Determining real $a$ for which $\int_{-1}^\infty\frac{1}{(a-x)^2}\,dx$ converges Can someone please help me?

Consider the integral
$$\int_{-1}^\infty\frac{1}{(a-x)^2}\,dx$$
where $a$ is a real constant. Carefully justify for which values of $a$ the integral converges, and calculate its value for those cases.

I used limit as $t$ goes to infinity and found that the limit is $1/(a+1)$, but I don't know what to do next.
 A: To change the limits of the integral from $-1$ to $\infty$ to $0$ to $\infty$, we used the substitution $u = a-x$. This substitution allows us to rewrite the integral in terms of the new variable $u$.
The substitution $u = a-x$ has the effect of shifting the limits of integration. The lower limit of the integral becomes $u = a-(-1) = a+1$, and the upper limit becomes $u = a-\infty = -\infty$.
To find the new limits of integration, we can substitute these values back into the original integral:
$$\int_{-1}^{\infty} \frac{1}{(a-x)^2} dx = \int_{a+1}^{-\infty} \frac{1}{u^2} du$$
We can then apply the standard rules for changing the limits of integration to find the new limits:
$$\int_{a+1}^{-\infty} \frac{1}{u^2} du = \int_{-\infty}^{a+1} \frac{1}{(-u)^2} (-du) = \int_{a+1}^{-\infty} \frac{1}{u^2} \, du
 $$
now since $\frac{1}{(-u)^2}$    isn't defined in $0$, when $$0 \geq a+1$$ we get infinity , since $$\begin{cases}
1/u = \infty & \text{when } u = 0 \
\end{cases}$$
so for every $a+1 > 0$ the integral converges and equal to $$\frac{1}{a+1}.$$
