TL; DR: Under which conditions can an improper integral be broken into an infinite sum of definite integrals (if any such condition exists)?

Background for the question: I need to estimate the following improper integral:

$$\int_0^\infty e^{-\alpha t}f(t)dt$$

where the function $f:\mathbb{R} \rightarrow \mathbb{R}$ is positive and $T$-periodic with $\int_0^T |f(t)|^2 dt < \infty$ (integrals taken in the Lebesgue sense).

My idea was to try to break the integral as:

$$\int_0^\infty e^{-\alpha t}f(t)dt = \sum_{n=0}^\infty \int_{nT}^{(n+1)T}e^{-\alpha t}f(t)dt$$

and then use some other conditions at hand to prove that this infinite sum converges. However, I couldn't find conditions which tell me whether I can break this integral as such.

  • 1
    $\begingroup$ Yes -- if an improper integral $I = \int_0^\infty g(t) \, dt$ exists then it means for $c \in [0,\infty)$, $I= \lim_{c \to \infty}\int_0^c g(t) \, dt = \lim_{m \to \infty} \int_0^{mT} g(t) \, dt = \lim_{m \to \infty} \sum_{n=0}^{m-1}\int_{nT}^{(n+1)T} g(t) \, dt $. The reverse is not always true. $\endgroup$ – RRL Mar 28 '18 at 20:25

Let's assume $\alpha \in \mathbb{R}$ for simplicity, but $\alpha \in \mathbb{C}$ is doable also (the conclusion will be on the real part of $\alpha$.)

We have, for all $u \in [0, T]$, because $u \mapsto e^{-\alpha u}$ is decreasing :

$\begin{equation*} \lvert e^{-\alpha u} f(u) \rvert \leq \lvert f(u) \rvert \end{equation*}$

As $f$ is integrable on $[0, T]$ because $f^2$ is integrable on $[0, T]$ and $[0, T]$ is a finite measure space.

It follows that $u \mapsto e^{-\alpha u} f(u)$ is integrable on $[0, T]$.

If $f$ is $T$-periodic, let be $n \in \mathbb{N}$:

$\begin{align*} \int_{nT}^{(n + 1)T} e^{-\alpha t} f(t) \textrm{d} t & = \int_{0}^{T} e^{-\alpha (u + nT)} f(u + nT) \textrm{d} u \\ & = \int_0^T e^{-\alpha(u + nT)} f(u) \textrm{d}u \\ & = e^{-\alpha n T} \int_0^T e^{-\alpha u} f(u) \textrm{d} u \\ \end{align*}$


$\begin{align*} \sum_{n=0}^{m} \int_{nT}^{(n + 1)T} e^{-\alpha t} f(t) \textrm{d} t & = \sum_{n=0}^{m} e^{-\alpha n T} \int_0^T e^{-\alpha u} f(u) \textrm{d} u \\ & = \left[\int_0^T e^{-\alpha u} f(u) \textrm{d} u\right] \sum_{n=0}^m (e^{-\alpha T})^n \\ \end{align*}$

Let's now distinguish whether $e^{-\alpha T} = 1$ or not:

First case : $e^{-\alpha T} = 1$, thus $\alpha T = 0$, thus: $\alpha = 0$ as $T > 0$.

We have:

$\begin{align*} \sum_{n=0}^m \int_{nT}^{(n + 1)T} e^{-\alpha t} f(t) \textrm{d} t = (m + 1)\int_0^T e^{-\alpha u} f(u) \textrm{d} u \end{align*}$

It is now clear that this sum diverges as $m \to +\infty$ if and only if $f$ is not the null function.

Thus, it cannot be possible for the integral to converge otherwise it would contradict the previous statement about the infinite sum.

It is clear that $\int_0^{+\infty} f(t) \textrm{d}t$ diverges now.

(And this is logical, no non-zero periodic function have a finite area under the curve from $0$ to $+\infty$.)

Conclusion : Either $f = 0$, either the integral diverges.

Second case : $e^{\alpha T} \neq 1$, i.e. $\alpha \neq 0$.

Then :

$\begin{align*} \sum_{n=0}^m \int_{nT}^{(n + 1)T} e^{-\alpha t} f(t) \textrm{d} t & = \left[\int_0^T e^{-\alpha u} f(u) \textrm{d} u\right] \dfrac{e^{-\alpha T (m + 1)} - 1}{e^{-\alpha T} - 1} \end{align*}$

As $m \to +\infty$, we have again two choices, either $\alpha < 0$, either $\alpha > 0$.

First sub-case : $\alpha < 0$.

Then, $\lim\limits_{m \to +\infty} \dfrac{e^{-\alpha T (m + 1)} - 1}{e^{-\alpha T} - 1} = +\infty$.

So, the sum diverges, thus the integral diverges as in the first case.

Second sub-case : $\alpha > 0$.

Then, $\lim\limits_{m \to +\infty} \dfrac{e^{-\alpha T (m + 1)} - 1}{e^{-\alpha T} - 1} = \dfrac{1}{1 - e^{-\alpha T}}$.

The sum converges and the integral converges, then.

Precisely, we have:

$\begin{align*} \int_0^{+\infty} e^{-\alpha t} f(t) \textrm{d} t & = \sum_{n=0}^{+\infty} \int_{nT}^{(n + 1)T} e^{-\alpha t} f(t) \textrm{d} t \\ & = \dfrac{1}{1 - e^{-\alpha T}} \int_0^T e^{-\alpha t} f(t) \textrm{d} t \end{align*}$

Conclusion : For all $\alpha > 0$, for all $f : \mathbb{R} \to \mathbb{R}$ $T$-periodic, $T > 0$, we have:

$\begin{equation*} \boxed{\int_0^{+\infty} e^{-\alpha t} f(t) \textrm{d} t = \dfrac{1}{1 - e^{-\alpha T}} \int_0^T e^{-\alpha t} f(t) \textrm{d} t} \end{equation*}$

If $\alpha \leq 0$ and $f \neq 0$, because $f$ is positive: $\begin{equation*} \int_0^{+\infty} e^{-\alpha t} f(t) = +\infty \end{equation*}$

Finally, if $f = 0$ and $\alpha = 0$, then: $\int_0^{+\infty} e^{-\alpha t} f(t) = 0$.

  • $\begingroup$ This is more or less what I did with my problem at hand, but the real question (really sorry if I worded it badly) was whether I could transform my improper integral into an infinite sum of definite integrals. I appreciate the help and the confirmation that my calculations were correct, though :) $\endgroup$ – AspiringMathematician Mar 28 '18 at 19:39
  • $\begingroup$ @AspiringMathematician As long as one of the two quantities exist in $\mathbb{R}$, AFAIK. $\endgroup$ – Raito Mar 28 '18 at 19:56
  • $\begingroup$ Oh, and I also had forgotten to say that indeed $\alpha > 0$. I've upvoted your answer, but I can't really accept it though. But please leave it as it is, since your answer for the original problem is quite deep and insightful! $\endgroup$ – AspiringMathematician Mar 29 '18 at 19:59

I myself have come up with a proof, after all. Please point out any errors if you find them.

If the integral converges (either to a finite or infinite value), we can break the integral as:

$$\int_0^\infty e^{-\alpha t}f(t)dt = \big(\sum_{n=0}^N \int_{nT}^{(n+1)T}e^{-\alpha t}f(t)dt\big) + \int_{(N+1)T}^\infty e^{-\alpha t}f(t)dt$$

So the problem at hand reduces to whether $\int_{NT}^\infty e^{-\alpha t}f(t)dt \rightarrow 0$ as $N \rightarrow \infty$. This can be further reduced, changing the variable to $s=t-NT$:

$$\int_{NT}^\infty e^{-\alpha t}f(t)dt = \int_{0}^\infty e^{-\alpha (s+NT)}f(s)ds = \int_{0}^\infty g_N(s)dt$$

With $g_N(s)=e^{-\alpha (s+NT)}f(s)$. Now we have:

  • $\lim_{N \rightarrow \infty} g_N = 0$ a.e.: Since $f$ is $T$-periodic and square-integrable over $[0,T]$, it is integrable (thanks Raito for pointing it out), and thus $f$ can only diverge in a set of measure zero in that interval. Since we can write $\mathbb{R}=\bigcup_{n\in\mathbb{Z}}[nT,(n+1)T]$, the same happens in $\mathbb{R}$. The claim follows from $e^{-\alpha (s+NT)} \rightarrow0$ as $N\rightarrow \infty$ for a given $s>0$.
  • For a given $s$, $g_0(s) \ge g_1(s) \ge g_2(s) \ge ... \ge 0$: Positiveness comes from hypothesis, and the inequalities come from the exponential.

Therefore by Lebesgue's Monotone Convergence Theorem,

$$\lim_{N \rightarrow \infty}\int_{NT}^\infty e^{-\alpha t}f(t)dt = \lim_{N \rightarrow \infty}\int_{0}^\infty g_N(s)dt =0$$

Interestingly enough, due to the periodicity of $f$, the converse seems to be true: If the infinite sum converges, the integral converges as well.

  • $\begingroup$ Sounds good. You can also ask the question of whether you can replace $[kT, (k + 1)T]$ segment by an arbitrary sequence $\{x_k\}$: $\int_0^{+\infty} e^{-\alpha t} f(t) \textrm{d}t = \sum_{n=0}^{+\infty} \int_{x_n}^{x_{n + 1}} e^{-\alpha t} f(t) \textrm{d} t$ assuming conditions on $\{ x_k \}$. Also, for all $k \in \mathbb{N}$, $g_k$ must be measurable functions in order to apply Lebesgue's monotone convergence theorem AFAIK. $\endgroup$ – Raito Mar 29 '18 at 20:48

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