Show that if $n \in \mathbb{N}, \int_{-1}^{1} \frac{1}{t^n}dt$ does not converge

Question

Show that if $n \in \mathbb{N}, \int_{-1}^{1} \frac{1}{t^n}dt$ does not converge.

I tried to divide it into $$\lim_{S\rightarrow0^-}\int_{-1}^{S} \frac{1}{t^n}dt + \lim_{R\rightarrow0^+}\int_{R}^{1} \frac{1}{t^n}dt $$ and for the second integral, I concluded that it diverges for $n > 1$, but I'm stuck in the first one.

Any help is appreciated!

Answer 1

\begin{align*} \int_{-1}^{S}\dfrac{1}{t}dt=\log(-t)\bigg|_{t=-1}^{t=S}=\log(-S)\rightarrow-\infty \end{align*} as $S\rightarrow 0^{-}$, and for $n\geq 2$, \begin{align*} \int_{-1}^{S}\dfrac{1}{t^{n}}dt=\dfrac{1}{-n+1}t^{-n+1}\bigg|_{t=-1}^{t=S}=-\dfrac{1}{n-1}\dfrac{1}{S^{n-1}}+\dfrac{1}{n-1}\dfrac{1}{(-1)^{n-1}} \end{align*} will tend to $\infty$ if $n$ is even and $-\infty$ when $n$ is odd as $S\rightarrow 0^{-}$.