As we all know:

  1. $\int_{a}^{\infty} \frac{1}{x^p}$ converges iff $p>1$ ($a>0$)

  2. $\int_{0}^{a} \frac{1}{x^p}$ converges iff $p<1$ ($p, a >0 $).

Is there exists any similar claims for $\int_{-\infty}^{a}$ in 1, and $\int_{-a}^{0}$ in 2?

In dirichlet theorem and the integral test of converging, in all sources (as I guess) they state the claim foe functions $f,g$ defined on $[a,\infty)$. Does it also hold if $f,g$ are defined on $(-\infty,a]$?


1 Answer 1


both are basically same. let, $x-a=y$,$dx=dy$ then our integral comes

  1. $\int_{-\infty}^{0} \frac{1}{(y+a)^p}$


so, it also satisfy the condition satisfied by the previous integral to be convergent.

  • 1
    $\begingroup$ They're not the same. If $a>0$ then the integral $\int_{-\infty}^ax^p\,dx$ is divergent for all $p\in\Bbb{R}$ (actually to avoid issues with branch cuts, let's just say all $p\in\Bbb{Z}$). So you should mention certain restrictions on $a$. $\endgroup$
    – peek-a-boo
    May 12, 2022 at 18:03
  • $\begingroup$ Yes sir thank you $\endgroup$
    – Nope
    May 13, 2022 at 0:45
  • $\begingroup$ @peek-a-boo can you plz add all the conditions of how changing the borders of the integral change the way it converges $\endgroup$
    – user864806
    May 17, 2022 at 6:57

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