This question has been answered Here.
I see now that holds true because f is monotonous. $$\int_{(n+1)\pi}^{(n+2)\pi}f(t)dt \le \pi f((n+1)\pi)$$
But why does $$\int_{(n+1)\pi}^{(n+2)\pi}f(t)dt \le \frac\pi C \int_{n\pi}^{(n+1)\pi}f(t)|\sin t|dt$$ stand true:
- The original question:
During our studies on improper integarls (Calculus II course) we were given the following question:
Let $f: [a, \infty) \rightarrow \mathbb{R}$ a non-negative, monotonous, continously >differentaible function s.t. $\int_a^\infty f(x)\sin(x)\ dx$ absoloutly converges. prove that >$\int_a^\infty f\ dx$ exists.
In a previous question I have shown:
Let $f: [a, \infty) \rightarrow \mathbb{R}$ a non-negative monotonous function s.t. >$\int_a^\infty f(x)\sin(x)\ dx$. Then $\displaystyle{\lim_{x\to\infty}f(x)=0}$.
- $\int_a^\infty f(t)dt$ converges.
So I do know that $\displaystyle{\lim_{x\to\infty}f(x)=0}$. The theorem I'm required to prove closely resembles Able's Convergence test for integrals. But it is the other way around. I also tried multiple times with Cauchy Criterion for convergence of integrals by trying a proof by contradiction.
Any elementary analysis by using convergence tests and Riemann integrals will be very helpful!