Exercise 8.4.16

Question

Prove the following analogue of the Weierstrauss M-Test for improper integrals: If $f(x,t)$ satisfies $\left|f(x,t)\right| \leq g(t)$ for all $x \in A$ and $\int_a^\infty g(t) dt$ converges, then $\int_a^\infty f(x,t) dt$ converges uniformly on $A$.

Ulisse Mini · Accepted Answer

We want, $\forall \epsilon> 0$, to find $M > a$ so
$$\left|\int_a^\infty f(x,t) dt - \int_a^b f(x,t) dt\right| <\epsilon$$
for all $b > M$. But
$$\left|\int_a^\infty f(x,t) dt - \int_a^b f(x,t) dt\right| = \left|\int_b^\infty f(x,t) dt\right| \leq \int_a^\infty  \left|f(x,t)\right| dt \leq \int_b^\infty g(t) dt $$
and since $\int_a^\infty g(t)$ converges, $\lim_{b \to \infty} \int_b^\infty g(t) dt  = 0$, which can easily be used to finish the proof.

Exercise 8.4.16

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Exercise 8.4.16

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