State whether the following improper integral converges, and, if it does, find its value.

Question

Hello, I have been struggling with the simplest question of this topic.

The integral of it is $\frac{-1}{(1+x)} dx$, and replacing infinity with the limit of $b \to \infty$, I found $\frac{-1}{(1+b)} +1/2$, and the limit goes to zero so the value must be $1/2$. However, the answer is $1$. Can you find the mistake I am making?

Answer 1

$$\lim_{z\to\infty}\int_0^z \frac{1}{(1+x)^2}dx = \lim_{z\to\infty} \frac{-1}{(1+x)} |_0^z = \lim_{z\to\infty}(\frac{-1}{(1+z)} - \frac{-1}{(1+x)}) = 0-\frac{-1}{(1+0)}= 1$$

Answer 2

By substitution, let $u=1+x$, so that $du=dx$ and our integral $$I=\int_0^{\infty}\frac{1}{u^2}du=\int_0^{\infty}u^{-2}du=\lim_{k\to \infty}[-\frac{1}{u}]_0^{k}=\lim_{x\to \infty}-\frac{1}{(x+1)} - \lim_{x\to 0}-\frac{1}{(x+1)}=0-(-1)=1.$$

Therefore, our integral converges to $1$.

Answer 3

Maybe you mistakenly integrated from $\mathbf 1$ to $\infty$ instead of from $\mathbf 0$ to $\infty$