# Regarding improper integrals

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 $$(-\infty,a]$$?

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

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

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

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

• 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 $a$. May 12, 2022 at 18:03
• Yes sir thank you
– Nope
May 13, 2022 at 0:45
• @peek-a-boo can you plz add all the conditions of how changing the borders of the integral change the way it converges
– user864806
May 17, 2022 at 6:57