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

Question

I've encountered this question in a physics textbook, it is not stated in the most accurate manner and the solution provided doesn't help me at all to understand it (paraphrasing the question):

Suppose $f:[0, \infty) \to \Bbb{R}$ is a funciton such that $\forall \alpha > 0$ $$ \lim_{t\to \infty}f(t)e^{-\alpha t} = 0 $$ Prove that $\int_0^{\infty} f(t)e^{-\alpha t} dt$ is convergent.

The book essentially claims that the conclusion follows immediately from the hypothesis without any further explanation, but I really can't see why...

I don't think this can work for any function $f$ in the above domain so lets assume it is $C^1$ in $[0, \infty)$. I believe the statement is true under these assumptions but not sure how to begin proving it for the general case. There may be counter examples and other conditions may need to be imposed in order for the statement to be true, but I am not entirely sure.
Thank you for any hints!

(This is only the first part of the question. The second part speaks about the analyticity of the fourier transform of $f$ in the upper plane, but I think it may be more straight forward)

Answer 1

Just a trial/suggestion. Putting $\alpha=2\alpha'$:

$\int_0^{\infty} f(t)e^{-\alpha t}dt=\int_0^{\infty} (f(t)e^{-\alpha' t})e^{-\alpha' t}dt$

Now we have that $|f(t)e^{-\alpha' t}|<1$ for $t>T$ and therefore:

$\int_0^{\infty} (f(t)e^{-\alpha' t})e^{-\alpha' t}dt<\int_0^{T} (f(t)e^{-\alpha' t})e^{-\alpha' t}dt+\int_T^{\infty} e^{-\alpha' t}dt<\int_0^T f(t)dt + \int_0^{\infty} e^{-\alpha' t}dt$

Now the second integral converges and if also we know that the first one is finite we should be done?

NB: I considered $f$ positive here, but I hope putting some modulus in a correct way the reasoning should still hold.

Answer 2

My answer fleshes out Thomas's answer, which gives the basic idea for the solution. Please upvote them.

Fix $\alpha >0$. We will show that $\int_0^\infty\left|f(t)e^{-\alpha t}\right|dt$ converges under the given condition on $f$, that is, $\int_0^\infty f(t)e^{-\alpha t}dt$ converges absolutely. This implies the convergence of $\int_0^\infty f(t)e^{-\alpha t}dt$; a proof of this result was left at the bottom.

Pick a $\beta>0$ that is less than $\alpha$. Since $\lim_{t\to\infty}f(t)e^{-\beta t}=0$, there is an $N\geq 0$ satisfying

$$|f(t)e^{-\beta t}|<1\text { for every }t\geq N$$

or what is the same,

$$|f(t)|<e^{\beta t}\text { for every }t\geq N$$

This implies

$$\left|f(t)e^{-\alpha t}\right|<e^{-(\alpha-\beta) t}\text { for every }t\geq N$$

and because we picked $\beta<\alpha$, the integral $\int_N^\infty e^{-(\alpha-\beta)}dt$ converges. It follows from comparison that $\int_N^\infty \left|f(t)e^{-\alpha t}\right| dt$ also converges, and thus so does

$$\int_0^\infty \left|f(t)e^{-\alpha t}\right|dt=\int_0^N \left|f(t)e^{-\alpha t}\right|dt+\int_N^\infty \left|f(t)e^{-\alpha t}\right|dt$$

Proof that absolute convergence implies ordinary convergence:

Suppose $\int_0^\infty |g(t)|dt$ converges. Denote the positive and negative parts of $g$ by $g^+$ and $g^-$, respectively. To be precise, $g^+$ and $g^-$ are defined as follows:

$$g^+ (t)=\begin{cases} g(t) &\text{ if } g(t)\geq 0\\ 0 &\text{ if } g(t)< 0\end{cases}$$ $$\text{and}$$ $$g^- (t)=\begin{cases} g(t) &\text{ if } g(t)\leq 0\\ 0 &\text{ if } g(t)> 0\end{cases}$$

It is then straightforward to show that for every $t\in [0,\infty)$, $g(t)=g^+(t) + g^-(t)$ and $|g(t)|=g^+(t) - g^-(t)$. This implies that for any $x>0$,

\begin{align*} \int_0^x g^+ (t) dt &= \int_0^x \left[|g(t)| + g^- (t)\right]dt\\ &= \int_0^x |g(t)|dt + \int_0^x g^- (t)dt\\ &\leq \int_0^\infty |g(t)|dt + 0 \text{ (since }g^-(t)\leq 0\text{ and }|g(t)|\geq 0\text{)}\\ &= \int_0^\infty |g(t)|dt \end{align*}

and

\begin{align*} \int_0^x g^- (t) dt &= \int_0^x \left[g^+ (t)-|g(t)|\right]dt\\ &= \int_0^x g^+(t)dt - \int_0^x |g(t)|dt\\ &\geq 0 - \int_0^\infty |g(t)|dt \text{ (since }g^+(t)\geq 0\text{)}\\ &= -\int_0^\infty |g(t)|dt \end{align*}

In particular, $\int_0^x g^+$ is monotone increasing (because $g^+$ is non-negative) and bounded above, whereas $\int_0^x g^-$ is monotone decreasing and bounded below. It follows that $\lim_{x\to\infty}\int_0^x g^+(t)dt$ and $\lim_{x\to\infty}\int_0^x g^-(t)dt$ both exist, so $\int_0^\infty g^+(t)dt$ and $\int_0^\infty g^-(t)dt$ both converge. Thus,

$$\int_0^\infty g(t)dt=\int_0^\infty g^+(t)dt+\int_0^\infty g^-(t)dt$$

converges.

Answer 3

One of two cases arise: either $\lim_{t\rightarrow\infty}{f(t)}$ is finite or is infinite ($\pm\infty$). In the first case things are simple, because we essentially have a horizontal asymptote, which makes all derivatives of $f$ approach $0$, so that $f^{(n)}(t)e^{-\alpha t}$ approaches $0$ for all natural numbers $n$. Then by applying IBP to $\int{f(t)e^{-\alpha t} dt}$ we get:

$$-{1\over\alpha}f(t)e^{-\alpha t}+{1\over\alpha}\int f'(t)e^{-\alpha t}dt$$

Clearly the first term goes to $0$ as $t\rightarrow\infty$ so convergence here is understood. The second term can follow the same process inductively, since applying IBP will just increase the number of derivatives of $f$ that come out of the integral, and we've already established that $f^{(n)}(t)e^{-\alpha t}$ goes to $0$, making the orignial integral convergent in this case.

The second case is slightly more complicated, but basically follows the same idea, except using l'Hosptal's Rule:

$$\lim_{t\rightarrow\infty}f(t)e^{-\alpha t}=\lim_{t\rightarrow\infty}{f(t)\over e^{\alpha t}}={1\over\alpha}\lim_{t\rightarrow\infty}{f'(t)\over e^{\alpha t}}$$

and now we can repeat the process for $\int f'(t)e^{-\alpha t}dt$ depending wether $f'(t)$ approaches a finite value or an infinite one. You can see how it just ends up being a big induction problem with by parts.

I hope that this helps.