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$e^{ax} \ge 1$ so for $f(x) \ge 0$ we have $\frac{f(x)}{e^{ax}} \le f(x)$ and by comparison test it is convergent. But what about negative f(x)? I also used integration by parts but it was confusing. If $g(x) = e^{-ax}$ then $g'(x) = -ae^{-ax}$ and if F'(x) = f(x) then $$e^{-ax}F(x) = \int e^{-ax}f(x) + \int -ae^{-ax}F(x)$$ on the other hand $$\int_0^{\infty}f(x) = \lim_{T\to\infty}\int_0^{T}f(x) = \lim_{T\to\infty}F(T) - F(0)$$ But how can I use these informations?

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    $\begingroup$ Negative $f$ is basically an identical comparison to positive $f$, you just turn some of the inequality signs around. More interestingly, what about $f$ that oscillate between positive and negative? $\endgroup$ – Arthur May 15 at 12:24
  • $\begingroup$ You should not solve the question using what you suggested, try to compare the given integrals $\endgroup$ – Achraf BOURASS May 15 at 12:26
  • $\begingroup$ Integration by parts with $F(x) = \int_x^\infty f(t) dt$ should do the trick. $\endgroup$ – Martin R May 15 at 12:28
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    $\begingroup$ If the integrals are in the Lebesgue sense then the sign of $f$ is of no consequence. $\endgroup$ – Kavi Rama Murthy May 15 at 12:30
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Define $F(x) = -\int_x^\infty f(t) \, dt$. Then $F'(x) = f(x)$ and $\lim_{x \to \infty} F(x) = 0$.

Integration by parts gives for $0 < x < y$ $$ \int_x^y e^{-at}f(t) \, dt = e^{-ay}F(y) - e^{-ax}F(x) +a \int_x^y e^{-at}F(t) \, dt \, . $$

With $M(x) = \max \{ |F(t)| : t \ge x \}$ we can now estimate $$ \left| \int_x^y e^{-at}f(t) \, dt \right| \le M(x) \left( e^{-ay} + e^{-ax} +a \int_x^y e^{-at} \, dt \right) \\ = 2 M(x) e^{-ax} \le 2 M(x) \to 0 \text{ for } x \to \infty $$ and that implies the convergence of $\int_0^\infty e^{-at}f(t) \, dt$.

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Claim: $$\left|\int_0^\infty \frac{f(x)}{e^{ax}}\right| \leq \sup_{b \geq 0} \left|\int_0^b f(x)\right|$$

Lemma: If $a_1, a_2, \cdots$ is a sequence of real numbers and $1 \geq b_1 \geq b_2 \geq \cdots$ is another sequence, then for every $n \geq 1$, $$\left|\sum_{k =1}^n a_kb_k \right| \leq \sup_{i \leq n} \left|\sum_{k = 1}^i a_k \right|$$

proof: Let $i$ be the smallest index such that $\left|\sum_{k =1}^i a_kb_k \right|$ is maximized. WLOG assume $$\sum_{k =1}^i a_kb_k \geq 0$$ We will show by induction that for every $0 \leq j \leq i$, $$\sum_{k = 1}^{j}a_kb_k + b_{j+1}\sum_{k = j+1}^i a_i \geq \sum_{k =1}^i a_kb_k$$ starting with $j = i$ and moving down.

The base case is given and the inductive step follows from inductive hypothesis and the fact that for every $j$, $\sum_{k = j+1}^ia_k \geq 0$ because otherwise we would have $$\sum_{k =1}^{j} a_kb_k > \sum_{k =1}^i a_kb_k$$ contradicting the definition of $i$.

Now, to prove the main claim it suffices to show for every, $C > 0$, $$\left|\int_0^C \frac{f(x)}{e^{ax}}\right| \leq \sup_{0 \leq b \leq C} \left|\int_0^b f(x)\right|$$ But this follows from the lemma and approximating the integral by Riemann Sums.

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  • $\begingroup$ $|\int_0^C g(x)dx | \le M$ for all $C$ does not imply the existence of $\int_0^\infty g(x) dx$. Or am I misunderstanding your argument? $\endgroup$ – Martin R May 15 at 13:47
  • $\begingroup$ You have break it into two parts. So $\int_0^\infty \frac{f(x)}{e^{ax}} = \int_0^C \frac{f(x)}{e^{ax}} + \int_C^\infty \frac{f(x)}{e^{ax}}$. The second is bounded by the $\sup_{b \geq C} \int_C^b f(x)$ which goes to $0$ as $C \rightarrow \infty$ so you don't get oscillation. $\endgroup$ – cha21 May 15 at 13:52

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