Convergence of $\int_{0}^{\infty}\frac{t^a}{1+t}\mathrm dt$

Question

Find the set of values of $a$ such that

$$\displaystyle\int_{0}^{\infty}\frac{t^a}{1+t}\mathrm dt$$

converges. I have managed to prove that the integral diverges for $a>0$.

Answer 1

Let $-b=a<0$. Then note

$$0\le \int_1^x{dt\over t^b+t^{1+b}}\,dt\le \int_1^x{dt\over t^{1+b}}\le {1\over b}$$

So the monotone sequence $\displaystyle\int_1^n{t^a\over 1+t}$ is bounded, hence converges, and of course this is exactly $\displaystyle\lim_{x\to\infty}\int_1^x{t^a\over 1+t}$, so this part of the improper integral converges. Now we turn to the other part to see if the whole thing converges or not:

$$\int_\epsilon^1{t^a\over 1+t}\,dt\ge {1\over 2}\int_\epsilon^1t^a\,dt={1\over 2}\left({1\over {a+1}}-{\epsilon^{a+1}\over a+1}\right)$$

so when $a<-1$ we get divergence, since then $a+1<0$ and we are dividing by something going to $0$.

For $-1< a < 0$ we see

$$\int_{\epsilon}^1{t^a\over 1+t}\,dt \le \int_{\epsilon}^1 t^a = {1\over a+1}-{\epsilon^{a+1}\over a+1}$$

which is finite. So for certain we get convergence for $-1

Finally the edge cases:

For $a=0$ we see

$$\log x=\int_1^x {dt\over t}<\int_0^x {dt\over 1+t}$$

so again taking limits we get divergence for $a=0$, without even having to consider the other integral, $\displaystyle\int_0^1{dt\over 1+t}$ we would need to converge to conclude convergence of the whole integral!

And for $a=-1$ we consider

$$\int_{\epsilon}^1{dt\over t+t^2}>\int_{\epsilon}^1{dt\over 2t}=-{1\over 2}\log\epsilon$$

which goes to infinity as $\epsilon\to 0$, showing divergence.