5

Let $f,g:[a,+\infty]\to\mathbb{R}$ be continuous and $K>0$ s.t. $$ \left|\int_c^df(x)\,dx\right|\leq K \quad \forall c,d\in [a,\infty). $$ If $g\in C^1$ is decreasing with $\displaystyle\lim_{x\to \infty}g(x)=0,$ prove that the limit $$\int_a^\infty f(x)g(x)\,dx=\lim_{x\to\infty} \int_a^x f(t)g(t)\,dt$$ exists.

I tried to use the Cauchy criterion, that is $\forall \epsilon>0, \exists A>a$ s.t. $A<c<d$ implies that $\left|\int_c^d f(t)g(t)\,dt\right|<\epsilon.$ However I could not get anywhere.

I appreciate any suggestions.

P.S. I already posted this question and as its expression was not completely right, I deleted it and asked it again here.

Adrian Keister
  • 10,099
  • 13
  • 30
  • 43
Majid
  • 3,357

2 Answers2

6

For $x \in [a,\infty)$, define $F(x) := \int^x_a f(t)\,dt$. Then $\lvert F(x) \rvert \le K$, and $F'(x) = f(x)$ by the fundamental theorem of calculus.

Now for $x,y \in[a,\infty)$, and wlog let $y \ge x$. By integrating by parts and using the triangle inequality, we see \begin{align*}\left\lvert \int^y_a f(t) g(t) \,dt - \int^x_a f(t)g(t)\,dt\right\rvert &= \left \lvert\int^y_x f(t) g(t)\, dt \right \rvert\\ &= \left \lvert \int^y_x F'(t)g(t) \,dt \right \rvert\\ &= \left \lvert [F(t)g(t)]^{t=y}_{t=x} - \int^y_x F(x)g'(x)\,dx \right \rvert\\ &\le \lvert F(y)g(y)\rvert + \lvert F(x)g(x) \rvert + \int^y_x \lvert F(t)g'(t) \rvert \,dt\\ & \le K\left(\lvert g(y) \rvert + \lvert g(x) \rvert + \int^y_x \lvert g'(t)\rvert \,dt \right). \end{align*} Since $g$ is decreasing, we have $g'(t) \le 0$, and so $\lvert g'(t)\rvert = -g'(t).$ Thus \begin{align*}\left\lvert \int^y_a f(t) g(t) \,dt - \int^x_a f(t)g(t)\,dt\right\rvert & \le K\left(\lvert g(y) \rvert + \lvert g(x) \rvert + \int^y_x \lvert g'(t)\rvert \,dt \right)\\ &= K\left(\lvert g(y) \rvert + \lvert g(x) \rvert - \int^y_x g'(t) \,dt \right)\\ &= K(\lvert g(y) \rvert + \vert g(x) \rvert + g(x) - g(y)) \\&\le 4K \text{max}(\lvert g(x) \rvert, \lvert g(y) \rvert). \end{align*} Since $\lvert g(t) \rvert \to 0$ as $t \to \infty$, for any $\epsilon > 0$, there is $M > 0$, such that $\lvert g(t)\rvert < \epsilon/4K$ for all $t > M$. But then for $x,y > M$, we have shown $$\left\lvert \int^y_a f(t) g(t)\, dt - \int^x_a f(t)g(t)\,dt\right\rvert < \epsilon.$$ Finally, take any sequence $\{x_n\}_{n=1}^\infty$ with $x_n \to \infty$, and this shows that $$\left\{\int^{x_n}_a f(t)g(t)\,dt \right\}_{n=1}^\infty$$ is a Cauchy sequence and hence converges. By the sequential criterion theorem, we conclude that $$\lim_{x\to\infty} \int^x_a f(t) g(t) \,dt$$ converges.

As an aside, this is completely analogous to the Dirichlet Test for infinite series, which is a generalization of the alternating series test.

Adrian Keister
  • 10,099
  • 13
  • 30
  • 43
User8128
  • 15,485
  • 1
  • 18
  • 31
0

Since it is given that $g$ is continuously differentiable, so $g'$ exists. Now the idea is to write $\displaystyle\int fg$ as $\displaystyle g\int f-\int \left(g'\int f\right)$.

Now using triangle inequality, $$\left|\int fg\right|\leq\left| g\int f\right|+\left|\int \left(g'\int f\right)\right|,$$ now I think you can complete it, but don't forget to use the fact that $\displaystyle\lim_{x\to{\infty}} g'(x)=0.$

Adrian Keister
  • 10,099
  • 13
  • 30
  • 43
Soumyadip Sarkar
  • 705
  • 5
  • 18
  • 1
    But we are not assuming that $g'(x) \to 0$, and indeed, it may not. – User8128 May 17 '19 at 14:58
  • 1
    You are right,actually i thought that asymptotically g is becoming constant and hence the derivative must go to $0$,but that's not true. – Soumyadip Sarkar May 17 '19 at 15:04
  • 3
    Exactly right: the prototypical counterexample is something like $g(x) = \frac{\sin(x^{10})}{1+x^2}$ – User8128 May 17 '19 at 15:05
  • 1
    @User8128 Note $g$ is decreasing. That still does not imply $g'\to0$, but you will need another counterexample: small jumps with large $|g'|$ such that the sum of the jumps is bounded, for instance, i.e. something like this: https://math.stackexchange.com/questions/788813/the-limit-of-the-derivative-of-an-increasing-and-bounded-function-is-always-0 – Jean-Claude Arbaut May 18 '19 at 11:39