Show that if $\int_a^{\infty}f(t) dt$ converges then $\lim_{x \to \infty} \int_x^{\infty} f(t)dt=0$.

Question

Here's my idea of proof.

Be $L=\int_a^{\infty} f(t)dt$, by definition,

$$\forall \epsilon > 0, \exists y \in R; x>y \Rightarrow \left|\int_a^x f(t)dt-L\right| < \epsilon$$

The inequality is equivalent to

$$ -L+\epsilon > \int_x^a f(t)dt > -L-\epsilon$$.

By definition, $$\left|\int_x^{\infty} f(t)dt\right| \leq \left| \int_x^a f(t)dt \right| + \left| \int_a^{\infty} f(t)dt \right| < -L + \epsilon + L = \epsilon, \qquad \forall x > y$$

Therefore, $\lim_{x \to \infty} \int_x^{\infty} f(t)dt=0$.

There is an error in the last set of inequalities, because i've shown that $\int_x^a f(t)dt < -L+\epsilon$, not $\left| \int_x^a f(t)dt \right| < -L+\epsilon$.

Answer 1

It's just $$\int_x^\infty f(t)\,dt = \int_a^\infty f(t)\,dt - \int_a^xf(t)\,dt \xrightarrow{x\to\infty} \int_a^\infty f(t)\,dt - \int_a^\infty f(t)\,dt = 0$$ by definition of $\int_a^\infty f(t)\,dt$.