
Received by the editors
September 23, 1965
Dnepropetrovsk Chemical-Technological
Institute