I was solving the problem by taking two cases i.e. for $n=1$ and $n \neq 1$ and reached to the point $$ \lim_{l \rightarrow a^{+}}\left\{\frac{1}{1-n} \left[ \frac{1}{(b-a)^{n-1}}- \frac{1}{(l-a)^{n-1}} \right] ,n \neq 1 \right. $$ After which according to me for n>1 the limit tends to $\infty$ and hence, integral diverges and for n<1 the limit tends to $\frac{1}{1-n} \left[ \frac{1}{(b-a)^{n-1}}\right]$ however the answer says the opposite for the values of n.
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3Does this answer your question? how do i check the convergence of $\int^{b}_{a}\frac{dx}{(x-a)^p}$ - found using an Approach0 search. Note the answer there basically agrees with yours, and I also agree, so your other answer is likely incorrect. Using $t=x-a$, the integral becomes $\int_{0}^{b-a}\frac{dt}{t^{n}}=\int_{0}^{b-a}t^{-n}dt$. ... – John Omielan Apr 11 '23 at 17:37
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(cont.) For $n=1$, we get $\ln(t)\rvert_{0}^{b-a}=\infty$, while for $n\neq 1$ gives $\left.\frac{t^{-n+1}}{-n+1}\right\rvert_{0}^{b-a}$ which, for $n\gt 1$, is $\infty$, while for $n\lt 1$, it's $\frac{(b-a)^{1-n}}{(1-n)}$, as you basically stated. Last, but not least, welcome to Math SE. – John Omielan Apr 11 '23 at 17:38