# Show If $\int_0^{\infty} f(x) dx$ is convergent, for $a \ge 0$, $\int_0^{\infty} \frac{f(x)}{e^{ax}}$ is convergent.

$$e^{ax} \ge 1$$ 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?

• 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? – Arthur May 15 at 12:24
• You should not solve the question using what you suggested, try to compare the given integrals – Achraf BOURASS May 15 at 12:26
• Integration by parts with $F(x) = \int_x^\infty f(t) dt$ should do the trick. – Martin R May 15 at 12:28
• If the integrals are in the Lebesgue sense then the sign of $f$ is of no consequence. – Kavi Rama Murthy May 15 at 12:30

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 $$\int_0^\infty e^{-at}f(t) \, dt$$.

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.

• $|\int_0^C g(x)dx | \le M$ for all $C$ does not imply the existence of $\int_0^\infty g(x) dx$. Or am I misunderstanding your argument? – Martin R May 15 at 13:47
• 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. – cha21 May 15 at 13:52