I previously asked about sufficient conditions to conclude that $$\lim_{s \to 0^{+}}\int_{0}^{\infty} e^{-sx} f(x) \, dx = \int_{0}^{\infty} f(x) \, dx$$ when $\int_{0}^{\infty} f(x) \, dx$ does not converge absolutely.

Daniel Fischer showed that a sufficient condition is if $\int_{0}^{\infty} e^{-sx} f(x) \, dx$ converges uniformly on $[0, \delta]$ for some $\delta >0$.

Recently I came across the following exercise:

Show that if $F(s) = \int_{0}^{\infty} e^{-sx} f(x) \, dx$ converges for $s=s_{0}$, then it converges uniformly on $[s_{0}, \infty)$.

The above excercise is exercise 27 in the first supplement to the textbook Introduction to Real Analysis by William F. Trench.

It's basically a stronger version of a theorem that states that if $f(x)$ is continuous on $[0, \infty)$ and $\int^{{\color{red}{x}}}_{0} e^{-s_{0}u} f(u) \, du$ is bounded for all $x \ge 0$, then $\int_{0}^{\infty} e^{-sx} f(x) \, dx$ will converge uniformly on $[s_{1}, \infty)$ for $s_{1} >s_{0}$. A proof of this theorem can be found on page 20 of the supplement.

But with the only condition being that $\int_{0}^{\infty} e^{-s_{0}x} f(x) \, dx$ must converge, it's hard to believe that there is not a counterexample.

Perhaps it has something to do with $e^{-sx}$ being monotonic in the parameter $s$.


By replacing $f(x)$ with $f(x) e^{-s_0 x}$, we can assume that $s_0 = 0$. For $t > 0$ define

$$R(t) := \int_t^{+\infty} f(x)\,dx.$$

Since the improper Riemann integral

$$\int_0^{+\infty} f(x)\,dx$$

exists, $R \colon (0,+\infty) \to \mathbb{R}$ is a continuous function with $\lim\limits_{t\to +\infty} R(t) = 0$. Let further

$$M(u) := \sup \{ \lvert R(t)\rvert : t \geqslant u\}.$$

Since we have only minimal regularity assumptions on $f$, it's easier to justify integration by parts from the other end: For $s > 0$ and $0 < a < b < +\infty$ we have

\begin{align} s\int_a^b R(x) e^{-sx}\,dx &= s\int_a^b R(b) e^{-sx}\,dx + s\int_a^b \bigl(R(x) - R(b)\bigr) e^{-sx}\,dx\\ &= R(b)\bigl(e^{-sa} - e^{-sb}\bigr) + s\int_a^b e^{-sx}\int_x^b f(u)\,du \,dx\\ &= R(b)\bigl(e^{-sa} - e^{-sb}\bigr) + \int_a^b f(u) s\int_a^u e^{-sx}\,dx\,du\\ &= R(b)\bigl(e^{-sa} - e^{-sb}\bigr) + \int_a^b f(u) \bigl(e^{-sa} - e^{-su}\bigr)\,du\\ &= R(b)\bigl(e^{-sa} - e^{-sb}\bigr) + e^{-sa}\bigl(R(a) - R(b)\bigr) - \int_a^b f(u) e^{-su}\,du\\ &= R(a) e^{-sa} - R(b)e^{-sb} - \int_a^b f(u)e^{-su}\,du. \end{align}

The change of order of integration is justified since $f$ is Riemann-integrable (and in particular bounded) on $[a,b]$. Since

$$\biggl\lvert s\int_a^b R(x)e^{-sx}\,dx\biggr\rvert \leqslant s\int_a^b \lvert R(x)\rvert e^{-sx}\,dx \leqslant M(a)s\int_a^b e^{-sx}\,dx \leqslant M(a) e^{-sa},$$

rearranging yields

$$\biggl\lvert \int_a^b f(x)e^{-sx}\,dx\biggr\rvert \leqslant \lvert R(a)\rvert e^{-sa} + \lvert R(b)\rvert e^{-sb} + M(a)e^{-sa} \leqslant 3\cdot M(a),\tag{$\ast$}$$

with a bound independent of $b$ and $s > 0$. Since $(\ast)$ also holds for $s = 0$, and $M$ is monotonically decreasing (nonstrictly, in general) with $\lim\limits_{u\to \infty} M(u) = 0$, this shows the uniform convergence of the improper Riemann integrals

$$\int_c^{+\infty} f(x) e^{-sx}\,dx \tag{1}$$

for $s \in [0,+\infty)$ for any fixed $c > 0$. If $f$ is bounded on $[0,\delta]$ for some $\delta > 0$, we can even take $c = 0$ here and are done. If $f$ is unbounded at $0$, so the integral $\int_0^{+\infty} f(x)\,dx$ is improper at both ends, a small modification of the argument using

$$P(t) := \int_0^t f(x)\,dx$$

and $N(u) := \sup \{ \lvert P(t)\rvert : t \leqslant u\}$ gives the estimate

$$\biggl\lvert \int_a^b f(x) e^{-sx}\,dx\biggr\rvert \leqslant 3\cdot N(b)$$

for $0 < a < b < +\infty$, and the uniform convergence of

$$\int_0^c f(x) e^{-sx}\,dx$$

for $s\in [0,+\infty)$ given any fixed $c > 0$. Together with $(1)$, we have the uniform convergence of

$$\int_0^{+\infty} f(x) e^{-sx}\,dx$$

for $s \in [0,+\infty)$.

  • $\begingroup$ Let's say that $\int_{0}^{\infty} f(x) g(x,s) \, dx$ exists as an improper Riemann integral for all values of $s$ in $[0,\delta]$ for some $\delta>0$. Are there any conditions we could place on $g(x,s)$ that would ensure that $\int_{0}^{\infty}f(x) g(x,s) \, dx $ converges uniformly on $[0,\delta]$? Having $g(x,s)$ be monotonic in $s$ on $[0,\delta]$ wouldn't be sufficient, would it? $\endgroup$ – Random Variable Aug 14 '17 at 19:44
  • $\begingroup$ Without picking up a pen to check anything, I'd think that $g$ being monotonic in both variables separately probably is sufficient. With monotonicity only in $s$, I suspect there might be some weird counterexample. But usually one works with rather nice $g$. What sort of $g$ are you looking at? $\endgroup$ – Daniel Fischer Aug 14 '17 at 19:58
  • $\begingroup$ I wasn't looking at any particular $g$, but one that comes to mind is $x^{s-1}$. This, however, might be a bit more complicated since whether $x^{s-1}$ is monotonically decreasing in $s$ or monotonically increasing in $s$ depends on $x$. $\endgroup$ – Random Variable Aug 14 '17 at 20:38
  • $\begingroup$ But that dependence is quite simple, so one can just look at $\int_0^1$ and $\int_1^{\infty}$. That ought to work out nicely. $\endgroup$ – Daniel Fischer Aug 14 '17 at 20:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.