Vtyjkyuk

I am just looking for someone to tell me if my reasoning is correct. Let's look at the integral:
$$int^infty_0 fracx^ax + 1dx$$
where $a$ is a real number. Is it valid to say that the integral will converge when the sum converges:
$$ int^infty_0 fracx^ax + 1dx sim sum^infty_n = 0 fracn^an + 1$$
So:

• for $a < 0$ the first term is undefined


• for $a=0$ the sum diverges


• for $a>1$ the sum diverges


• for $0<a<1$ the sum converges

Is it valid to conclude that the integral converges only for $0<a<1$? Does this reasoning work for other integrals of this type, is there something I should pay attention to?

No it is not a valid method in general, for the convergence we need to show that for some convergent $sum a_n$

$$0le int^infty_0 fracx^ax + 1dx lesum^infty_n = 0 a_n$$

As an alternative note that for $age0$

$$fracx^ax + 1sim x^a-1=frac1x^1-a$$

and for $a<0$ let $b=-a>0$

$$fracx^ax + 1=frac1x^b(x + 1)$$

and

• as $xto 0^+ implies frac1x^b(x + 1) sim frac1x^b$




• as $xto infty implies frac1x^b(x + 1) sim frac1x^b+1$

then refer to limit comparison test to conclude that the given integral converges for $ain(-1,0)$.

No, you cannot assert that the integral converges for $0 < a < 1$. Using comparison test you would notice that
$$
frac x^a1+x sim frac 1x^1-a quad [n to infty],
$$
and $int_A^infty (1/ x^p) mathrm dx ;[A>0]$ converges iff $p > 1$. Hence the integral converges when and only when $colorreda < 0$ [WRONG HERE, see UPDATE 2].

Meanwhile, the series is also diverges by similar testing method.

There is a test for convergence of series: integral test. However this requires the function be decreasing. So generally you cannot use the correspondent series to test the convergence for this kind of improper integrals [although I do not know the counterexample when $f$ is not decreasing. Any comments about this are welcome].

UPDATE

Thanks to @RobertIsrael, the integral test is applicable to your question since $f(x) = x^a /(x+1)$ is monotonic for sufficiently large $x$.

UPDATE 2

I was wrong in the comparison test above. When $a < 0$, the point $0$ would also be questionable. Combining the asymptotic behavior as $x to 0^+$ and $x to +infty$, the conclusion should be: the integral converges iff $boldsymbol 0 < a < 1$.